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Topics in Modern Algebra
Introduction to Algebraic Geometry
a.mustata
March 12, 2020
Some Context
Figure 1: A rough comparison of areas studies geometric shapes
1 Plane Curves
Let K be a field. In this chapter we will look at the cases K = R (real curves) and K = C (complex curves).As dimR(C) = 2, the complex curves will actually correspond to real surfaces.
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Definition 1.1. An algebraic curve C ⊂ K2 is the solution set of a polynomial equation P (x, y) = 0 where
P (x, y) =N∑
i,j=0
pijxiyj ∈ K[x, y].
2 Drawing Algebraic Curves
Figure 2: Smooth curves defined by degree 3 equations
Here is how the same process works for degree 3 curves:
2
Figure 3: Curves defined by degree 2 equations
For higher degree curves of general equation∑N
i,j=0 pijxiyj = 0, it may be more difficult to complete the
description as above, but later in the course we will discover neater ways of describing them, for examplethe genus (= ”number of holes”) is given by the degree-genus formula for smooth complex curves:
g =(d− 1)(d− 2)
2.
We will prove this formula later in the course. Here the degree is defined by
Definition 2.1. The degree of a plane curve given by the equation∑N
i,j=0 pijxiyj = 0 is the largest integer
d such that there exist i, j with i+ j = d and pij 6= 0.
Parametric Curves
The examples below are relatively straightforward because y can be easily written in terms of variable x,even is this works only locally (on separate branches of the curve). We say that such curves are locallyparametrisable, in the sense that they can be described locally in terms of the parameter x.
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Figure 4: Algebraic curves - from Frances Kirwan: Complex Algebraic Curves
4
The simplest curves to plot are those which admit a parametrisation, namely those for which thecoordinates can be written as functions of one parameter:
x = x(t) and y = y(t).
These can be sketched by simply choosing sufficiently many values of t and plotting the resulting pointsin the plane.
Example 2.2. Affine Line. Consider the line L given by equation 3y − 2x + 5 = 0 in the plane K2.Then L can be parametrised by
φ : K1 → K2 given by φ(t) = (t,2
3t− 5
3).
Indeed, this is obtained algebraically be setting x = t in the equation 3y − 2x+ 5 = 0 and solving for y.
Example 2.3. The Graph of a Polynomial Function. Given the polynomial function f(x) ∈ K[x],the graph of the function is the plane curve Γf in K2 given by the equation y = f(x). This is a parametrisedfunction with parameter x.
Not all algebraic curves admit polynomial parametrisation; in fact, very few curves do. However, there is
Theorem 2.4. The Implicit Function Theorem. Given a differential (or analytic/holomorphic) func-tion P (x, y) and a point (x0, y0) such that P (x0, y0) = 0 and ∂P
∂y (x0, y0) 6= 0, then there exist open in-tervals Ux0 and Uy0 around x0 and y0 respectively, and a differential (or analytic/holomorphic) functionϕ : Ux0 → Uy0 such that for all (x, y) ∈ Ux0 × Uy0
P (x, y) = 0 ⇐⇒ y = ϕ(x).
In other words x can be chosen as a local parameter for the curve P (x, y) = 0 whenever ∂P∂y (x0, y0) 6= 0
(e.g. the parabola y = x2). Similarly y can be chosen as a local parameter for the curve P (x, y) = 0whenever ∂P
∂x (x0, y0) 6= 0 (e.g. the parabola x = y2).
The IFT fails at the points where both partial derivatives are 0. We call these singular points:
Definition 2.5. Given an algebraic curve C ⊂ K2 defined by a polynomial equation P (x, y) = 0 we saythat a point A0 = (x0, y0) is singular if
∂P
∂x(A0) =
∂P
∂y(A0) = 0.
Otherwise we say that a point A0 = (x0, y0) is non-singular. Namely, for non-singular points
∂P
∂x(A0) 6= 0 or
∂P
∂y(A0) 6= 0.
We say that a curve is non-singular or smooth if it has no singular points. Otherwise the curve is calledsingular.
Singularities of algebraic curves were studied by Newton around 1700.
Here are some more interesting examples of curves which can be parametrised at all but a few isolatedpoints:
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Example 2.6. The cuspidal cubic. The curve of equation y2 = x3 in R2 can be written as the imageof the map φ : K→ K2 given by φ(t) = (t2, t3), in other words it admits a parametrisation
x = t2 and y = t3.
This corresponds to the ring homomorphism
φ∗ : R[x, y] −→ R[t]
p(x, y) −→ p(t2, t3)
which has Kernel ker(φ∗) = 〈y2 − x3〉 and hence defines a ring isomorpshim
R[x, y]
〈y2 − x3〉−→ Im(φ∗) = {g(t) ∈ R[t] ; g′(0) = 0}.
This is called the coordinate ring of the curve C, that is, the ring of polynomial functions defined on C withvalues in the field R. The line R has coordinate ring R[t], and the coordinate ring of C being isomorphicto Im(φ∗) = {g(t) ∈ R[t] ; g′(0) = 0} indicates that the curve C is not diffeomorphic to the line R, but ithas a special point 0 where it makes a ”sharp turn”.
Example 2.7. A nodal cubic. The curve of equation y2 = x3 + x2 in R2 admits a parametrisation
x = t2 − 1 and y = t3 − t.
which means that it can be written as the image of the map φ : R→ R2 given by φ(t) = (t2 − 1, t3 − t).
The corresponding ring homomorphism
φ∗ : R[x, y] −→ R[t]
p(x, y) −→ p(t2 − 1, t3 − t)
has Ker(φ∗) = 〈x3 + x2 − y2〉 and Im(φ∗) = {p(t) ∈ R[t]; p(1) = p(−1)}. The coordinate ring of C beingisomorphic to Im(φ∗) = {p(t) ∈ R[t]; p(1) = p(−1)} again tells us that the curve described here is notdiffeomorphic to R, but it is the image of a map R→ C which identifies the points 1 and −1.
Example 2.8. Stereographic Projection and its inverse. Consider the circle S1R = V(x2 +y2−1) ⊂R2. Then S1R admits a partial parametrisation with parameter t:
φ : R1 → S1R
t →(
1− t2
1 + t2,
2t
1 + t2
)which is the inverse of the stereographic projection φ−1 given by t = φ−1(x, y) = y
1+x = 1−xy . The corre-
spondence is constructed geometrically as follows: for every point (x, y) 6= (−1, 0) on the circle, we denoteby (0, t) the intersection of the line connecting (x, y) and (−1, 0) with the vertical axis. This gives us theexpected formulas for x and y.
To get a full parametrisation for the circle, we need to enlarge K1 by adding a point at ∞. In otherwords, we need to replace the affine line K1 by the projective line KP1.
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Moreover, if we work in the complex projective plane, we can apply the same method to get a parametri-sation for every conic - application to solving homogeneous quadratic diophantine equations - findingrational points on quadrics.
In all of the examples above, the parametrisation was obtained by the same procedure
i) choose a point a0 = (x0, y0) on the curve,
ii) construct a function
ψ : { lines through a0} −→ C
l = line through a0. −→ ψ(l) = the intersection point in l⋂C \ {a0}
iii) if ψ is well defined and bijective at all but finitely many points, then the curve C is parametrised byφ = ψ ◦ h where h is an injective function
K→ { lines through a0}.
This last function h can consist in either identifying each line with its slope (like in the cases ofthe cuspidal and nodal cubic) or with its Ox or Oy-intercept (like in the case of the stereographicprojection).
Note that in these cases, ψ is well defined only becase every line through A0 intersects the curve C atexactly 2 points, one of which is A0. This motivates the study the intersection of lines through A0 = (x0, y0)with the curve C given by P (x, y) = 0.
Lemma 2.9. Taylor expansion around A = (x0, y0). Any polynomial P (x, y) =∑d
i,j=0 pijxiyj can be
written as
P (x, y) =d∑
i,j=0
∂i+jP
∂xi∂yj(A)
(x− x0)i(y − y0)j
i!j!
= P (A) +∂P
∂x(A)(x− x0) +
∂P
∂y(A)(y − y0) +
+∂2P
∂x2(A)
(x− x0)2
2+
∂2P
∂x∂y(A)(x− x0)(y − y0) +
∂2P
∂y2(A)
(y − y0)2
2+ ...
Corollary 2.10. Intersection of a line with a curve. A line lt of slope t through the point A = (x0, y0)has equation (y − y0) = t(x − x0). Plugging this in to the equation P (x, y) we get the equation for theintersection lt
⋂C:
0 = Pt(x) = P (x, y0 + t(x− x0)) and y = y0 + t(x− x0)
where
Pt(x) = P (x, y0 + t(x− x0)) =d∑
i,j=0
∂i+jP
∂xi∂yj(A)
ti(x− x0)i+j
i!j!
= P (A) + (x− x0)
[∂P
∂x(A) + t
∂P
∂y(A)
]+
(x− x0)2
2
[∂2P
∂x2(A) + 2t
∂2P
∂x∂y(A) + t2
∂2P
∂y2(A)
]+ ...
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(This discussion does not include the lines of slope t =∞. This case can be treated similarly by replacingt with s = t−1 and considering instead the line equation (x− x0) = s(y − y0) in all of the above.)
Definition 2.11. Let A = (x0, y0) be a point at the intersection of the curve C given by P (x, y) = 0 andthe line lt. Then we define the multiplicity of intersection of the curve C with the line lt at A by
mC⋂lt(A) = the order of the zero x0 of the polynomial Pt(x) = P (x, y0 + t(x− x0)),
namely mC⋂lt(A) = the highest power m such that (x− x0)m divides the polynomial Pt(x).
Note that if A = (x0, y0) is a point at the intersection of the curve C given by P (x, y) = 0 and the linelt then (x− x0) is already a factor of the polynomial Pt(x) above, so mC
⋂lt(A) ≥ 1.
Definition 2.12. Tangent lines and tangent space. We say that a line lt is tangent to the curve Cat the point a if mC
⋂lt(A) ≥ 2.
The union of all the line tangent to a curve C at a point forms the tangent space to C at that point.
Proposition 2.13. Equation of a Tangent Line (Affine Case) Given an algebraic curve C ⊂ K2
defined by a polynomial equation P (x, y) = 0 and A = (x0, y0) a on C then
a) If A is a singular point, then all the lines through A are tangent to C.
b) If A is non-singular point, then the unique tangent line to C at A = (x0, y0) has equation
(x− x0)∂P
∂x(A) + (y − y0)
∂P
∂y(A) = 0.
Proof. Indeed, the condition mC⋂lt(A) ≥ 2 is equivalent to (x − x0)2|Pt(x). Looking at the formula for
Pt(x) above we get that (x− x0)2|Pt(x) iff
∂P
∂x(A) =
∂P
∂y(A) = 0 or
∂P
∂x(A) + t
∂P
∂y(A) = 0.
. In the first case, a0 is a singular point and then all the lines intersect C with multiplicity order ≥ 2. Thesecond case makes the equation for lt into:
(x− x0)∂P
∂x(A) + (y − y0)
∂P
∂y(A) = 0.
The case (∂P∂x (A) 6= 0) works similarly.
Example 2.14. The cubic y2 = x3 + 3x contains a rational point A0 = (1, 2) ∈ Q2.The polynomial P (x, y) = y2 − x3 − 3x has partial derivatives
∂P
∂x(A) = −3x2 − 3 |x=1 = −6 ,
∂P
∂y= 2y |y=2 = 4
hence tangent line to the given cubic at the point A0 is
−6(x− 1) + 4(y − 2) = 0, ⇐⇒−6x+ 4y − 2 = 0, ⇐⇒
y =3x+ 1
2.
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We can use this to find a new rational point on the cubic, namely the other intersection point of the cubicwith the given tangent line. Indeed, subbing back into the original equation y2 = x3 + 3x we get
y =3x+ 1
2and
(3x+ 1
2
)2
= x3 + 3x.
The cubic equation in x simplifies to
4x3 − 9x2 + 6x− 1 = (x− 1)2(4x− 1)
hence A1 = (14 ,
78). We can continue this process to get more rational points on the cubic, at each step n
intersecting the cubic with the tangent line at An.
Remark 2.15. The intersection of a line with a curve. Assume that K is an algebraically closedfield (e.g. C). Given a degree d plane curve C as above and a line lt in the plane, then unless the line lt iscontained inside C. we have ∑
A∈C⋂lt
mC⋂lt(A) = degPt(x) ≤ d
Example 2.16. If C is the nodal cubic y2 = x3 − λx2, then the point 0 = (0, 0) is a singular point of Chence mC
⋂lt(0) ≥ 2 for all t ∈ K. Since the curve C is of degree 3, it follows that each line has at most one
other intersection point with C. Indeed, the lines through 0 have equations y = xt hence their intersectionwith C is given by
x2(t2 − x+ λ) = 0.
Hence lt⋂C consists of 0 with multiplicity 2 and one other point (x, y) = (λ + t2, t(λ + t2)). This yields
the parametrisation
φ : K −→ { lines through 0} −→ C
t −→ lt −→ (λ+ t2, t(λ+ t2)) ∈ lt⋂C.
Note that the case t2 = λ gives φ(t) = (0, 0) = 0. This corresponds to lines lsqrtλ and l−sqrtλ which intersectC with multiplicity 3 at 0. The map φ fails to be injective exactly at these two points. The map φ isbijective when λ = 0, the case of the cuspidal curve.
Proposition 2.17. Tangents for non-singular parametric curves. Let K = R or C and let φ : K→K2 with φ(t) = (x(t), y(t)) be a differentiable map giving a parametrisation of a plane curve C. Assumethat ∇φ(t0) 6= 0. Then the tangent line at the point (x0, y0) = (x(t0), y(t0)) is given as the limit directionof chords passing through the points (x0, y0) and (x(t), y(t)) when t→ t0.
Proof Indeed, the tangent vector to C at (x0, y0) is limit direction of a chord passing through the points(x0, y0) and (x(t), y(t)) is
limt→t0
φ(t)− φ(t0)
t− t0=
(dx
dt(t0),
dy
dt(t0)
)hence the parametric equation of the tangent line is
(x, y) = (x0, y0) + λ
(dx
dt(t0),
dy
dt(t0)
).
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Eliminating λ give the implicit equation
(y − y0)dx
dt(t0)− (x− x0)
dy
dt(t0) = 0.
On the other hand, differentiating P (x, y) = P (x(t), y(t)) = 0 at t0 gives
∂P
∂x(x0, y0) · dx
dt(t0) +
∂P
∂y(x0, y0) · dy
dt(t0) = 0.
Combining the two equations above gives the equation of the tangent line.
Proposition 2.18. Reducible plane curves. Let C = C1⋃C2 be a reducible plane curve where Ci is
given by the equation Pi(x, y) = 0 and hence P (x, y) = P1(x, y)P2(x, y).
If a ∈ C1⋂C2 then a is a singular point of C.
Proof. Apply the product rule to ∂P∂x and ∂P
∂y .
The multiplicity of intersection of lines with curves allows us describe how ”bad” a singularity is:
Definition 2.19. Let a be a singular point of an algebraic curve C. We define the multiplicity (order) ofthe singularity at a to be
m(A) = min{mC⋂lt(A); lt line through a}.
Definition 2.20. The ”tangent cone” to a curve at a singular point. If a is a singular point of analgebraic curve C, then the ”tangent cone” of C at a is made of all the lines lt for which
mC⋂lt(A) > m(A).
After recalling the formula for the polynomial
Pt(x) =d∑
i,j=0
∂i+jP
∂xi∂yj(A)
ti(x− x0)i+j
i!j!
the condition that (x− x0)m(A)|Pt(x) gives
Proposition 2.21. Let C be a plane curve given by the equation P (x, y) = 0 and let A = (x0, y0) be asingular point of C. Then
m(A) = min{m;∂mP
∂xi∂yj(A) 6= 0 for some i, j with i+ j = m.}
The tangent cone to C at A is the set of points in the plane defined by the equation
∑i+j=m(A)
∂m(A)P
∂xi∂yj(A)
(x− x0)i(y − y0)j
i!j!= 0
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Example 2.22. The curve given by the equation
(x4 + y4 − x2 − y2)2 = 9x2y2
has a singular point at the origin O = (0, 0) with order of singularity m(O) = 4. The equation of thetangent cone at the origin is found by keeping all the degree 4 monomials from the original equation (afterexpanding the squared bracket):
(x2 + y2)2 − 9x2y2 = 0
which factors into
(x2 + y2 − 3xy)(x2 + y2 + 3xy) = 0 ⇐⇒
(y − 3 +√
5
2x)(y − 3−
√5
2x)(y − −3 +
√5
2x)(y − −3 +
√5
2x) = 0
the union of 4 lines.
Note. All the monomials (x− x0)i(y − y0)j in the equation of the tangent cone have the same totaldegree i + j = m. Such polynomial expressions are known as homogeneous in degree m. Homogeneouspolynomials of degree m > 1 in two variables can always be factorised in an algebraically closed field:
Proposition 2.23. Let K be an algebraically closed field. Let C be a plane curve in K2 given by theequation P (x, y) = 0 and let A = (x0, y0) be a singular point of C.
The tangent cone consists of a union of lines through A, which approximates the shape of C close to A.
Proof. We will prove that for special values of t, the line of equation (y − y0) = t(x − x0) is inside thetangent cone above. To do that, we want to substitute (y − y0) = t(x− x0) in the equation of the tangentcone. We get the polynomial equation
(x− x0)m
∑i+j=m
∂mP
∂xi∂yj(A)
tj
i!j!
= 0.
After simplifying by (x− x0)m, the resulting polynomial equation in the variable t in general is of degreem and has m roots t1, ..., tm which are the slopes of the lines in the tangent cone.
A special case is when ∂mP∂ym (A) = 0, which leads to an equation of degree < m. However, in this case
the line of slope t =∞ accounts for the missing roots. Indeed, in order to allow for the case of slope t =∞we consider t = u
v , where v could be 0. Then the equation of the line could be rewritten parametrically as{x = x0 + vλ
y = y0 + uλfor all λ ∈ K
Substituting this in the equation of the tangent cone we get
λm
∑i+j=m
∂mP
∂xi∂yj(A)
viuj
i!j!
= 0 for all λ ∈ K.
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After simplifyind by λm, the resulting homogeneous polynomial in u, v factors into m linear terms:
∂mP
∂xi∂yj(A)
viuj
i!j!=
m∏i=1
(uvi − vui) = 0
for some constants ui, vi, including the case v = 0 which corresponds to t = ∞. Indeed, as seen in thecase of conics, there is a bijection between homogeneous equations in u, v and equations in the variablet = u
v .
3 Projective lines and plane
Let O = (x0, y0) be a point in the plane K. In the examples above we have used a map
slope : { lines through O in the plane K2} −→ K⋃{∞}.
For simplicity, let O = (0, 0). In order to allow for the case of infinite slope t = ∞, we write t = u2u1
anddescribe the line lt by parametric equations{
x = u1λ
y = u2λfor all λ ∈ K
but the direction vector of the line (u1, u2) is not uniquely determined: In fact, any other point (λu1, λu2)would work as well. This leads bijective maps
K⋃{∞} ∼= { lines through O in the plane K2} ∼=
K2 \ {O}∼
=: P1
where ∼ is the equivalence relation given by (u1, u2) ∼ (λu1, λu2) for all λ ∈ K∗. We call this set theprojective line P1, and we call [u1 : u2] the homogeneous coordinates of the point lt ∈ P1. Note thathomogeneous coordinates are defined up to multiplication by a non-zero constant λ.
The definition of a projective line can be generalised to higher dimensions as follows:
Definition 3.1. The n-dimensional projective space KPn, or in short Pn is defined as the set of lines inthe (n+ 1)-dimensional space Kn+1 passing through the origin O = (0, 0, · · · , 0):
Pn = {l; l line in Kn+1 such that O ∈ l}.
For each such line, l \ {0} is the orbit of the multiplicative group K∗ acting on Kn+1:
K∗ ×Kn+1 −→ Kn+1
(λ, (x0 : x1 : · · · : xn)) −→ λ(x0, x1, · · · , xn) := (λx0, λx1, · · · , λxn).
Hence Pn can be described as a quotient
Pn =Kn+1 \ {0}
K∗or equivalently, Pn =
Kn+1 \ {0}∼
,
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whose points are equivalence classes of (n + 1)-dimensional vectors (x0, x1, · · · , xn) with respect to theequivalence relation
(x0, x1, · · · , xn) ∼ λ(x0, x1, · · · , xn),
where λ ∈ K∗. These (n+ 1)-tuples are known as homogeneous coordinates in Pn are are well defined onlyup to multiplication by a non-zero constant. A point in Pn with the homogeneous coordinates above willbe denoted [x0 : x1 : ... : xn].
Remark 3.2. The projective space CPn is compact as it can be viewed as the quotient of the sphereSn = {(x0, ..., xn) ∈ Cn+1;
∑i |xi|2 = 1} which is a compact subset of Cn+1. For this we write CPn = Sn
S0 .Similarly RPn is a compact space.
Consider the hyperplane H1 given by the equation x0 = 1 in Kn+1. Most lines through 0 in Kn+1
intersect this hyperplane at exactly one point. The only exceptions are the lines parallel to H1. These arethe lines in the hyperplane H0 given by the equation x0 = 0. Hence there is a one-to-one correspondence:
{ lines through 0 in Kn+1 \H0} ←→ { points in H1}{[x0 : x1 : ... : xn] ∈ Pn ; x0 6= 0} ←→ Kn;
[x0 : x1 : ... : xn] −→(x1
x0, ...,
xnx0
)[1 : u1 : ... : un] ←− (u1, ..., un)
Indeed, the two morphisms above are inverses of each other as
[x0 : x1 : ... : xn] =
[1 :
x1
x0: ... :
xnx0
]The correspondence above identifies U0 = {[x0 : x1 : ... : xn] ∈ Pn ; x0 6= 0} with Kn. Note that thecomplement of U0 in Pn is the set {[0 : x1 : ... : xn] ∈ Pn} = Pn−1. Thus
Pn = Kn⊔
Pn−1
where the points in Pn−1 are called points at ∞ and Pn−1 is called a hyperplane at ∞.
Remark 3.3. Since Pn is compact and Pn = Kn⊔Pn−1, we can say that Pn is the completion of Kn to a
compact space (or compactification, in short).
Note that there is nothing special in the choice of the coordinate x0 in the above construction. Wecould just as well identify Ui = {[x0 : x1 : ... : xn] ∈ Pn ; xi 6= 0} = Kn making {[x0 : x1 : ... : xn] ∈Pn ; xi = 0} = Pn−1 into the hyperplane at the infinity. (Like the horizon line - which points are at infinitydepends the viewer’s perspective.)
Exercise: Prove that Pn with the cover Pn =⋃ni=0 Ui is a manifold and that the transition maps are
algebraic, namely are fractions of polynomials in the given variables.
3.1 Completion of an affine curve C ⊂ K2 to a projective curve C ⊂ P2.
For a fixed decomposition P2 = K2⊔P1, there is a unique way to complete an affine curve to a projective
curve.
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Considering P2 with homogenous coordinates [X : Y : Z] and the inclusion
K2 −→ P2
(x, y) −→ [x : y : 1],
then [x : y : 1] = [X : Y : Z] =[XZ : YZ : 1
]leads to the substitution x = X
Z and y = YZ . The equation
P (x, y) = 0 becomes P(XZ ,
YZ
)= 0 which after clearing denominators becomes
P (X,Y, Z) = ZdP
(X
Z,Y
Z
)= 0.
Definition 3.4. The solution set of the polynomial equation P (X,Y, Z) = 0 is a projective curve C calledthe completion of the affine curve C.
Example 3.5. Conics. Consider the following affine curves in K2:
a) The circle C1 given by the equation x2 + y2 = 1.
b) The parabola C2 of equation y = x2.
c) The hyperbola C3 of equation xy = 1.
Applying the substitution x = XZ and y = Y
Z to the 3 equations above and clearing denominators weget:
a) The equationX2+Y 2 = Z2. The solution set C1 of this equation consists of the circle C1 together withthe points at infinity given by Z = 0 and X2+Y 2 = Z2. Hence C1 = C1
⋃{[1 :
√−1 : 0], [
√−1 : 1 : 0]}
in the case K = C.
b) The equation X2 = Y Z with solution set C2 = C2⋃{[0 : 1 : 0]}. Note that we can rewrite the
equation as X2 +(Y−Z
2
)2=(Y+Z
2
)2which is the same as (a) after a change of variables.
c) The equation XY = Z2 with solution set C3 = C3⋃{[0 : 1 : 0], [1 : 0 : 0]}. Note that the equations
from (b) and (c) differ only by a permutation of variables.
We can conclude that the projective curves Ci are isomorphic, their equations differ only by somechanges of variables. These are examples of projective varieties.
If P (x, y) is a degree d polynomial, we note that the polynomial P (X,Y, Z) satisfies
P (λX, λY, λZ) = λdZdP
(λX
λZ,λY
λZ
)= λdP (X,Y, Z).
Definition 3.6. A polynomial f ∈ K[x0, x1, · · · , xn] is homogeneous of degree d if
f(λx0, λx1, · · · , λxn) = λdf(x0, x1, · · · , xn)
for all λ ∈ C∗.
14
Note: The definition insures that for a homogeneous polynomial,
f(x0, x1, · · · , xn) = 0 ⇐⇒ f(λx0, λx1, · · · , λxn),
hence the condition f(x0, x1, · · · , xn) = 0 is independent on the choice of representatives in the class[x0 : x1 : ... : xn].
Let K[x0, x1, · · · , xn]k denote the subset of K[x0, x1, · · · , xn] whose elements are all the homogeneouspolynomials of degree k in variables x0, x1, · · · , xn.
3.2 Projective lines in the projective plane.
A projective line P1 in the projective plane P2 can be written in two ways:
Implicit Homogeneous Equation. As the completion of the affine line ax + by + c = 0, by takingx = X
Z and y = YZ , leading to the homogeneous degree 1 equation
aX + bY + cZ = 0.
Proposition 3.7. Any two distinct lines in P2 intersect at exactly 1 point.
Proof. If [a : b : c] 6= [a′ : b′ : c′] then solving the simultaneous equations
aX + bY + cZ = 0 and a′X + b′Y + c′Z = 0
yields a unique solution in P2, namely
[X : Y : Z] =
[∣∣∣∣ b cb′ c′
∣∣∣∣ :
∣∣∣∣ c ac′ a′
∣∣∣∣ :
∣∣∣∣ a ba′ b′
∣∣∣∣] .Note that at least one of the determinants above must be non-zero since [a : b : c] 6= [a′ : b′ : c′], anddepending on which of these is non-zero you can arrive at the formula above after solving the system byelimination.
A quick explanation for the formula above comes from the row-expansion of determinants
0 =
∣∣∣∣∣∣a b ca b ca′ b′ c′
∣∣∣∣∣∣ =
∣∣∣∣∣∣a′ b′ c′
a b ca′ b′ c′
∣∣∣∣∣∣ .
Parametric Equation. The projective line through the points A1 = [X1 : Y1 : Z1] and A0 = [X0 : Y0 :Z0] is obtained as the quotient of the two dimensional vector space generated by the vectors (X1, Y1, Z1)and (X0, Y0, Z0):
LA1A0 =〈(X1, Y1, Z1), (X0, Y0, Z0)〉 \ {0}
∼= {[u0X0 + u1X1 : u0Y0 + u1Y1 : u0Z0 + u1Z1] ; (u0, u1) ∈ K2 \ {0}}.
The parametric equation is equivalent to giving an isomorphism
φ : P1 −→ LA1A0 ⊂ P2
[u0 : u1] −→ [u0X0 + u1X1 : u0Y0 + u1Y1 : u0Z0 + u1Z1].
15
Lemma 3.8. Parametric to Implicit Equations for Lines. If A1 6= A0 ∈ P2 then the implicitequation for the line LA1A0 described above can be written as∣∣∣∣∣∣
X Y ZX0 Y0 Z0
X1 Y1 Z1
∣∣∣∣∣∣ = 0
Proof. Indeed, the equation above is equivalent to saying that the first row is a linear combination of theother two, which is equivalent to the parametric equations above.
The Intersection of a Projective Curve with a Projective line. Let C be a projective curvedefined by the homogeneous polynomial P (X,Y, Z), let A0 = [X0 : Y0 : Z0] and be a point on C andconsider a projective line
LA1A0 = {[u0X0 + u1X1 : u0Y0 + u1Y1 : u0Z0 + u1Z1] ; (u0, u1) ∈ K2 \ {0}}.
The intersection of the curve C with the line LA1A0 is given by the equation in [u : v]:
PA1A0(u0, u1) = P (u0X0 + u1X1 : u0Y0 + u1Y1 : u0Z0 + u1Z1) = 0.
For u0 6= 0 we will use the affine coordinate u = u1u0
to get a polynomial equation in one variable
PA1A0(u) = P (X0 + uX1 : Y0 + uY1 : Z0 + uZ1) = 0.
We note that the intersection point A0 corresponds to the root u = 0 of this polynomial. Just like in theaffine case, the multiplicity of intersection of C and LA1A0 at A0 is defined as
mC⋂LA1A0
(A0) = the order of the root u = 0 of PA1A0(u)
= min{m; tm|PA1A0(u)}.
Proposition 3.9. The intersection of a projective line with a projective curve. Assume thatK is an algebraically closed field (e.g. C). Given a degree d projective curve C ⊂ P2 and a line l in theplane not contained inside C, then the intersection C
⋂l consists of exactly d points (when counted with
multiplicity): ∑A∈C
⋂l
mC⋂l(A) = d
Proof. Indeed, for the line l = A0A1 the points in the intersection C⋂l are given in homogeneous coordi-
nates [u0 : u1] on l by the equation
PA1A0(u0, u1) = P (u0X0 + u1X1 : u0Y0 + u1Y1 : u0Z0 + u1Z1) = 0.
If P (X,Y, Z) is a homogeneous polynomial of degree d in variables X,Y, Z then PA1A0(u0, u1) is a homo-geneous polynomial of degree d in the variables u0, u1. In terms of affine coordinate u = u1
u0:
PA1A0(u0, u1) = ud0PA1A0(1, u)
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If K is algebraically closed, then the polynomial f(u) = PA1A0(1, u) is a product of linear factors:
f(u) = PA1A0(1, u) = c(u− r1) · (u− rk) hence
PA1A0(u0, u1) = ud0f(u) = ud0c(u− r1) · · · (u− rk)= ud−k0 c(u1 − r1u0) · · · (u1 − rku0),
with roots [u0 : u1] = [0 : 1] (with multiplicity (d− k)) and [1 : r1], ..., [1 : rk].
The Taylor Expansion for Homogeneous Polynomials: For A = [X : Y : Z] we have
PAA0(u) = P (uX +X0, uY + Y0, uZ + Z0) =d∑
i,j,k=0
∂i+j+kP
∂Xi∂Y j∂Zk(A0)
XiY jZk
i!j!k!ui+j+k
= P (A0) + u
[X∂P
∂X(A0) + Y
∂P
∂Y(A0) + Z
∂P
∂Z(A0)
]+
+u2
2
[PXX(A0)X2 + 2PXY (A0)XY + 2PXZ(A0)XZ + PY Y (A0)Y 2 + 2PY Z(A0)Y Z + PZZ(A0)Z2
]+ ...
where PXX = ∂2P∂X2 , PXY = ∂2P
∂X∂Y , PXZ = ∂2P∂X∂Z etc.
Hence the following definitions are the projective versions of their affine counterparts:
Definition 3.10. Let A0 = [X0 : Y0 : Z0] be a point on the projective curve C defined by the homogeneouspolynomial equation P (X,Y, Z) = 0. We say that A0 is a singular point of C iff
∂P
∂X(X0, Y0, Z0) =
∂P
∂Y(X0, Y0, Z0) =
∂P
∂Z(X0, Y0, Z0) = 0.
Otherwise we say that the point A0 is non-singular. A curve is called non-singular iff all its points arenon-singular.
In other words, if A0 = [X0 : Y0 : Z0] is a singular point of the curve if mC⋂LAA0
(A0) ≥ 2 for all linesLAA0 through A0.
Definition 3.11. Let A0 = [X0 : Y0 : Z0] be a non-singular point on the projective curve C. The tangentline to C at a is defined by the equation
X∂P
∂X(A0) + Y
∂P
∂Y(A0) + Z
∂P
∂Z(A0) = 0.
If A0 = [X0 : Y0 : Z0] then the tangent to C at A0 is the unique line LAA0 such that mC⋂LAA0
(A0) ≥ 2.
Similarly, the multiplicity at a singular point A0 = [X0 : Y0 : Z0] of the projective curve C can bedefined as the smallest integer m such that
∂mP
∂Xi∂Y j∂Zk(A) 6= 0 for some i, j, k such that i+ j + k = m
and is the same as the minimum intersection multiplicity mC⋂LAA0
(A0) among all lines AA0.
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Lemma 3.12. Euler’s Formula Suppose P (X,Y, Z) is a homogenous function of degree n. Then thefollowing relation holds nP = X ∂P
∂X + Y ∂P∂Y + Z ∂P
∂Z
Proof. As P is homogenous, for all λ, we have P (λX, λY, λZ) = λnP (X,Y, Z). Differentiating both sideswith respect to λ gives, via the chain rule,
∂P
∂X(λX, λY, λZ)
∂λX
∂λ+∂P
∂Y(λX, λY, λZ)
∂λY
∂λ+∂P
∂Z(λX, λY, λZ)
∂λZ
∂λ= nλn−1P (X,Y, Z)
λn−1
(∂P
∂XX +
∂P
∂YY +
∂P
∂ZZ
)= nλn−1P (X,Y, Z)
for all λ. The desired result is obtained for λ = 1.
Proposition 3.13. Consider a projective curve C defined by a homogenous polynomial P (X,Y, Z). LetA0 = [X0 : Y0 : Z0] be a point on the projective curve C. Assume that Z0 6= 0 so that the point A0 ison the affine curve CZ = C
⋂UZ where UZ = (Z 6= 0). Then A0 is a non-singular point of C iff it’s a
non-singular point of CZ and in this case the tangent line to C at A0 is the completion of the tangent lineto CZ at A0.
Proof. The intersection of C with the affine plane UZ = (Z 6= 0) is the affine curve CZ given by thepolynomial PZ(x, y) = P (x, y, 1). Hence
∂PZ∂x
(x0, y0) =∂P
∂X(x0, y0, 1) and
∂PZ∂y
(x0, y0) =∂P
∂Y(x0, y0, 1).
Now A0 = [X0 : Y0 : Z0] = [x0 : y0 : 1] is a singular point of CZ iff
0 = PZ(x0, y0) =∂PZ∂x
(x0, y0) =∂PZ∂y
(x0, y0)
Using Euler’s formula this is equivalent to
0 = P (x0, y0, 1) =∂P
∂X(x0, y0, 1) =
∂P
∂Y(x0, y0, 1) =
∂P
∂Z(x0, y0, 1).
Furthermore Euler’s Formula gives
0 = nP (x0, y0, 1) = x0∂P
∂X(x0, y0, 1) + y0
∂P
∂Y(x0, y0, 1) +
∂P
∂Z(x0, y0, 1).
Using this to substitute ∂P∂Z (x0, y0, 1) in the equation of the projective tangent line
X∂P
∂X(x0, y0, 1) + Y
∂P
∂Y(x0, y0, 1) + Z
∂P
∂Z(x0, y0, 1) = 0
with [X : Y : Z] = [x : y : 1] we get exactly the equation of the tangent line to CZ :
(x− x0)∂P
∂X(x0, y0, 1) + (y − y0)
∂P
∂Y(x0, y0, 1) = 0.
Similarly the multiplicity of intersection of a curve with a line at a point is the same whether definedfor affine curves or their completions.
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3.3 Conics in the Projective Plane
Proposition 3.14. Let C be a plane conic in P2C given by the degree two homogeneous equation P (X,Y, Z) =
0 where
P (X,Y, Z) = q11X2 + 2q12XY + 2q13XZ + q22Y
2 + 2q23Y Z + q33Z2
=(X Y Z
)Q
XYZ
for Q =
q11 q12 q13
q12 q22 q23
q13 q23 q33
.
If det(Q) 6= 0 then C is a non-singular conic, isomorphic to the circle X2 + Y 2 = Z2 and hence with P1.
If det(Q) = 0 then C is the union of two intersecting lines (possibly the same line twice).
Proof. Let A0 = [X0 : Y0 : Z0]. Then(∂P∂X (A0) ∂P
∂Y (A0) ∂P∂Z (A0)
)= 2
(X0 Y0 Z0
)Q.
Hence there exists a singular point A of C iff the equation(X0 Y0 Z0
)Q =
(0 0 0
)has non-trivial
solutions, i.e. det(Q) = 0. If this is the case, then considering any other point B on C, the line AB intersectsC at A with multiplicity ≥ 2 and at B with multiplicity ≥ 1. Since C is of degree 2, this is only possible ifthe line AB is inside C. Hence C must be a union of two lines.
If C is non-singular, then the symmetric matrix can be diagonalised and after a change of variables, theequation of C can be brought to the form X2 + Y 2 +Z2 = 0 or letting Z = iU , we get X2 + Y 2 = U2.
Exercise. Assuming that det(A) = 0, find a way to factorise the polynomial P (X,Y, Z).
Remark 3.15. Given a conic C and a point A0 = [X0 : Y0 : Z0] on C as in the proposition above, let
PX =∂P
∂X, PY =
∂P
∂Y, PZ =
∂P
∂Z.
If A0 is non-singular then the equation of the tangent line to C at the point A0 = [X0 : Y0 : Z0] is
0 =(X Y Z
) PX(A0)PY (A0)PZ(A0)
= 2(X Y Z
)Q
X0
Y0
Z0
.
Definition 3.16. Inflection Point. Let A0 = [X0 : Y0 : Z0] be a point on the projective curve C definedby P (X,Y, Z) = 0. We say that A0 is an inflection point of C iff there exists a line LAA0 such thatmC
⋂LAA0
(A0) ≥ 3.
By comparing with the Taylor expansion for homogeneous polynomials we see that this conditioninvolves both 1st and second order derivatives. However, with the insights gained from conics, this can bereduced to a simpler condition involving the determinant of a matrix:
Definition 3.17. Hessian. Let P (X,Y, Z) be a homogeneous polynomial of degree d. The Hessian HessPof P is the polynomial defined by
HessP = detHP , where HP =
PXX PXY PXZPY X PY Y PY ZPZX PZY PZZ
.
19
Here HP denotes the Hessian matrix with entries
PX =∂P
∂X, PXX =
∂2P
∂X2, PXY =
∂2P
∂X∂Yetc.
Proposition 3.18. Let A0 = [X0 : Y0 : Z0] be a point on the projective curve C defined by a homogeneouspolynomial P (X,Y, Z). Then A0 is an inflection point of C iff
HessP (X0, Y0, Z0) = 0.
Proof. Let P (X,Y, Z) be a homogeneous polynomial of degree n. From the Taylor expansion for homoge-neous polynomials we see that mC
⋂A0A(A0) ≥ 3 iff the point A = [X : Y : Z] is both on the tangent line
to C at A0, and on the conic CP,A0 associated to the matrix HP (A0) :
(X Y Z
) PX(A0)PY (A0)PZ(A0)
= 0 and(X Y Z
)HP (A0)
XYZ
= 0 (3.1)
On the other hand, Euler’s relations for PX , PY and PZ give
HP (A0)
X0
Y0
Z0
= (n− 1)
PX(A0)PY (A0)PZ(A0)
and hence the line tangent to the curve C is also tangent to the conic CP,A0 :
(X Y Z
) PX(A0)PY (A0)PZ(A0)
= 0 ⇐⇒(X Y Z
)HP (A0)
X0
Y0
Z0
= 0.
Thus eq.(3.1) actually says that the lineAA0 is included inside the conic CP,A0 (since otherwisemCP,A0
⋂AA0
(A0)+mCP,A0
⋂AA0
(A) = 2 + 1 = 3).
This is equivalent to the conic CP,A0 being degenerate, which from the reducibility criterion for conicsis equivalent to HessP (X0, Y0, Z0) = 0.
3.4 Elliptic Curves.
Example 3.19. Let Eλ denote the cubic curve given by equation
Y 2Z = X(X − Z)(X − λZ).
Prove that Eλ is non-singular iff λ 6= 0, 1. Prove that the line X = 0 is tangent to Eλ at A = [0 : 0 : 1]that line Z = 0 is tangent to Eλ at the inflection point B = [0 : 1 : 0].
Indeed, let P (X,Y, Z) = Y 2Z −X(X − Z)(X − λZ). Then
PX(X,Y, Z) = −X(X − Z)−X(X − λZ)− (X − Z)(X − λZ) = −3X2 + 2(1 + λ)XZ − λZ2;
PY (X,Y, Z) = 2Y Z;
PZ(X,Y, Z) = Y 2 + (1 + λ)X2 − 2λXZ.
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We note that PX = PY = PZ = 0 iff Y = X = λ = 0 or Y = 0, λ = 1 and Z = X, hence the onlysingular curves in the family happen when λ = 0, 1. At a = [0 : 0 : 1] we have PX(a) = −λ, PY (a) = 0,PZ(a) = 0 hence the tangent line has equation X = 0. At b = [0 : 1 : 0] we have PX(b) = PY (b) andPZ(b) = (1 + λ) 6= 0. Hence the equation of the tangent line at b is Z = 0. The intersection with Eλhas equation X3 = 0 of multiplicity 3 at X = 0, hence b is an inflection point. Alternatively, the Hessianmatrix is
HP =
PXX PXY PXZPY X PY Y PY ZPZX PZY PZZ
=
−6X + 2(1 + λ)Z 0 2(1 + λ)X − 2λZ0 2Z 2Y
2(1 + λ)X − 2λZ 2Y 2λX
.
Subbing in b makes the first row and columns equal to 0.
Assuming that each elliptic curve has an inflection point, we can prove that all non-singular cubics canbe written as members of a 1-parameter family of cubics (a family containing one cubic for each value ofthe parameter λ.)
Theorem 3.20. Elliptic Curves in Weierstrass Standard Form. Every non-singular plane cubic(which contains an inflection point B and a tangent line BA at A)∗ can be written in the form Eλ givenby equation
Y 2Z = X(X − Z)(X − λZ)
with λ 6= 0, 1, for a suitable choice of coordinates X,Y, Z.
∗Note: we will soon prove that every non-singular cubic contains at least an inflection point B and atangent line BA at A. This will come as a consequence of Bezout’s Theorem.
Proof. Let B be an inflection point and let lB be the tangent to the cubic at B. Let A be another pointon the cubic such that AB is tangent to E at A. We can always choose homogeneous coordinates on P2
such that A = [0 : 0 : 1], B = [0 : 1 : 0] and the tangent line lB is given by Z = 0. Let
P (X,Y, Z) =∑
i+j+k=3
pijkXiY jZk
be the equation of the cubic in these coordinates.
The line AB through A = [0 : 0 : 1] and B = [0 : 1 : 0] has points of the form [0 : u1 : u2] = [0 :u1
u2: 1]
when u2 6= 0. Let u =u1
u2. Since the line AB intersects E with multiplicity 2 at A = [0 : 0 : 1], it follows
that
u2 | P (0, u, 1) =∑i+k=3
p0ikui =⇒ p003 = p012 = 0.
Similarly the line lB of equation Z = 0 has points of the form [u : 1 : 0] with u ∈ K. Since the line lBintersects E with multiplicity 3 at B = [0 : 1 : 0], it follows that
u3 | P (u, 1, 0) =∑i+j=3
pij0ui =⇒ p210 = p120 = p030 = 0.
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Given the above coefficients, we get
P (X,Y, Z) = p300X3 + p201X
2Z + p102XZ2 + p021Y
2Z + p111XY Z.
After rewriting the quadratic p201X2 + p021Y
2 + p111XY as (aX + bY )2 − cX2, changing variables so thataX + bY becomes Y , we can factor out the polynomial in X, Z to get
P (X,Y, Z) = Y 2Z −X(X − αZ)(X − βZ).
Finally, we can re-scale Z by β to get
P (X,Y, Z) = Y 2Z −X(X − Z)(X − λZ).
3.5 Parameter spaces for projective curves in the projective plane.
A degree d projective curve in the plane is given by a homogeneous equation
P (X,Y, Z) =∑
i+j+k=d
pijkXiY jZk
The number of terms in the sum above is given by the number of non-negative integer triplets (i, j, k) withi+ j + k = d. By a ”Stars and Bars” argument, there are
(d+2
2
)such triplets.
It is often convenient to think about the set of all plane curves in P2 of a given degree. Interestingly,such sets also have a structure of projective spaces.
Proposition 3.21. The set of all degree d curves in P2 is a projective space of dimension(d+2
2
)− 1.
Proof. Let S denote the set of all degree d homogeneous polynomials in the variables X,Y, Z and let Pdenote the set of all degree d curves in P2. Then S is a vector space with basis {XiY jZk ; i+j+k = d} andS = (P \ {0})/K∗, as multiplying a polynomial by a constant in K∗ does not modify its solution set.
Example 3.22. The set of lines in P2 forms a projective plane P2 This is called the dual plane of P2 andit’s sometimes denoted as (P2)V . Indeed, the line given by implicit equation AX + BY + CZ = 0 gives apoint [A : B : C] ∈ (P2)V and reciprocally.
For a given point p0 = [X0 : Y0 : Z0] ∈ P2, the set of all lines in P2 passing through p0 formsa line in (P2)V . Indeed, the coordinates [A : B : C] ∈ (P2)V of such lines must satisfy the equationAX0 +BY0 +CZ0 = 0. Thought of as an equation in homogeneous variables A,B,C with given coefficientsX0, Y0, Z0, this is the equation of a line in (P2)V .
Example 3.23. The set of conics in P2 forms a projective space of dimension(
2+22
)− 1 = 5. Indeed, the
conic given by implicit equation
p200X2 + p110XY + p101XZ + p020Y
2 + p011Y Z + p002Z2 = 0
gives a point [p200 : p110 : p101 : p020 : p011 : p002] ∈ P5 and reciprocally, every 5-tuple of homogeneouscoordinates defines a conic as above.
22
The set of conics in P2 passing through a fixed point forms a P4, the set of conics passing through 2fixed points forms a P3 and the set of conics passing through 3 non-collinear points forms a P2. The set ofconics passing through 4 points no 3 of which are collinear forms a projective line P1. Given 5 points, no3 of which are collinear, there exists a unique conic passing thorough those points.
Indeed, given a point P1 = [X1 : Y1 : Z1] ∈ P2, then the set of conics passing through that point is thesolution set of the linear equation
p200X21 + p110X1Y1 + p101X1Z1 + p020Y
21 + p011Y1Z1 + p002Z
21 = 0
where [p200 : p110 : p101 : p020 : p011 : p002] are the unknowns and X1, Y1, Z1 are known fixed quantities.Hence the solution set is a codimension one linear subset in P5, hence a P4.
Each new point which must be on the conics imposes an additional linear equation on the set of conics.If the equations are linearly independent, then the dimension of the solution set decreases by 1 with eachnew equation. The condition that the points are distinct and in general position (no 3 are collinear) ensuresthat the resulting equations are linearly independent. Algebraically, for conics passing through k pointsthis is equivalent to the matrix
X21 X1Y1 X1Z1 Y 2
1 Y1Z1 Z21
X22 X2Y2 X2Z2 Y 2
2 Y2Z2 Z22
· · · · · · · · · · · · · · · · · ·X2k XkYk XkZk Y 2
k YkZk Z2k
having maximum rank, which may be tedious to check algebraically. We can replace this by a geometricargument as follows: Assuming that the k-th row is a linear combination of the others, this would implythat all the conics passing through the points Pi = [Xi : Yi : Zi] for i = 1, 2, ..., k − 1, also pass throughthe point Pk = [Xk : Yk : Zk]. However, if k ≤ 5 and no 3 of the points are collinear, we can always finda conic passing though P1, ..., Pk−1 but not Pk, For example when k = 5, we can take our conic to be theunion of lines P1P2
⋃P3P4.
Consider two non-degenerate conics given by the quadratic polynomials P (X,Y, Z) and Q(X,Y, Z). ByBezout’s theorem, these conics intersect at 4 points (counted with multiplicity). By the discussion above,the set of conics passing through these points forms a projective line P1. Hence any conic passing throughthose 4 points must be given by a polynomial λP (X,Y, Z) + µQ(X,Y, Z), with λ, µ ∈ K. Indeed, such aconic does pass through the 4 points of intersection, and it corresponds uniquely to a point [λ : µ] ∈ P1.
Example 3.24. The set of cubics in P2 forms a projective space of dimension(
2+32
)− 1 = 9. Indeed, the
cubic given as the zero set of the polynomial
p300X3 + p210X
2Y + p201X2Z + p120XY
2 + p111XY Z + p102XZ2 + p030Y
3 + p021Y2Z + p012Y Z
2 + p003Z3
gives a point [pijk]i+j+k=3 ∈ P9 and reciprocally, every 10-tuple of homogeneous coordinates defines a cubicas above.
For 1 ≤ k ≤ 8, a set of points P1, ..., Pk in the plane are said to be in general position if no 3 arecollinear and no 6 of them are on the same conic. For 1 ≤ k ≤ 8, given k such points P1, ..., Pk in generalposition, the set of cubics passing through all k points forms a projective space P9−k. In particular, thereexists a line P1 of cubics passing through 8 points in general position. All the cubics on that line have anadditional 9th point in common.
23
Indeed, each point gives a new linear equation for the coordinates pijk of the cubic. To check thatthe conditions are linearly independent, it is enough, for each k points in general position, to find a cubicpassing though P1, ..., Pk−1 but not Pk. When k = 8, we can take our cubic to be the union of the conicthrough P1, ..., P5 with the line P6P7.
4 Resultant and the intersection of two plane curves
4.1 The Resultant of Two Polynomials
Working in the projective plane over an algebraically closed field allows for consistent count-ing of intersection points.
Let K be an algebraically closed field.
Let f, g ∈ K[x, y] be two polynomials. We will denote by V(f) the plane curve given by f = 0 andsimilarly by V(g) the plane curve given by g = 0, while V(f, g) will represent the intersection of the twocurves.
Similarly if f, g ∈ K[X,Y, Z] are homogeneous polynomials. We will denote by V(f) the projectiveplane curve given by f = 0 and similarly by V(g) the projective plane curve given by g = 0, while V(f, g)will represent the intersection of the two curves. We would like to study the intersection set V(f, g).
Example 4.1. Let f(x, y) = 2y + x− 5 and g(x, y) = 3y − 2x− 11.
Then in order to find the set V = V(f, g) we can first eliminate y as follows:
3f(x, y)− 2g(x, y) = 6y + 3x− 15− 6y + 4x+ 22 = 7x+ 7
Hence f = g = 0 implies x = −1 and then solving f(−1, y) = 0 we get y = 3. We have found V ={(−1, 3)} ⊂ K2.
Example 4.2. Let f(x, y) = y − 2x− 1 and g(x, y) = xy − 1.
Then V = V(f, g) is a set in the plane K2 which can be found by first eliminating y as follows:
−xf(x, y) + g(x, y) = −xy + 2x2 + x+ xy − 1 = 2x2 + x− 1 = (x+ 1)(2x− 1)
Hence f = g = 0 implies x = −1 or x = 1/2 and then solving f(−1, y) = 0 we get y = −1 and solvingf(1/2, y) = 0 we get y = 2. We have found V = {(−1,−1), (1/2, 2)} ⊂ K2.
Remark 4.3. Eliminating y corresponds to projecting π : K2 → K with π(x, y) = x.
Building on the previous example, consider f, g ∈ K[x, y] and V = V(f, g) ⊆ K2. We consider theprojection π : K2 → K1 given by π(x, y) = x. Our goal is to eliminate y and thus find an equation forπ(V ). Just like before, we would like to find a combination
R(x) = s(x, y)f(x, y) + t(x, y)g(x, y) ∈ 〈f, g〉⋂
K[x]
which would give us a necessary (but not always sufficient) equation for the projection π(V ).
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One-variable case
If we think of x as a constant then we write the equation above as
R = s(y)f(y) + t(y)g(y).
where s(y) =∑k
i=0 uiyi, f(y) =
∑di=0 fiy
i, t(y) =∑h
i=0 viyi, g(y) =
∑ei=0 fiy
i in K[y].
To make our life as easy as possible, we assume deg s(y) < deg g(y) and deg t(y) < deg f(y).
By identifying the coefficients of yi on LHS and RHS of the equation above we get
R = f0s0 + g0t0 and 0 =
i∑j=0
(fjsi−j + gjti−j) for i = 1, 2, ...deg(f) + deg(g)− 1.
We can rewrite this in terms of matrices:
f0 g0
f1 f0 g1 g0
f2 f1 f0 g2 g1 g0
· · · · · · · ·· · · · · · · · · g0
fd · · · · · · · · · g1
fd · · · · f0 · · · · ·· · · · f1 ge · · · ·· · · · ge · · ·· · · · · ·· · · ·
· fd ge
s0
s1
s2
···
se−1
t0t1t2·
td−1
=
R00········0
(4.1)
The square matrix has size d+ e = deg f(y) + deg g(y) and is known as the Sylvester matrix Syl(f, g).
Definition 4.4. Given a field K and polynomials f(y), g(y) ∈ K[y], define the resultant
R(f, g) = det(Syl(f, g))
where Syl(f, g) denotes the Sylvester matrix of f and g.
Proposition 4.5. Given a field K and polynomials f(y), g(y) ∈ K[y], there exist polynomials s(y), t(y) ∈K[y] with deg s(y) < deg g(y) and deg t(y) < deg f(y) such that the resultant R = R(f, g) satisfies
R = s(y)f(y) + t(y)g(y).
Moreover, R as well as the coefficients of s and t are polynomials of fi and gi-s with integer coefficients.
Proof. In the matrix equation (4.1) above with R = R(f, g), either R 6= 0 hence the Sylvester matrix isinvertible, or R = 0 and then the Sylvester matrix has determinant 0, so the d + e equations are linearlydependent. In the first case the system has a unique solution (s0, ..., se−1, t0, ..., td−1). In the second casethe system has infinitely many solutions.
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To prove the last statement, rewrite equation (4.1) in the form ST = L where S is the Sylvester matrix,T = (s0, ..., se−1, t0, ..., td−1)T and L = (R, 0, ..., 0)T . Then in the first case
T = S−1L =1
detSS∗L =
1
RS∗L
where the entries of S∗ are minors of S and hence integer polynomials in fi, gi-s, while L is a multiple of Rwhich cancels 1/R leaving only integer coefficients. In the second case, there are infinitely many solutionsof the form T = λTp where Tp is a particular solution and λ is a free parameter. Calculating Tp by standardlinear algebra methods gives fractions whose numerator and denominator are both integer polynomials infi, gi-s. We can choose λ to be such that it cancels all denominators in Tp, leaving only integer polynomialsin fi, gi-s.
Corollary 4.6. Let K be a field and let f, g ∈ K[y].The resultant R(f, g) = 0 in K if and only if gcd(f, g) is a non-constant polynomial in y.If K is algebraically closed, then R(f, g) = 0 iff f and g have a common root in K.
Proof. Recall that gcd(f, g) is the non-zero polynomial of smallest degree which can be written in the forms(y)f(y) + t(y)g(y). If R(f, g) 6= 0 then the equation
R(f, g) = s(y)f(y) + t(y)g(y) ⇐⇒ 1 =s(y)
R(f, g)f(y) +
t(y)
R(f, g)g(y) ⇐⇒ gcd(f, g) = 1 in K[y],
and the coefficients s(y)R(f,g) ,
t(y)R(f,g) ∈ K[y] can be calculated by Euclid’s algorithm.
If R(f, g) = 0 then s(y)f(y) = −t(y)g(y) and since K[y] is a UFD, every irreducible factor of g(y)must divide s(y) or f(y). Since deg s(y) < deg g(y), then some irreducible factor of g(y) divides f(y) andreciprocally.
The Resultant for Plane Curves.
In the case of plane curves defined by the polynomials f(x, y) and g(x, y), we can find the x-coordinatesof their intersection points by calculating the resultant as a function of x:
R(x) = u(x, y)f(x, y) + v(x, y)g(x, y) ∈ 〈f, g〉⋂
K[x].
In this case we write f(x, y) = fx(y) =∑d
i=0 fiyi and g(x, y) = gx(y) =
∑ei=0 giy
i, with coefficientsfi, gi ∈ K[x]. Then calculate R(x) = R(fx, gx) as in Definition 4.4. Moreover as in the 1 variable case, theresultant R(fx, gx) is constant 0 as a polynomial in x iff f and g have a common factor in K[x, y] which isnot constant in y.
Proposition 4.7. Zeroes of the Resultant. Let f, g ∈ K[x, y] be two polynomials of degrees deg(f) = dand deg(g) = e such that such that
f(x, y) =d∑i=0
fiyi with fi ∈ K[x] and 0 6= fd ∈ K,
g(x, y) =e∑i=0
giyi, with gi ∈ K[x] and 0 6= ge ∈ K.
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Then the resultant of f and g is a polynomial R(x) and
R(x0) = 0⇐⇒ ∃y0 ∈ K such that f(x0, y0) = g(x0, y0) = 0.
Proof. Indeed, from the definition of the resultant
R(x) = u(x, y)f(x, y) + v(x, y)g(x, y) for every x ∈ K,
and in particular for x0. The conditions 0 6= fd ∈ K and 0 6= ge ∈ K insure that deg f(x, y) = deg f(x0, y)and deg g(x, y) = deg g(x0, y) hence R(x0) is the resultant of the polynomials f(x0, y) and g(x0, y) ∈ K[y].The proposition follows from the property of the resultant in 1 variable.
Remark 4.8. The proposition above says that the projection π : K2 → K given by π(x, y) = x restricts toa surjective map
π0 : { common zeroes of f, g} −→ { zeroes of resultant R}.
Hence the zeroes of R can be used to count the common zeroes of f and g.
4.2 Bezout’s Theorem
For consistent counting, we need to work in the projective plane. As well, we need to choose coordinatesso that π0 is a one-to-one correspondence.
Note that we can extend π0 to a projection π : P2 → P1 such that π[X : Y : Z] = [X : Z]. This is theprojection of centre [0 : 1 : 0], namely π[X : Y : Z] = [X : Z] can be thought of as the intersection of theline connecting [0 : 1 : 0] and [X : Y : Z] with the line Y = 0. Hence
Definition 4.9. Let O = [0 : 1 : 0]. A Suitable Choice of Coordinates satisfies the conditions:
(a) O is not on the same line with two common zeroes of f and g.
(b) O is not on a line tangent to V(f) or V(g) at a common zero of f and g.
(c) f(O) 6= 0 6= g(O).
Conditions (a) and (b) insure that π0 is injective and no multiple counting of roots is required, while(c) is equivalent to the condition on leading coefficients from the previous Proposition. Note that one canalways find a suitable choice of coordinates, as most points of P2 satisfy the conditions (a)-(c) above.
In the case of homogeneous polynomials f(X,Y, Z) and g(X,Y, Z) we calculate the resultant
R(X,Z) = u(X,Y, Z)f(X,Y, Z) + v(X,Y, Z)g(X,Y, Z) ∈ K[X,Z].
For a suitable choice of coordinates, the roots [X0 : Z0] of the resultant are in one-to-one correspondencewith common roots [X0 : Y0 : Z0] of f(X,Y, Z) and g(X,Y, Z).
Proposition 4.10. Let f(X,Y, Z) and g(X,Y, Z) be homogeneous polynomials of total degrees deg(f) = dand deg(g) = e. Then the resultant R(X,Z) is a homogeneous polynomial of degree deg(R) = de.
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Proof. With the notations from the Sylvester matrix, we have deg(fi) = d − i and deg(gj) = e − j. Theentries of the Sylvester matrix are
sij =
{fi−j if 1 ≤ j ≤ egi−j+e = j − i if e+ 1 ≤ j ≤ e+ d
Hence R =∑
σ∈Sd+esign(σ)
∏d+ei=1 siσ(i) has summands of degree
deg(
d+e∏j=1
sσ(j)j) =
d+e∑j=1
deg(sσ(j)j)
=
e∑j=1
deg(fσ(j)−j) +
e+d∑j=e+1
deg(gσ(j)−j+e)
=
e∑j=1
(d− σ(j) + j) +
e+d∑j=e+1
(−σ(j) + j) = de
Definition 4.11. Let A0 = [X0 : Y0 : Z0] be a common root of f(X,Y, Z) and g(X,Y, Z). Then for asuitable choice of coordinates, we define the multiplicity of intersection of the curves V(f) and V(g) at a
mf ,g(A0) = the order of the zero [X0 : Z0] of R(X,Z)
Here the order of a zero [X0 : Z0] of the homogeneous polynomial R(X,Z) is the highest power of (XZ0 −ZX0) which divides R(X,Z).
Note: Ideally, one should prove that this definition does not depend on the choice of coordinates, aslong as it’s a suitable choice. (see for example Kirwan’s book).
Theorem 4.12. Bezout’s Theorem Let K be an algebraically closed field. Let f(X,Y, Z) and g(X,Y, Z)be homogeneous polynomials of total degrees deg(f) = d and deg(g) = e. Then the set of common zeroesof f(X,Y, Z) and g(X,Y, Z), when counted with multiplicity, has exactly d · e elements:∑
A∈V(f ,g)
mf ,g(A) = deg(f) deg(g)
Proof. This is a direct consequence of the definition and proposition above. Indeed, R(X,Z) is a degree dehomogeneous polynomial over an algebraically closed field has exactly de roots (when counted with theirorders).
Proposition 4.13. Reducible plane curves. If C = C1⋃C2 is a reducible plane curve in P2, then C
has at least deg(C1) · deg(C2) singular points.
Proof. Let C = C1⋃C2 be a reducible plane curve where Ci is given by the equation Pi(x, y) = 0 and
hence P (x, y) = P1(x, y)P2(x, y). If a ∈ C1⋂C2 then a is a singular point of C. Indeed, apply the product
rule to get ∂P∂x (a) = 0 and ∂P
∂y (a) = 0.
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Corollary 4.14. If C ⊂ P2 is an irreducible curve of degree d, then C has at most(d−1
2
)= (d−1)(d−2)
2singular points.
Proof. If C is of degree 1, C is a line; it does not have singular points. If C is an irreducible conic thenC is smooth. Hence it enough to consider the case d ≥ 3. Suppose that C has
(d−1
2
)+ 1 singular points
P1, P2, · · · , P(d−12 )+1
. Choose d − 3 other random points on C, say Q1, · · · , Qd−3, so that together(d−1
2
)+ 1 + (d− 3) = d2
2 −d2 − 1 =
(d2
)− 1 points have been fixed on C.
There exists a curve C ′ of degree d − 2 which passes through all d2
2 −d2 − 1 points given above. This
is because the space of curves in P2 of degree d − 2 is a projective space PN with N =(d2
)− 1, and the
condition that such a curve passes through a given point gives one linear equation in PN .
Note that C and C ′ intersect at the (d−3) points Qj , and the(d−1
2
)+1 points Pi which have intersection
multiplicity of at least 2 as they are singular points of C.
=⇒ #(C ∩ C ′) ≥ d− 3 + 2(
(d− 1
2
)+ 1) = d− 1 + (d− 1)(d− 2) = d2 − 2d+ 1
=⇒ #(C ∩ C ′) > deg(C) · deg(C ′),
and so C and C ′ have a common component by Bezout’s theorem. However, C is irreducible and deg(C) >deg(C ′), so C and C ′ cannot have a common component. This is a contradiction, so the initial assumptionthat C has
(d−1
2
)+ 1 singular points must be false.
4.3 Elliptic Curves.
Another application of Bezout’s Theorem involves the count of inflection points on a curve.
Corollary 4.15. Inflection Point Count. Let P (X,Y, Z) be a homogeneous irreducible polynomial ofdegree d ≥ 3 defining the projective curce C in P2. Then C has at least 1 and at most 3d(d− 2) points ofinflection.
Proof. As deg(PXX) = deg(PXY ) = deg(PY Y ) = ... = (d − 2), we have degHessP ≤ 3(d − 2). ApplyingBezout’s theorem we get at least 1 and at most 3d(d− 2) points of inflection, unless P |HessP .
Exercise Let P (X,Y, Z) be a homogeneous irreducible polynomial. Then P |HessP iff d = 1.
[Hint: Assume P (0, 0, 1) = 0 6= PY (0, 0, 1) and use the Implicit Function Theorem.]
Recall that in the previous section we have used the existence of an inflection point to deduce
Theorem 4.16. The Standard Form of Elliptic Curves. Every non-singular plane cubic can bewritten in the form Eλ given by equation
Y 2Z = X(X − Z)(X − λZ)
with λ 6= 0, 1, for a suitable choice of coordinates X,Y, Z.
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Indeed, in Theorem 3.20 we have shown how to bring to standard form any cubic which contains aninflection point B and a tangent BA at A. Inflection points exist as a corollary to Bezout’s theorem andthe Hessian characterisation. Moreover, given an inflection point B = [X1 : Y1 : Z1] as above, a pointA = [X : Y : Z] such that BA is tangent to the conic is given by the symultaneus equations
X1∂P
∂X(A) + Y1
∂P
∂Y(A) + Z1
∂P
∂Z(A) = 0 and P (A) = 0.
By Bezout’s theorem, solutions always exist.
Theorem 4.17. Cayley- Bacharach Theorem. Every cubic curve C1 on an algebraically closed fieldthat passes through a given set of eight points in general position P1, ..., P8 also passes through a certain(fixed) ninth point P9, counting multiplicities.
Proof. As discussed in the section on families of curves, C1 is an element in a P1 of cubics passing throughthe 8 given points. Let C2 be another element of that P1, that is, another cubic passing through the8 points. According to Bezout’s theorem, C1 and C2 intersect at exactly 9 points (when counting withmultiplicity).
Theorem 4.18. The group structure of an elliptic curve. Every non-singular cubic plane curve Ewith inflection point OE has a special group structure (E ,+) where + is defined such that
A+B + C = OE (4.2)
whenever A,B,C are 3 collinear points on the curve E. Here OE is the identity element for +.
Group structures on elliptic curves over finite fields are particularly useful in cryptography, due to theunusual nature of the operation which makes it non-trivial to solve equations like nX = Y in this group.
Proof. Note that OE being an inflection point means that there is a tangent line intersecting E at OE withmultiplicity 3. This is compatible with the equation 4.2 as in this case 3OE = OE is a natural property ofthe identity element.
Let A,B ∈ E and let C be the third point of intersection of the curve E with the line AB. Theline through C and OE intersects the curve at another point D. We define A + B = D. Indeed, this iscompatible with equation 4.2 applied to the collinear points A,B,C and C,OE , D respectively, as
A+B + C = OE =⇒ C = −(A+B) and OE +D + C = OE =⇒ D = −C = A+B.
We note that the operation is commutative, as the construction only depended on the line AB which isthe same as line BA.
By applying the previous construction to A and OE we find A + OE = A for all A ∈ E . As well, theinverse −A is the third point of intersection of the line AOE with E . It remains to prove the associativity:
(A+B) + C = A+ (B + C) or equivalently, − ((A+B) + C) = −(A+ (B + C))
This is equivalent to proving that the line through A and (B + C) intersects the elliptic curve E for thethird time at the exact same point as the line through (A+B) and C.
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Figure 5: The associativity low for cubics
To prove this, let C1 denote the reducible cubic curve given as union of the line AB, the line throughOE and (B + C), and the line through C and (A+B). Hence by construction we have
C1
⋂E = {OE , A,B,C, (A+B),−(A+B), (B + C),−(B + C),−((A+B) + C)}
Let C2 denote the reducible cubic curve given as union of the line BC, the line through OE and (A+B),and the line through A and (B + C). Hence
C2
⋂E = {OE , A,B,C, (A+B),−(A+B), (B + C),−(B + C),−(A+ (B + C))}.
Hence all 3 cubics E , C1 and C2 pass through the 8 points
OE , A,B,C, (A+B),−(A+B), (B + C),−(B + C),
However, like in the Cayley-Bacarach theorem, there is a P1 of cubics passing through these 8 points andall the cubics in this pencil pass through a fixed 9th point of intersection of any 2 distinct cubics in thepencil. From above we find that this point must be −(A+ (B + C)) = −((A+B) + C).
Note: like with the Cayley-Bacarach theorem, for every 7 points among the 8 above, we have to find acubic which passes through them but not through the 8th point. This in not hard considering that the 8points are on a 3×3 grid given by the lines in C1 and C2. For every 7 points as above, there must be 3 whichare on one of the 3 lines in C1 (by the Pigeonhole Principle). One can then choose a non-degenerate conicpassing through the other 4 points. The union of these line and conic forms a cubic which is guaranteednot to pass through the 8th point.
Theorem 4.19. Pascal’s theorem Let X ⊂ P2 be a conic (the zero locus of a homogeneous polynomialof degree 2) and suppose A,B,C,D,E, F ∈ X are the vertices of a hexagon inscribed on this conic. ThenP = AB ∩DE, Q = BC ∩ EF , and R = CD ∩ FA are collinear.
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A
B
C
DEX
FP
R
Q
Proof. Consider two reducible cubics, X1 = AB ∪ CD ∪ EF and X2 = BC ∪DE ∪ FA, with associatedequations f1 = 0 and f2 = 0 respectively. Both equations are cubic (of degree 3), so by Bezout’s theoremX1 and X2 meet at nine points (as deg(X1 ∩X2) = deg(X1) · deg(X2) = 9). Six of these points are clearlyA, B, C, D, E, and F , so P , Q, and R must be the remaining three points by their original definitions.
Now choose another point S on X so that S does not coincide with any of the previously mentionedpoints. Then there exist a, b ∈ K such that (a · f1 + b · f2)(S) = 0. Hence S is a point on the cubic X3
given by the equation a · f1 + b · f2 = 0. We note that X3 intersects the conic X in at least seven points:A, B, C, D, E, F , and S. However, because deg(X3) · deg(X) = 6 < 7, it follows that X3 and X mustshare a common component by Bezout’s theorem. Therefore X3 = X ∪ L, where L is a line.
P , Q, and R lie on X1 and X2, and they are solutions of f1 = 0 and f2 = 0. Therefore they are alsosolutions of a · f1 + b · f2 = 0, implying P , Q, R ∈ X
⋃L. However, P , Q, R 6∈ X, so P , Q, R ∈ L; they
are collinear.
5 Topological Invariants
We have started our discussion of plane curves from the observation that certain plane curves can beparametrised, which greatly simplifies the task of drawing them. In order to parametrise a curve C itis enough to find a map φ : C → P1 which is one-to-one (almost) everywhere. Some curves cannot beparametrised at all, however, this is not easy to prove directly. For example, even if we can prove that acertain map φ : C → P1 is not one-to-one, there may exist other one-to-one maps which we just haven’tfound yet. Still, the situation is not as dreary as we might think. We will find that considering d-to-1maps φ : C → P1 may be useful in determining whether C can be parametrised or not. The idea is toassociate a number to each geometric object C, such that for two isomorphic objects C and C ′, the numbersassociated to them are the same. Such numbers are called invariants. (More generally, an invariant maynot necessarily be a number, it may also be a group, a ring or some other algebraic structure, but allinvariants have the properties that they are the same for all isomorphic objects).
5.1 The Euler Number and Topological Genus.
As an example of invariant, we will consider the Euler Number. We will define it for every 2-dimensionalreal surface. Note that a non-singular complex curve is made of coordinate patches diffeomorphic to C (aconsequence of the Implicit Function Theorem). Since C =2, we can think of non-singular complex curvesas real two-dimensional manifolds (i.e. surface). Complex projective curves are compact real surfaces, andmoreover the complex structure gives them a natural orientation.
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Definition 5.1. Every 2-dimensional real surface can be covered by a set of triangles ∆, namely subsetswhich are homeomorphic with a given triangle in R2. Such covering by triangles is called triangulationof the surface C if every point of C is either a vertex, or it is a point on exactly one edge or in exactly onetriangle in the triangulation.
Definition 5.2. For a closed real surface C and a triangulation of C,
e(C) = V − E + F
is the Euler characteristic of C, where V is the number of vertices in the triangulation, E is the numberof edges, and F is the number of faces.
Remark 5.3. Two isomorphic surfaces have the same Euler characteristic. Indeed, an isomorphism sendsa triangulation of the first surface into a triangulation of the second surface with the same numbers ofvertices, edges and faces.
Lemma 5.4. The Euler Characteristic is independent of choice of triangulation.
Proof. To see this, we must introduce the concept of a refinement of a triangulation. Given a triangulationof a surface S, we can construct a triangulation with more, smaller triangles without changing the EulerCharacteristic. There are three such elementary refinements:
• Suppose we have a triangle T . By taking a point inside T , and joining it to the vertices of T , wereplace T with three smaller triangles. So we add 2 triangles to our triangulations, 1 vertex, and 3edges, which will not affect the Euler Characteristic.
• Between two triangles sharing an edge, we can select a point on this edge, and join it to the oppositevertex, replacing 2 triangles with 4. So we add 2 triangles, 1 vertex, and 3 edges again, again notchanging the Euler Characteristic.
• One may do the same thing but to a boundary of S instead of a shared edge, resulting in an additionaltriangle, 2 edges, and 1 vertex, again not changing the Euler Characteristic.
These three operations are called elementary refinements, and do not change the Euler Characteristic.A general refinement is the composition of several elementary refinements, and can not therefore changethe Euler Characteristic.
Now, given two triangulations of a single surface S, to see they have the same Euler Characteristic, itis enough to show that they share a common general refinement, as refinements do not change the EulerCharacteristic. To see that they do, imagine the two triangulations, T1 and T2, super imposed onto thesame copy of S. In general, this super-imposed decomposition will not be a triangulation, as there will betriangles overlapping each other, resulting in regions not homeomorphic to a triangle. However, we canmake it one by adding in edges to ”misaligned” vertices of T2 and joining them to ”neighboring” verticesof T1, via the first elementary refinement, and resolving intersecting edges, or edges and boundaries byturning these intersections into vertices via the second or third elementary refinement.
Using this, we see that this union of triangulations can be resolved to be a triangulation itself, as inthe figure above, and note that this new triangulation is a refinement of both triangulations. It followsthat both triangulations give rise to the same Euler characteristic.
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Figure 6: A refinement(blue) of a misaligned triangle(red) on a different triangulation(black).
Definition 5.5. A related number is the topological genus g(C), which can be defined by the formula
2− 2g(C) := e(C) or equivalently, g(C) =2− e(C)
2.
The genus is a measure of how many ”holes” the surface has.
Example 5.6. For CP1 ∼= RS2 the 2-dimensional real sphere, we have e(CP1) = 2 and hence g(CP1) = 0.
Indeed, we can cover the sphere with 2 triangles so that V − E + F = 3− 3 + 2 = 2.
Example 5.7. For Eλ a complex elliptic curve, we have seen earlier that Eλ has the structure of a 2-dimensional real torus. Based on the triangulation below we can check e(Eλ) = 1 − 3 + 2 = 0 and henceg(Eλ) = 1.
Figure 7: Triangulation of the sphere (left), of the torus (center), and of the torus again, and the planarrepresentation of the torus as a rectangle with the opposite sides identified.
Now a genus g surface can be obtained by gluing together a string of g tori in (g − 1) places. To gluetwo triangulated surfaces S1 and S2 together, we glue a triangle of the first surface onto a triangle of thesecond surface; we erase the faces of the two triangles and merge each pair of corresponding vertices/edgesinto one vertex/edge. Thus the resulting new surface S1
∨S2 has Euler characteristic
e(S1
∨S2) = e(S1) + e(S2)− 2.
34
In particular, gluing together a string of g tori yields a surface S with e(S) = −2g + 2 and g(S) = g.
5.2 Riemann-Hurwitz Formula.
Now, suppose we have a polynomial map between two complex plane curves (i.e. compact RiemannianSurfaces). If we understand the map, and know the genus of the first surface, we would like to be able tocompute the genus of the second surface.
Our prototype map f : C → CP1 will be the projection from a degree d complex curve C ⊂ CP2 toa line CP1 ⊂ CP2. To construct such a map, consider a point S 6∈ C. For every point p ∈ C, we defineq = f(p) to be the intersection of the line sp with the line CP1 Hence for every q ∈ CP1, the preimage
f−1(q) = {p ∈ C ; f(p) = q} = sq⋂C,
a set of d points (possibly counted with multiplicity. We say that f is a d-to-1 map. The points p wherethe intersection multiplicity msq
⋂C(p) > 1 are called branching points of the map f and the multiplicity
of intersection at these points is called ramification index:
rf (p) = msq⋂C(p).
Counting the intersection points of line sq with C we get d =∑
p∈f−1(q) rf (p).
q′
S
q
p2
p1
p′
Definition 5.8. Let f : X → Y be a d-to-1 map between complex projective curves. We say that a pointp ∈ X is a branching point for f with ramification index rf (p) = r if f(p) = q and for every sequence{qn}n ⊂ Y \{q} with limn→∞ qn = q, there exist r sequences {pn(1)}n, ..., {pn(r)}n ⊂ X \{p} with distinctterms pn(i) 6= pn(j) for i 6= j, and such that
f(pn(i)) = qn for all n and limn→∞
pn(i) = p for all i.
Example 5.9. If
35
Theorem 5.10. The Riemann-Hurwitz Formula. Let f : X → Y be a d-to-1 algebraic map betweentwo complex projective curves. Then
2g(X)− 2 = d(2g(Y )− 2) +∑p∈X
(rf (p)− 1),
where rf (p) is the ramification index of f at p.
Proof. Let R denote the set of ramification points for f . Take a triangulation of Y such that every pointin f(R) is a vertex. Let this triangulation give VY vertices, EY edges, and FY triangles. We can lift thisvia f to a triangulation in X, with VX vertices, EX edges, and FX triangles. As the map f is d − to − 1outside the vertices, we have
EX = dEY , FX = dFY .
Now consider how many vertices there are. Fix one vertex, q in Y . We have
d =∑
p∈f−1(q)
rf (p) =⇒ |f−1(q)| =∑
p∈f−1(q)
1 = d−∑
p∈f−1(q)
(rf (p)− 1)
Therefore the total number of vertices VX is
VX =∑
q vertex of Y
|f−1(q)| =∑
q vertex of Y
(d−
∑p∈f−1(q)
(rf (p)− 1))
= dVY −∑
p ramification point of f
(rf (p)− 1)
Since rf (p)− 1 = 0 when p is not a ramification point, we can replace the sum above with
VX = dVY −∑p∈X
(rf (p)− 1)
In the diagram below, R is a ramified point with ramification index rf (R) = 4.
C
C’
d-to-1
R′
R′′
R
36
Putting these together:
e(X) = VX − EX + FX = dVY −∑p∈X
(rf (p)− 1)− dEY + dFY =
= d · e(Y )−∑p∈X
(rf (p)− 1).
After substituting e = 2− 2g we get 2g(X)− 2 = d(2g(Y )− 2) +∑
p∈X(rf (p)− 1).
Let us consider a non-singular complex projective curve C ⊂ CP2, which does not contain O = [0 : 1 : 0],and the map π : C → CP1, the projection from [0 : 1 : 0] to the line CP1 = {[X : Y : Z]|Y = 0} viaπ([X : Y : Z]) = [X : Z].
Lemma 5.11. Let C be a degree d non-singular complex projective curve in CP2, defined by the homoge-neous polynomial P (X,Y, Z) and not containing [0 : 1 : 0]. Then projection map π : C → CP1 given byπ([X : Y : Z]) = [X : Z] is a d− to− 1 map with ramification points given by the equations
P (X,Y, Z) =∂P
∂Y(X,Y, Z) = 0.
Proof. If we consider the inclusion CP1 ↪→ CP2 given by [X : Z]→ [X : 0 : Z], the map π can be rewrittenas π([X : Y : Z]) = [X : 0 : Z]. Hence for A′ = [X : 0 : Z] we have
π(A) = A′ ⇐⇒ A′ ∈ OA⋂
CP1 ⇐⇒ A ∈ OA′⋂C.
The ramification points of π occur when mOA⋂C ≥ 2, namely exactly at the tangency points of OQ with
C. For a point P ∈ C, the line OP is tangent to C at P iff
O = [0 : 1 : 0] satisfies X∂P
∂X(A) + Y
∂P
∂Y(A) + Z
∂P
∂Z(A) = 0 ⇐⇒ ∂P
∂Y(A) = 0.
In addition, A is a point on C iff P (A) = 0. Hence the ramification points are given by P (A) = ∂P∂Y (A) = 0.
This is the intersection of a degree d with a degree (d− 1) curve, hence this leads to d(d− 1) solutions(when counted with multiplicity). (Note that P (0, 1, 0) 6= 0 implies that the coefficient of Y d in P (X,Y, Z)is non-zero, hence ∂P
∂Y is a non-trivial polynomial of degree (d−1). Also note that P (X,Y, Z) is irreducible,hence the two equations have no common factor.)
Note: It can be shown that mOA⋂C(A) = 2 implies mP,PY
(A) = 1 – see for example Kirwan’s book.
Theorem 5.12. Let C be a smooth curve of degree d in the complex projective plane CP2. Then
e(C) = d(d− 3) and g(C) =(d− 1)(d− 2)
2=
(d− 1
2
).
37
Proof. Given a curve C as above, we have seen that there are only a finite number of inflection points andhence finitely many tangent lines through these inflection points. Take a point O which is not on any suchtangent line and let π denote the projection of C on a line l centered at O.
The choice of O insures that π has exactly d(d − 1) ramification points, with ramification indicesrπ(A) = mOA
⋂C(A) = 2, since otherwise A would be an inflection point with tangent OA.
By the Riemann-Hurwitz formula from above,
=⇒ e(C) = d e(P1)− d(d− 1)
=⇒ 2− 2g = 2d− d(d− 1)
=⇒ g =d2 − 3d+ 2
2=
(d− 1)(d− 2)
2=
(d− 1
2
)
6 Affine Varieties in Kn
Here K denotes a field - for example Q,R,C, or finite fields Kpn for p= prime. when we draw diagrams wewill take K = R, when we prove most theorems we’ll take K = C or other algebraically closed field, whileK = Q or Kpn are used to get Number Theory results from Algebraic Geometry.
Multivariable polynomial: f(x1, x2, ..., xn) ∈ K[x1, x2, ..., xn] is a sum of products written as
f(x1, x2, ..., xn) =∑
j1,j2,...,jn
fj1,j2,...,jnxj11 x
j22 · · ·x
jnn =
∑J=(j1,j2,...,jn)
fJxJ
where fJ = fj1,j2,...,jn ∈ K and xJ is just short for xj11 xj22 · · ·x
jnn .
Algebraic variety: V(f1, f2, ..., fr) ⊆ Kn is defined as the solution set of the polynomial equationsfi(x1, x2, ..., xn) = 0 in Kn for i = 1, ..., r.
Note. Conventions differ according to authors. Some authors use the terminology of algebraicsets for solution sets of algebraic equations and reserve the name of varieties for algebraic sets whichare irreducible, namely which cannot be written as finite union of other algebraic sets. For us, algebraicvarieties are solution sets of algebraic equations, and irreducible varieties will be called domains.
Examples:
• circle, hyperbola, other curves in the plane;
• intersection of two curves in the plane.
• paraboloid and cone in 3D
• two circles in 3D which can be presented either as the intersection of a sphere with two planes, or asphere with a cylinder, or a sphere with a cone.
• SLn, On, in Rn2, Un in R(2n)2 .
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• Uf = {(x1, ..., xn) ∈ Kn; f(x1, ..., xn) 6= 0} ∼= V(f(x1, ..., xn)xn+1 − 1) ⊂ Kn+1. In particular, GLnis an affine variety in Kn2+1.
Note: V(x2 + y2 + z2− 25, xy + y2− 16) = V(z2− 9, x2 + y2− 16) Hence the need for an ideal, not justsome finite set of equations:
The ideal of an affine variety V ⊆ Kn is
I(V ) = {f ∈ K[x1, x2, ..., xn] ; f(x1, x2, ..., xn) = 0 for all (x1, x2, ..., xn) ∈ V }.
Exercise. Prove I(V ) is an ideal in K[x1, x2, ..., xn]. (For simplicity denote p = (x1, x2, ..., xn) ∈ V ).
The affine variety of an ideal I ⊆ K[x1, x2, ..., xn] is
V(I) = {p ∈ Kn ;f(p) = 0 for all f ∈ I}.
Theorem 6.1. Hilbert’s Basis Theorem: Every ideal I of K[x1, ..., xn] has a finite set of generators:I = 〈f1, ..., fr〉. Hence every solution set of an ideal is actually determined by a finite number of equations.
Proof. We will prove this by induction on n. When n is 1, we know that K[X] is a Euclidean domain hencea PID, hence every ideal is generated by just one element.
Assume that we have proven the theorem for n = m and consider an ideal I ⊂ K[x1, ..., xm, y]. Forevery polynomial f in K[x1, ..., xm, y], we will denote by lc(f) the coefficient of the highest power of y in f .Note that lc(f) ∈ K[x1, ..., xm]. Assuming that I is not finitely generated, we can find an infinite sequence
〈f0〉 ⊂ 〈f0, f1〉 ⊂ 〈f0, f1, f2〉 ⊂ · · · ⊂ I
such that at each step, fk+1 is a polynomial of minimum degree in I \ 〈f0, ...fk〉. Then the ideal
〈lc(f0), lc(f1), ...〉 =∞⋃k=0
〈lc(f0), ..., lc(fk)〉
in K[x1, ..., xm] must be finitely generated. It follows that
〈lc(f0), lc(f1), ...〉 = 〈lc(f0), ..., lc(fN )〉
for some N . This means that lc(fN+1) =∑
i<N ci · lc(fi) where ci ∈ K[x1, ..., xm]. Then the element
g =∑i<N
ciydeg(fN+1)−deg(fi)fi ∈ 〈lc(f0), ..., lc(fN )〉
satisfies lc(g) = lc(fN+1). Hence f−g ∈ I\〈lc(f0), ..., lc(fN )〉 and deg(fN+1−g) < deg(fN+1) contradictingthe construction of fN+1. Hence our assumption was impossible and I must be finitely generated.
Note: We say that the ring K[x1, x2, ..., xn] is Noetherian.
Note: The ideal of a variety gives conditions (equations) that the points of the variety must satisfy.The more conditions, the fewer points satisfy them. The fewer conditions, the more points satisfy them.
Questions:
39
• How big is the affine variety defined by a given ideal (When is it ∅?; when is it a finite set of points?how can we define the dimension of a variety?)
• How do the properties of an ideal (e.g. prime, maximal, principal) determine those of the corre-sponding affine variety?
7 The Ideal - Affine Variety Correspondence
Each row in the table below represents a property of the affine variety corresponding to a specific ideal orcombination of ideal. The proof of each property is an exercise.
Assumption on the Ideal I in K[x1, x2, ..., xn] Conclusion on the Affine Variety V(I) in Kn
large small
small large
I1 ⊆ I2 V(I1) ⊇ V(I2)
I = {0} V({0}) = Kn
I = K[x1, x2, ..., xn] V(K[x1, x2, ..., xn]) = ∅
I = 〈x1 − a1, ..., xn − an〉 V(x1 − a1, ..., xn − an) = {(a1, ..., an)}maximal ideal A = (a1, ..., an) = one point.
I = 〈f〉 V(I) = a hypersurface (by definition)(principal ideal generated by f) (a variety of dimension (n− 1) in Kn)
I = I1 + I2 V(I1 + I2) = V(I1)⋂V(I2)
I = I1⋂I2 or I = I1 · I2 V(I1 · I2) = V(I1)
⋃V(I2) = V(I1
⋂I2)
Solution for V(I1⋂I2) = V(I1 · I2) = V(I1)
⋃V (I2): we have I1 · I2 ⊆ I1
⋂I2 ⊆ Ij hence V(I1 · I2) ⊇
V(I1⋂I2) ⊇ V(I1)
⋃V(I2). On the other hand, for p ∈ V(I1 · I2), either p ∈ V(I1) or there is f ∈ I1 such
that f(p) 6= 0. But then for every g ∈ I2 we have fg ∈ I1 · I2 and so fg(p) = 0 hence g(p) = 0. Thusp ∈ V(I2).
Example V(xz, yz) = V(〈z〉〈x, y〉) = V(x, y)⋃V(z) the union between a line and a plane.
Corollary The intersections and finite unions of affine varieties are also affine varieties.
These are exactly the properties satisfied by closed sets in a topology. (Equivalently, unions and finiteintersections of open sets should be open). Hence, Kn can be endowed with a topology structure such thatthe affine varieties correspond to closed sets, and their complements to open sets. This is called the Zariskitopology.
40
8 The Affine Variety - Radical Ideal Correspondences
Recall that for every affine variety V ⊆ Kn there is an ideal of polynomials I(V ) in K[x1, x2, ..., xn], hencewe might ask if the table above can be read backwards: do the given properties of the varieties implycorresponding propertied of the ideals?
Indeed, if V1 ⊂ V2 then I(V1) ⊃ I(V2) (large varieties have small ideals). However not all the propertiesin the table are preserved as well. This correspondence is not perfect:
{ Algebraic Varieties in Kn} −→ { Ideals in K[x1, x2, ..., xn]}V −→ I(V )
V(I)←− I
As it stands, the map I → V(I) is not injective.
Example 8.1. Lack of real solutions when K = R: The ideals I1 = 〈1〉 = R[x, y] and I2 = 〈x2 +y2 +1〉satisfy V(I1) = V(I2) = ∅ as the equation x2 + y2 + 1 = 0 has no solution in R2.
This problem can be easily fixed by requiring that the field K is algebraically closed. By definition,this guarantees that every polynomial equation in K[x1, ..., xn] has solutions in Kn for every n > 0.
From now on we will consider the case when K is algebraically closed. The main exampleis K = C. With this in mind, we can now start to reverse the columns in the Ideal-Variety table. Forexample, one of the first correspondence we’d like to establish is:
Theorem 8.2. (Weak Nullstellensatz). If I ⊆ K[x1, ..., xn] is an ideal and its corresponding affinevariety V(I) = ∅, then I = K[x1, ..., xn].
The German Nullstellensatz translates as theorem about the places of the zeroes of functions (Null=Zeroes,Stellen=Places, Satz=Theorem).
Even this first property is not easy to prove. To prove it, we will need to get a better handle on how tomanipulate the equations of a variety and perform operations like slicing and projecting. We will dedicatethe next 2 sections to its proof. For the time being, for the rest of this section we will assume the WeakNullstellensatz true and use it to get a clearer picture on the Variety-Ideal correspondence.
However, even with the Weak Nullstellensatz in place, the ideal-variety correspondence is not perfect.For example, if f is any polynomial, then the ideals 〈f〉 and 〈f2〉 have the same solution sets: V(f) = V(f2).In other words:
Proposition 8.3. Ideal-Variety Correspondence. For any variety V ∈ Kn and any ideal I ⊂K[x1, ..., xn] we havea) V(I(V )) = V .b) I ⊆ I(V(I)). However there exist ideals I for which I 6= I(V(I)).
Part (a) can be reformulated as p ∈ V ⇐⇒ f(p) = 0 for all f ∈ I(V ).
Part (b) states that the polynomials in I are zero at the points in V(I), but there might be some otherpolynomials with the same property which are not in I. Here are some examples:
41
• Take I = 〈x3〉 in K[x, y]. Then V(I) is the Oy-axis, given by the equation x = 0. Hence I(V(I)) = 〈x〉.
• Take I = 〈x2 +y2−4, x−2〉 in R[x, y]. Then V(I) = {(0, 2)} in R2 and I(V(I)) = 〈x−2, y〉. However,y 6∈ I, even though y2 = x2 + y2 − 4− (x+ 2)(x− 2) ∈ I.
In both cases we’ve seen that the difference between I and I(V(I)) comes from some element f ∈ I(V(I))such that f 6∈ I but fm ∈ I.
Let R be any commutative ring and let I be an ideal of R.
The radical of an ideal I ⊆ R is√I = {f ∈ R ; fm ∈ I for some positive m ∈ Z}.
A radical ideal is an ideal I ⊆ R such that√I = I.
Example 8.4.√〈x2 + y2 − 4, x− 2〉 = 〈x− 2, y〉.
Proposition 8.5. For any ideal I, the set√I is a radical ideal.
Proof. We first have to prove√I is an ideal. Assume f, g ∈
√I. Then fa, gb ∈ I for some positive powers
a, b. Hence
• (rf)a = rafa ∈ I hence rf ∈√I for all r ∈ R.
• Then (f + g)a+b =∑a+b
k=0
(a+bk
)fkga+b−k ∈ I since either k ≥ a or a+ b− k ≥ b. Hence f + g ∈
√I.
To prove that√I is radical, let fm ∈
√I. Then fmn ∈ I for some n hence f ∈
√I.
The problem found in the previous proposition can be fixed by the strong Hilberts Nullstellensatz:
Theorem 8.6. Hilbert’s Nullstellensatz. Let K be an algebraically closed field and I ⊆ K[x1, ..., xm].
Then I(V(I)) =√I, the radical ideal of I.
Proof. Part I.√I ⊆ I(V(I)) Let f ∈
√I. Then fm ∈ I hence fm(x) = 0 for all x ∈ V(I). But then f(x) = 0
for all x ∈ V(I) hence f ∈ I(V(I)).
Part II. I(V(I)) ⊆√I. Assume I = 〈f1, ..., fr〉 and let f ∈ I(V(I)). Hence f(x) = 0 whenever
f1(x) = ... = fr(x) = 0. We wish to prove that fm ∈ I for some positive integer m.
This can be reduced to the Weak Nullstellensatz by an ingenious trick: We embedd our affine spaceKm in Km+1 by adding an extra variable y, and consider the ideal J = 〈f1, ..., fr, yf −1〉 in K[x1, ..., xm, y].Then V(J) = ∅ in Km since yf(x)− 1 = 0− 1 = −1 whenever f1(x) = ... = fr(x) = 0. Hence by the WeakNullstellensatz we have 1 ∈ J. We can thus write
1 = g1(x, y)f1(x) + ...+ gr(x, y)fr(x) + g(x, y)(yf(x)− 1) in K[x, y]. (8.1)
This is true for any (x, y) ∈ Km+1. To eliminate y from this equation, we work with (x, y) ∈ V(yf(x)− 1).Note that any x ∈ Km\V(f) admits a y = 1/f(x) such that (x, y) ∈ V(yf(x)−1). Hence we can substitutey = 1/f(x) in the equation (8.2) to get
1 = g1(x,1
f(x))f1(x) + ...+ gr(x,
1
f(x))fr(x) in K(x). (8.2)
42
Since all expressions in the equation above are polynomials, we can get a common denominator f(x)m
and then multiply both sides by f(x)m to get f(x)m as a linear combination of fi(x) with polynomialcoefficients in K[x]. This proves fm ∈ 〈f1, ..., fm〉 ∈ I so f ∈
√I. We have thus proven I(V(I)) ⊆
√I.
Corollary 8.7. There is an (order reversing) one-to-one correspondence:
{ Algebraic Varieties in Kn} −→ { Radical Ideals in K[x1, x2, ..., xn]}V −→ I(V )
V(I)←− I
Proof. By Hilbert’s Nullstellensatz, I(V(I)) =√I for every ideal I in K[x1, x2, ..., xn]}. If I is a radical
ideal, then√I = I so in that case I(V(I)) = I. Together with Proposition on Ideal-Variety Correspondence,
this implies that V and I as defined in this Corollary are inverses of each other.
We can now reverse the columns in the Ideal-Variety table constructed earlier. We this get a Variety-Radical Ideal table and we can now prove each row in the table below, using Hilberts Nullstellensatz whenneeded.
Assumptions on Affine Variety V in Kn Conclusions on Ideal I(V ) in R = K[x1, x2, ..., xn]
large/small small/large
V1 ⊂ V2 I(V1) ⊃ I(V2)
V = Kn I(Kn) = {0}
V = ∅ I(∅) = R = K[x1, x2, ..., xn]
V = { one point } I( one point) = maximal ideal
V = a hypersurface I(V ) = 〈f〉(a variety of dimension (n− 1) in Kn) (principal ideal generated by 1 polynomial f)
V = V1⋂V2 I(V1
⋂V2) =
√I(V1) + I(V2)
V = V1⋃V2 I(V1
⋃V2) = I(V1)
⋂I(V2) =
√I(V1) · I(V2)
V = irreducible I(V ) = prime
Note: In general I1 · I2 ⊆ I1⋂I2 but not necessarily the other way around but
√I1 · I2 =
√I1⋂√
I2.
Exercise: Prove that Hilbert Nullstellenzats holds true in K1.
Assuming the Nullstellensatz known, we can prove the following
Proposition 8.8. Maximal ideals. Every maximal ideal M ⊂ K[x1, x2, ..., xn] is of the form M =〈x1 − a1, x2 − a2, ..., xn − an〉 for some fixed (a1, ..., an) ∈ Kn.
43
Proof. For every maximal ideal M ⊂ K[x1, x2, ..., xn], its variety V(M) must consist of a single pointa. Indeed, given any a ∈ V(M) we have M ⊆ I({a}) and I({a}) 6= K[x1, x2, ..., xn] as {a} 6= ∅ henceM = I(a) = 〈x1 − a1, x2 − a2, ..., xn − an〉.
Corollary 8.9. If a = (a1, ..., an) is a point of intersection of two varieties V and W in Kn, thenI(V )
⋂I(W ) ⊂ 〈(x1 − a1)m1 , (x2 − a2)m2 , · · · , (xn − an)mn〉 for some positive integers mi.
Basic operations with varieties and ideals
We can reduce the study of high dimensional spaces to lower dimensional spaces by two operations: sec-tions (slicing) and projections (squashing). However, sometimes we may also need to carefully choose thedirections in which we choose to slice or project. This may involve some linear changes of variables.
Cutting Sections.
Consider an affine variety V ⊆ Kn and consider the linear subspace L ≡ Km ⊂ Kn given by xm+1 =cm+1, ..., xn = cn for some fixed constants ci.
Lemma 8.10. V⋂L is an affine variety in Km and its ideal in K[x1, ..., xm] is
{f(x1, ..., xm, cm+1, ..., cn) ; f ∈ I(V )}.
Cutting the section V⋂L through V corresponds to fixing the variables xm+1, ..., xn.
Projections
Let π : Kn → Km be the projection which forgets the last n −m coordinates π(x1, ..., xn) = (x1, ..., xm).Then π(V ) might not be an affine variety.
Example 8.11. Let V = V(xy − 1) in R2 and consider the projection π : R2 → R given by π(x, y) = x.
Then π(V ) = R∗ = R \ {0} is not an affine variety in R.
Notation . For every set W ⊂ Kn, we define by W the smallest variety in Kn which contains W .
This notation is consistent with the definition of closed sets in the Zariski topology.
Since π(V ) might not be an affine variety as in the previous example, we cannot talk about the idealof π(V ). So we will look at the next best thing, that is the ideal of π(V ).
Proposition 8.12. The ideal of π(V ) is I(π(V )) = I(V )⋂K[x1, ..., xm].
Projecting on L roughly corresponds to eliminating the variables xm+1, ..., xn.
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Proof. By definition I(π(V )) = {g ∈ K[x1, ..., xm]; g ≡ 0 on π(V )}.
Let f ∈ K[x1, ..., xm]. We can also think of f as a polynomial in K[x1, ..., xm, y] of degree 0 in y. Hencefor any q = (a1, ..., am, b) ∈ Km+1 and p = (a1, ..., am) = π(q) ∈ Km, we have f(p) = 0 as a polynomialin K[x1, ..., xm] if and only if f(q) = 0 as a polynomial in K[x1, ..., xm, y]. Thus f ∈ I(π(V )) iff f ∈I(V )
⋂K[x1, ..., xm].
Example 8.13. Consider the two cases when n = 3 and m = 2 and for V = 〈z − x2 − y2〉 for V =V (〈y − x2, z − x3〉).
Decomposing a variety into irreducible pieces
We might also need to reduce the study of finite unions of affine varieties to the study of their parts.We’ve seen that V(I1 · I2) = V(I1)
⋃V(I2). Working through this backwards for V = V(I1)
⋃V(I2) and
W = V(I1), then V \W = V(I2) which might help make sense of the following:
Exercise. Given two affine varieties V and W in Kn, prove that
I(V \W ) = I(V ) : I(W )
where by definition I(V ) : I(W ) = {f ∈ R ; f · I(W ) ⊆ I(V )}.
9 Resultants and Elimination of Variables
Recall the Proposition
Proposition 9.1. Let π : Kn → Km be the projection which forgets the last n−m coordinates
π(x1, ..., xn) = (x1, ..., xm).
The ideal of π(V ) is I(π(V )) = I(V )⋂K[x1, ..., xm].
Note: Let Im = I(V )⋂K[x1, ..., xm]. The polynomials in Im are still elements of I(V ) but do not contain
the variables xm+1, ..., xn hence the process of finding Im is called elimination of variables. This is a verycommon approach to solving simultaneous equations.
Building on the examples from curves, consider the case when I = 〈f, g〉 ⊂ K[x1, ..., xm, y] and V =V(f, g) ⊆ Km+1. We consider the projection π : Km+1 → Km given by π(x1, ..., xm, y) = (x1, ..., xm).
Our goal is to eliminate y and thus find an equation for π(V ). To simplify notations, we will denotex = (x1, ..., xm) ∈ Km so that the projection map π : Km+1 → Km is simply given by π(x, y) = x.
Just like before, the resultant can be defined as the unique combination
R(x) = u(x, y)f(x, y) + v(x, y)g(x, y) ∈ I⋂
K[x]
where degy u(x, y) < degy g(x, y) and degy v(x, y) < degy f(x, y)
We can reduce this problem to the 1 variable case in two different ways:
45
(a) For each a ∈ Km, evaluate the polynomials at a and then compute the resultant:Let fa(y) = f(a, y) and ga(y) = g(a, y) ∈ K[y], then calculate R(fa, ga) as in Definition 4.4.
(b) Write f(x, y) = fx(y) =∑d
i=0 fiyi and g(x, y) = gx(y) =
∑ei=0 giy
i, with coefficients fi, gi ∈ K[x].Then calculate R(fx, gx) as in Definition 4.4, by working with coefficients in the field of fractionsK(x), which is the smallest field containing K[x].
Note: Even though we work with coefficients in K(x), Proposition 4.5 implies R(fx, gx) ∈ K[x].
Corollary 9.2. Consider the projection π : Km+1 → Km which forgets y and let f, g ∈ K[x, y].
• In general, R(fx, gx) ∈ 〈f, g〉⋂K[x]. Hence R(fx, gx) takes zero values on the projection π(V (f, g)).
• The resultant R(fx, gx) is constant 0 as a polynomial in x iff f and g have a common factor in K[x, y]which is not constant in y.
Proof. By definition R(fx, gx) ∈ K[x] since it is an integer polynomial in fi, gi-s all of which are in K[x]by Proposition 4.5. The same proposition also implies that R(fx, gx) = s(y)fx(y) + t(y)gx(y) where thecoefficients si, ti ∈ K[x]. Hence s(y), t(y) ∈ K[x, y] and hence R(fx, gx) ∈ 〈f, g〉.
Assume now that the resultant R(fx, gx) is the constant polynomial 0. By working with coefficientsin the field F = K(x), Corollary 4.6 implies that f and g have a common factor in K(x)[y] which isnon-constant in y. However, since f, g ∈ K[x][y], their factors are also in K[x][y] (this is part of Gauss’sTheorem which states that K[x][y] is a UFD). Hence f, g have a common factor in K[x][y] = K[x, y] whichis not constant in y.
Note: In general, we might expect that evaluating R(fx, gx) at x = a would give R(fa, ga), howeverthis is not always the case. The reason is that fx and fa might have different degrees as polynomials iny, hence the corresponding Sylvester matrices would have different sizes. This is the case iff fd(a) = 0. Asimilar problem arises when ge(a) = 0.
Corollary 9.3. With the notations from above, assume that fd(a)ge(a) 6= 0 for some a ∈ Km. Then
R(fx, gx)|x=a = R(fa, ga).
Proof. In this case fx and fa have the same degrees as polynomials in y, and similarly for gx and ga. Hencethe Sylvester matrix Syl(fa, ga) is obtained by evaluating Syl(fx, gx) at x = a, and the same is true ofresultants.
Proposition 9.4. For any two polynomials f, g ∈ K[x, y], we preserve the notations from above. Then
V(R(fx, gx)) = π(V(f, g))⋃
V(fd, ge),
Here fd, ge are the coefficients of the highest powers of y in fx, gx.
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Proof. Part I: ”⊇”. Let a ∈ Km. Note that in the case fd(a) = ge(a) = 0, then the last row of the Sylvestermatrix Syl(fx, gx) becomes 0 when evaluated ata, and hence taking determinant R(fx, gx)(a) = 0. HenceV(fd, ge) ⊆ V(R(fx, gx)). Also, π(V(f, g)) ⊆ V(R(fx, gx)) by 9.2. Hence
V(R(fx, gx)) ⊇ π(V(f, g))⋃
V(fd, ge).
Part II: ”⊆”. To prove the inclusion in the opposite sense, consider a ∈ V(R(fx, gx)) and assumea 6∈ V(fd, ge). We will prove that a ∈ π(V(f, g)).
Since a 6∈ V(fd, ge), it follows that fd(a) 6= 0 or ge(a) 6= 0.
Case 1: If both fd(a) 6= 0 and ge(a) 6= 0, then by the previous corollary we have
0 = R(fx, gx)(a) = R(fa, ga).
Then by Corollary 4.6 it follows that fa, ga have a common zero y = b and hence f(a, b) = g(a, b). Hencea = π(a, b) ∈ π(V(f, g)).
Case 2: On the other hand if ge(a) = 0 while fd(a) 6= 0, then we can replace g by g = g + yef , whosehighest power of y has coefficient gB+A = fd. Then Ex Set 2 shows that
R(fx, gx) = (−1)dfdR(fx, gx − yefx) = (−1)dfdR(fx, gx).
Hence iff R(fx, gx)(a) = 0 iff R(fx, gx)(a) = 0. Since gB+A(a) = fd(a) 6= 0, then case 1 implies a ∈π(V(f, g)). Moreover V(f, g) = V(f, g) hence a ∈ π(V(f, g)).
In fact we only need a milder version which we have already proven in the case of 2 variables:
Corollary 9.5. Consider the projection π : Km+1 → Kn given by π(x, y) = x. Let f, g ∈ K[x, y] be twopolynomials such that f(x, y) =
∑di=0 fiy
i and g(x, y) =∑e
i=0 giyi, with coefficients fi, gi ∈ K[x], and
such that the leading coefficients fd and ge are non-zero constants in K. Then the image underprojection π(V(f, g)) is an algebraic variety in Km and is given by
π(V(f, g)) = V(Rf,g),
where the resultant is computed wrt y.
Proof. Indeed in this case V(fd, ge) = ∅ hence the above equality holds. Since the RHS is an algebraicvariety, then so is the LHS.
This can be extended to any number of polynomials as follows:
Proposition 9.6. The Ideal of a Closed Projection: Consider the projection π : Km+1 → Kn given
by π(x, y) = x. Let I = 〈f1, ..., fr〉 where fj ∈ K[x, y] are polynomials such that fj(x, y) =∑dj
i=0 fj,iyi
with coefficients fj,i ∈ K[x], and such that the leading coefficients of all fj for the variable y arenon-zero constants in K. Then π(V(I)) is closed and π(V(I)) = V(I
⋂K[x]).
Proof. Wlog we can let the generators of the ideal I be written in increasing order of their degree in y:
d1 < d2 < ... < dr.
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Indeed, if two generators fi and fj have equal degree d, we can always replace fi by fi = fi + ydfj sothat deg(fi) > d while 〈fi, fj〉 = 〈fi, fj〉. Note that the leading coefficients of y continue to be non-zeroconstants.
To reduce our problem to the case of 2 generators, let
f = f1,
g = u2f2 + u3f3 + ...+ urfr
Note that Rf,g ∈ K[x]⋂〈f, g〉 by Corollary 9.2 and 〈f, g〉 ⊂ I hence Rf,g ∈ I
⋂K[x] for all u2, ..., ur ∈ K.
Moreover, for ur 6= 0 we have Rf,g(a) = R(fa, ga) for all a ∈ Km.
We have already shown π(V(I)) ⊆ V(I⋂K[x]). We have to show V(I
⋂K[x]) ⊆ π(V(I)).
Let a ∈ V(I⋂
K[x]) be such that 0 = Rf,g(a) = R(fa, ga) for all u2, ..., ur ∈ K not all 0.
We have reduced the number of generators to 2, but have introduced extra variables ui. Hence fromnow on we will work over F = K(u2, ..., ur) which contains K[u2, ..., ur].
Then by Corollary 4.6, the polynomials fa(y), ga(y) have a common factor d(y) ∈ K(u2, ..., ur)[y] whichis non-constant as a polynomial in y. However, ga(y) ∈ K[u2, ..., ur, y] which is a UFD and hence its factord(y) ∈ K[u2, ..., ur, y]. On the other hand d(y)|fa(y) and fa(y) ∈ K[y] (it involves no terms ui) hence itsfactor d(y) ∈ K[y]. This is non-constant as a polynomial in y and K is an algebraically closed group,so d(y) has a root y = b. Hence fa(b) = f(a, b) = 0 and ga(b) = g(a, b) = 0 .The last equality implies
u2f2(a, b) + u3f3(a, b) + ...+ urfr(a, b) = 0
for all u2, u3, ..., ur ∈ K not all 0. Hence
f2(a, b) = f3(a, b) = fr(a, b) = 0
and hence (a, b) ∈ V(I). We have proven a ∈ π(V(I)).
The previous proposition gives us a neat description for the projection of a variety in Km+1 on ahyperplane Km, however it depends on a strong condition on the equations of the variety. Next we will seethat we can always choose our hyperplane Km in such a way that the required conditions are satisfied.
To achieve that, instead of organising our polynomials by the degree in the variable y only, we will usethe total degree:
deg(yn∏i
xnii ) = n+
∑i
ni.
Then for every polynomial f(x1, ..., xm, y) ∈ K[x1, ..., xm, y], we will denote by lpf(x1, ..., xm, y) the leadingpolynomial of f , that is the sum of monomials with highest total degree in f .
Example 9.7. For f(x, y) = x3 − 2x2y + y2 + 5xy + 2 we havedeg(f) = 3 and lpf(x, y) = x3 − 2x2y, the degree 3 part of f .
Proposition 9.8. Closed Projection: Let K be an algebrically closed field. For every algebraic varietyV ⊂ Km+1, we can find a hyperplane Km ⊂ Km+1 with a projection map π : Km+1 → Km such that π(V)is an affine variety in Km.
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The name Closed projection comes from the Zariski topology, whose closed sets correspond exactly toaffine varieties.
Proof. Let V = V(I) where I = 〈f1, ..., fr〉 ⊂ K[x1, ..., xm, y]. Consider the linear change of variables:
y = y,
xi = xi − aiy
for some suitable constant a ∈ Km. Then xi = xi+aiy and every polynomial f(x1, ..., xm, y) ∈ K[x1, ..., xm, y]is rewritten in the new variables as
f(x1, ..., xm, y) = f(x1, ..., xm, y) = f(x1 + a1y, ..., xm + amy, y)
= lpf(a1y, ..., amy, y) + terms with lower powers of y =
= lpf(a1, ..., am, 1)ydeg(f) + terms with lower powers of y.
It remains to make sure that our polynomials fj in the new variables satisfy the conditions of Propositionon Ideal of Closed Projections. For this it is enough to choose a such that lpfj(a, 1) 6= 0 for all j = 1, ..., r.It is always possible to find such an a ∈ Km, since the polynomials lpfj(x, 1) are not identically 0 byconstruction.
Example 9.9. Consider V = V(xy − 1) ⊂ C2. Then taking x = s+ t and y = s− t changes the equationxy = 1 into s2 − t2 = 1. The projection π : C2 → C given by π(s, t) = t yields π(V ) = C because for everyt ∈ C, the equation s2 = t2 + 1 has some solutions s ∈ C. This is still true if we replace C by R.
Theorem 9.10. Weak Nullstellensatz. Let K be an algebraically closed field and I ⊆ K[x1, ..., xn].
If V(I) = ∅, then I = K[x1, ..., xn].
Proof. Recall that I = K[x1, ..., xn] is equivalent to I = 〈1〉 or 1 ∈ I. Intuitively, we say that V(I) = ∅ isbecause V(I) must satisfy the impossible equation 1 = 0.
We will prove this by induction on n. When n = 1, the ring K[x] is an Euclidean domain hence a PID.Thus every ideal is of the form I = 〈f〉 for some polynomial f ∈ K[x]. But every non-constant polynomialhas solutions in K. Hence V(f) = ∅ iff f = non-zero constant f = c. Since all non-zero constants areinvertible in K, we have 1 = cc−1 ∈ I so I = K[x].
Let us now assume that the statement is true in the case n = m. To prove the case n = m+ 1, let I bean ideal in K[x1, ..., xm, y] such that V(I) = ∅. By the Propositions on Closed Projection we can always findπ : Km+1 → Kn such that π(V(I)) = V(I
⋂[x]). Then by the Proposition on Ideals of Closed Projection we
have V(I⋂
[x1, ..., xm]) = π(∅) = ∅. Hence by the induction hypothesis, 1 ∈ I⋂
[x1, ..., xm]) hence 1 ∈ I.
10 Coordinate Rings and Maps between Varieties
We have managed to describe each affine variety algebraically by its ideal, however this description dependson the choice of an ambient space Kn. The same geometric object can be placed in different ambient spacesin different ways:
49
Definition 10.1. Let V1 ⊆ Km and V2 ⊆ Kn be affine varieties. A regular map from V1 to V2 is a map
ϕ : V1 → V2 ⊆ Kn
x → ϕ(x) = (ϕ1(x), ..., ϕn(x))
for every x ∈ V1 ⊆ Km, where ϕi(x) ∈ K[x] = K[x1, ..., xm are polynomials in m variables.If ϕ is bijective (injective and ϕ(V1) = V2), then we say that ϕ is an isomorphism. We say that V1 andV2 are isomorphic – we may think of them as the same abstract variety V living in two different ambientspaces Km and Kn.
Example 10.2. Affine Line. For example the line K1 can exist
a) as an abstract line in itself,
b) as a line inside a plane V1 = V(3y − 2x+ 5) ⊂ K2,
c) as a twisted cubic V2 = V(y − x2, z − x3) ⊂ K3, etc.
To compare the case (b) with (a), it is enough to construct an injective morphism ϕ : K1 → K2 suchthat ϕ(K1) = Vi. In case (c), we want ϕ : K1 → K3.
Case (b) Let ϕ : K1 → K2 given by ϕ(t) = (t, 23 t−
53). Indeed, this is obtained algebraically be setting
x = t in the equation 3y− 2x+ 5 = 0 and solving for y. In other words, we have set a ring homomorphism
ϕ∗ : K[x, y] → K[t]
x → ϕ∗(x) = t
y → ϕ∗(y) =2
3t− 5
3
p(x, y) → ϕ∗(p(x, y)) = p(t,2
3t− 5
3)
and we can check that ker(ϕ∗) = 〈3y − 2x + 5〉, hence by the Fundamental Theorem of Isomorphism we
have K[x,y]〈3y−2x+5〉
∼= K[t].
Similarly in case (c) we may let x = t and then from the equations of V2 we get y = t2 and z = t3
giving a map ϕ : K1 → K3 by ϕ(t) = (t, t2, t3). This corresponds to the homomorphism of rings
ϕ∗ : K[x, y, z] → K[t]
p(x, y, z) → = p(t, t2, t3).
We can check that ker(ϕ∗) = 〈y−x2, z−x3〉, hence by the Fundamental Theorem of Isomorphism we haveK[x,y]
〈y−x2,z−x3〉∼= K[t].
Note: Parametric descriptions like above are very useful in graphing affine varieties. Unfortunatelly, notevery affine variety has such a nice parametric description. Indeed, a parametric description is equivalentlyto describing a variety as isomorphic to an affine space Km, given by parameters t1, ..., tm.
Still, it is always true that the same variety V can lie in two different ambient spaces Kn and Kn′ .Such a variety would have different ideals I(V ) ⊂ K[x1, ..., xn] and I′(V ) ⊂ K[x1, ..., xn′ ]. The fact that itis the same variety V in both cases is reflected by the isomorphism of quotient rings: K[x1, ..., xn]/I(V ) ∼=K[x1, ..., xn′ ]/I′(V ).
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Definition 10.3. Given an affine variety V ⊆ Kn, the quotient K[x1, ..., xn]/I(V ) is called the coordinatering of V and denoted as K[V ].
In particular K[x1, ..., xn] is called the coordinate ring of Kn. Its elements can be thought of as poly-nomial functions f : Kn → K. It contains the coordinate functions xi and all polynomial expressionsgenerated by them.
We may think of K[V ] = {n− variable polynomial functions g : V → K}. Then the restriction function
j∗ : K[x1, ..., xn] → K[V ]
f → f|V
has kernel ker(j∗) = I(V ) hence the isomorphism K[x1, ..., xn]/I(V ) ∼= K[V ].
Remark 10.4. Just like affine varieties in Kn are in one-to-one correspondence with radical ideals inK[x1, x2, ..., xn], for every affine variety V we have a one-to-one correspondence
{ Subvarieties of V } ←→ { Radical Ideals in K[V ]}
Note that every ideal J of K[V ] = K[x1,...,xn]I(V ) is itself a quotient J = J ′
I(V ) where J ′ is an ideal of K[x1, ..., xn]
containing I(V ). In other words, for W = V(J ′), we have I(V ) ⊂ J ′ = I(W ) which translates into W ⊂ V ,and the natural isomorphism
K[V ]
J=
K[x1, ..., xn]/I(V )
I(W )/I(V )∼=
K[x1, ..., xn]
I(W )= K[W ]
corresponds to presenting W either as a subvariety of V ⊂ Kn or directly as a subvariety of Kn.
Algebraic properties of the coordinate ring K[V ] reflect properties of the variety V :
Properties of the affine variety V Properties of the coordinate ring K[Y ]
V is irreducible K[V ] is an Integral Domain.
V = {p} is a point K[V ] ∼= K is a field.
V is an irreducible curve K[V ] is a PID.
V =⊔i Vi is disjoint union of varieties K[V ] ∼= ×iK[Vi]
Indeed, V is irreducible iff I(V ) is prime iff K[V ] = K[x1,...,xn]I(V ) is an Integral Domain. For reducible
varieties V =⊔i Vi made of pairwise disjoint components, we have I(V ) = I(V1)
⋂...⋂I(Vn) and
K[V ] =K[x1, ..., xn]
I(V1)⋂...⋂I(Vn)
∼=K[x1, ..., xn]
I(V1)× ...× K[x1, ..., xn]
I(Vn)= K[V1]× ...×K[Vn]
f → (f|V1 , ..., f|Vn)
(by the Chinese Remainder Theorem, provided that I(Vi)+I(Vj) = K[x1, ..., xn], i.e. Vi and Vj are disjoint.)In the case V =
⋃i Vi when the components Vi are not pairwise disjoint, there is still an injective ring
51
homomorphism
Φ : K[V ] =K[x1, ..., xn]
I(V1)⋂...⋂I(Vn)
−→ K[x1, ..., xn]
I(V1)× ...× K[x1, ..., xn]
I(Vn)= K[V1]× ...×K[Vn]
f −→ (f|V1 , ..., f|Vn)
however it is not surjective. Indeed,
Im(Φ) = {(f1, ..., fn) ∈ K[V1]× ...×K[Vn]; fi = fj on Vi⋂Vj .}
Definition 10.5. Given two affine varieties V ⊆ Km and W ⊆ Kn, a map φ : V →W is called regular (orequivalently polynomial) if there are polynomials φ1, ..., φn ∈ K[x1, ..., xm] such that φ(p) = (φ1(p), ..., φn(p))for every p ∈ V.
Example 10.6. Consider the twisted cubic V = V(z − x3, y − x2) ⊂ K3. Then φ : V → K2 given by
φ(x, y, z) =(y3
z ,zy
)is a regular map on V . Even though y3
z ,zy are not polynomials in K[x, y, z], on V
we have y = x2 and z = x3 hence we can rewrite y3
z = x6
x3= x3 and z
y = x3
x2= x hence we can write
φ = (φ1, φ2) for the polynomials φ1(x, y, z) = x3 and φ2(x, y, z) = x.
Proposition 10.7. Regular maps of varieties/Homomorpshims of coordinate rings. Considertwo affine varieties X ⊆ Km and Y ⊆ Kn. There is a one-to-one correspondence:
{ regular maps φ : V →W} ←→ { ring homomorphisms φ∗ : K[W ]→ K[V ]}.
Moreover, this correspondence gives a contravariant functor:Given φ : V →W and ψ : W → T , then (ψ ◦ φ)∗ = φ∗ ◦ ψ∗.
Proof. For every regular map φ : V → W there are polynomials φ1, ..., φn ∈ K[x1, ..., xm] such thatφ = (φ1, ..., φn) on V . Then we can define
φ∗ : K[W ] =K[y1, ..., yn]
I(W )−→ K[V ] =
K[x1, ..., xm]
I(V )
[p(y1, ..., yn)] −→ [(p ◦ φ)(x)] = [p(φ1(x), ..., φn(x))]
Note that φ∗ is well defined: If p(y1, ..., yn) ∈ I(W ), then p(φ(x)) = 0 for all x ∈ V , since φ(x) ∈W . Thus(p ◦ φ)(x1, ..., xm) ∈ I(V ).
Forgetting the variables, the definition can be written in short notation as φ∗(p) = p◦φ, which is easierunderstood if we think of p ∈ K[W ] as polynomial functions p : W → K. Then we have the commutativediagram
V
φ∗(p)=p◦φ
φ //W
p
��K
Conversely, given a ring homomorphism Φ : K[W ]→ K[V ], we let φi(x) = Φ(yi) ∈ K[V ] for each i = 1, ..., n(uniquely defined mod I(V ). This gives a morphism φ = (φ1, ..., φn) : V → W , and Φ = φ∗: Indeed, sinceΦ is a ring homomorphism, it commutes with operations in a polynomial, hence
Φ([p(y1, ..., yn)]) = [p(Φ(y1), ...,Φ(yn))] = [p(φ1(x), ..., φn(x))] = [(p ◦ φ)(x)] = φ∗(p(y1, ..., yn)]).
52
The final statement is checked as follows: given φ : V →W and ψ : W → T , then
(ψ ◦ φ)∗(p) = p ◦ ψ ◦ φ = φ∗(p ◦ ψ) = φ∗(ψ∗(p)) = (φ∗ ◦ ψ∗)(p).
Example 10.8. The map φ : K→ K3 given by φ(t) = (t, t2, t3) corresponds to ring homomorphism
φ∗ : K[x, y, z] −→ K[t]
p(x, y, z) −→ p(t, t2, t3)
with φ∗(x) = t, φ∗(y) = t2, φ∗(z) = t3.
Furthermore φ∗ has Kernel ker(φ∗) = 〈z − x3, y − x2〉 and hence defines a ring homomorphism
ψ∗ : K[V ] =K[x, y, z]
〈z − x3, y − x2〉−→ K[t]
[p(x, y, z)] −→ p(t, t2, t3)
corresponding to the map ψ : K1 → V into the twisted cubic V = V(z − x3, y − x2). Let i : V → K3 be theinclusion map. Then we have
K1
ψ��
φ
!!V
i// K3
←→ K[t]
K[V ]
ψ∗
OO
K[x, y, z]i∗
oo
φ∗ee
where i∗ : K[x, y, z]→ K[V ] = K[x,y,z]〈z−x3,y−x2〉 is the quotient map. Note that i is injective while i∗ is surjective.
The map ψ is bijective and correspondingly φ∗ : K[V ] = K[x,y,z]〈z−x3,y−x2〉 → K[t] is a ring isomorphism.
Definition 10.9. A regular map which is bijective and whose inverse is also regular is called an isomor-phism of algebraic varieties.
Corollary 10.10. An isomorpshim of algebraic varieties f : V → W gives an isomorpshim of coordinaterings f∗ : K[W ]→ K[V ].
Proof: Indeed, according to the Proposition on Regular Maps, the inverse f−1 : W → V of f gives(f−1)∗ : K[V ]→ K[W ] which is the inverse of f∗.
Example 10.11. The map φ : K→ K2 given by φ(t) = (t2, t3) corresponds to ring homomorphism
φ∗ : K[x, y] −→ K[t]
p(x, y) −→ p(t2, t3)
with φ∗(x) = t2 and φ∗(y) = t3. The map φ∗ has Kernel ker(φ∗) = 〈y2 − x3〉 and hence defines a ringhomomorphism
ψ∗ : K[V ] =K[x, y]
〈y2 − x3〉−→ K[t]
[p(x, y)] −→ p(t2, t3)
53
corresponding to the map ψ : K1 → C into the cuspidal cubic V = V(y2 − x3). Let i : C → K2 be theinclusion map. Then we have
K1
ψ��
φ
!!C
i// K2
←→ K[t]
K[C]
ψ∗
OO
K[x, y]i∗
oo
φ∗dd
In this case ψ is reqular and bijective, but it is not an isomorphism of algebraic varieties, because its inverseis not a regular map (whether we define ψ−1(x, y) = 3
√y or ψ−1(x, y) = y/x when x 6= 0 and 0 when x = 0
etc). Correspondingly, the ring homomorphism ψ∗ : K[C] = K[x,y]〈y2−x3〉 → K[t] is injective but not surjective:
Im(ψ∗) = {g(t) ∈ K[t]; g′(0) = 0} 6= K[t].
Properties of regular map φ : X → Y Properties of the ring homomorphism φ∗ : K[Y ]→ K[X]
φ isomorphism of affine varieties φ∗ isomorphism of rings.
φ inclusion of varieties φ∗ surjective
φ dominant i.e. Im(φ) = Y φ∗ injective.
Affine varieties are not the be all and end all of algebraic geometry. At best, they are the buildingblocks of more general objects called varieties, which are obtained by gluing affine varieties along open setssimilarly to the construction of manifolds in differential geometry.
Earlier we have seen some examples of affine varieties which can be described by parameters (i.e.as images of injective maps from Km): lines, planes, the twisted cubic. On the one hand, parametricdescriptions are very useful in computer generated drawings of varieties. On the other hand we note thatthe set of parametrisable affine varieties is quite small, if the only parametric maps we allow are regularmaps. We would first like to enlarge the set of maps we can use for parametrising subvarieties.
The coordinate ring of a principal open set in Kn. Localisation. Given f(x1, ..., xn) ∈K[x1, ..., xn], the set Uf = {(x1, ..., xn) ∈ Kn; f(x1, ..., xn) 6= 0} is called principal open set. The projec-tion π : Kn+1 → Kn restricts to a bijective map
V(f(x1, ..., xn)xn+1 − 1)→ Uf
thus giving Uf a structure of affine variety with K[Uf ] = K[x1, ..., xn,1f ] = K[x1, ..., xn]f .
Definition 10.12. Consider two affine varieties V1 ⊆ Km and V2 ⊆ Kn. A rational morphism is a map
ϕ : V1 → V2 ⊆ Kn
x → ϕ(x) = (ϕ1(x), ..., ϕn(x))
where for every x ∈ V1 ⊆ Km we have ϕi(x) = fi(x)gi(x) = ratio of polynomials in m variables, with
gi(x) 6= 0. Note that by definition fi(x1,...,xm)gi(x1,...,xm) ∈ K(x) = K(x1, ..., xm), the field of fractions of K[x1, ..., xn].
A rational map ϕ : V1 99K V2 is defined similarly with a rational morphism, except that in this case V1
might have points where gi(x) 6= 0 so ϕ is not well defined.
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Let’s take the simple example of the circle and its stereographic projection:
Example 10.13. Stereographic Projection and its inverse. Consider the circle S1R = V(x2 + y2 −1) ⊂ R2. Then S1R admits a partial parametrisation with parameter t:
φ : R1 → S1R
t →(
1− t2
1 + t2,
2t
1 + t2
)which is the inverse of the stereographic projection φ−1 given by t = φ−1(x, y) = y
1+x = 1−xy . The corre-
spondence is constructed geometrically as follows: for every point (x, y) 6= (−1, 0) on the circle, we denoteby (0, t) the intersection of the line connecting (x, y) and (−1, 0) with the vertical axis. This gives us theexpected formulas for x and y.
We say this is a partial parametrisation because φ is injective but not bijective as (−1, 0) is not in itsimage. In other words, φ : R1 → S1R is a rational morphism, but φ−1 : S1R 99K R1 is a rational mapbecause neither y
1+x nor 1−xy is well defined at the point (−1, 0).
When we replace R with C, then φ : C1 → S1C is just a rational map as it is not well defined at thevalues t = ±i.
To get a full parametrisation for the circle, we need to enlarge K1 by adding a point at ∞. In otherwords, we need to replace the affine line K1 by the projective line KP1.
Moreover, if we work in the complex projective plane, we can apply the same method to get a parametri-sation for every conic - application to solving homogeneous quadratic diophantine equations - findingrational points on quadrics.
11 Projective space
Recall the n-dimensional projective space Pn defined as the set of lines in the (n + 1)-dimensional spaceKn+1 passing through the origin 0 = (0, 0, · · · , 0):
Pn = {l; l line in Kn+1 such that 0 ∈ l} =Kn+1 \ {0}∼
,
with to the equivalence relation (x0, x1, · · · , xn) ∼ λ(x0, x1, · · · , xn), where λ ∈ K∗. These (n + 1)-tuplesare known as homogeneous coordinates in Pn are are well defined only up to multiplication by a non-zeroconstant. A point in Pn with the homogeneous coordinates above will be denoted [x0 : x1 : ... : xn].
11.1 Projective Varieties and Homogeneous Ideals
Definition 11.1. A polynomial f ∈ K[x0, x1, · · · , xn] is homogeneous of degree d if
f(λx0, λx1, · · · , λxn) = λdf(x0, x1, · · · , xn)
for all λ ∈ C∗.
Let K[x0, x1, · · · , xn]k denote the subset of K[x0, x1, · · · , xn] whose elements are all the homogeneous poly-nomials of degree k in variables x0, x1, · · · , xn.
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Note: The definition insures that for a homogeneous polynomial,
f(x0, x1, · · · , xn) = 0 ⇐⇒ f(λx0, λx1, · · · , λxn),
hence the condition f(x0, x1, · · · , xn) = 0 is independent on the choice of representatives in the class[x0 : x1 : ... : xn].
Definition 11.2. A projective variety is a set of points in Pn given by a fixed set of homogeneous polynomialequations in x0, x1, · · · , xn−1, xn.
Example 11.3. A hyperplane is the solution set of a linear equation a0x0 + a1x1 + · · ·+ anxn = 0 in Pn.
A hyperplane in Pn is isomorphic to Pn−1. Indeed, there must exist a non-zero coefficient ai 6= 0 andthen the projection
V(∑i
aixi) −→ Pn−1
[x0 : · · · : xi : · · · : xn] −→ [x0 : · · · : xi : · · · : xn] has inversex0 : · · · : −∑j 6=i
ajaixj : · · · : xn
←− [x0 : · · · : xi−1 : xi+1 : · · · : xn] .
Example 11.4. By induction, the solution set of k linear homogeneous equations gives an (n − k)–dimensional linear projective subspace Pn−k ⊂ Pn, provided that the k equations are linearly independent.In other words, k hyperplanes in general positions intersects at a linear projective subspace Pn−k.
Exercise. Prove that two linear projective subspaces Pk and Ph intersect at a space Pn−k−h in Pn.
Example 11.5. A degree d hypersurface in Pn is V(f) where f is a homogeneous polynomial of degree d.
The ideal of a projective variety V is defined as for affine varieties
I(V ) = {f(x0, ..., xn) ∈ K[x0, ..., xn]; f(a0, ..., an) = 0 for all [a0 : ... : an] ∈ V }.
Every polynomial f(x0, ..., xn) ∈ K[x0, ..., xn] can be written as a sum
f(x0, ..., xn) =M∑k=0
fk(x0, ..., xn)
where fk is the homogeneous part in degree k of f . Then
f(λx0, ..., λxn) =
M∑k=0
fk(λx0, ..., λxn) =
M∑k=0
λkfk(x0, ..., xn)
and the equation f(λx0, ..., λxn) = 0 is true for all λ ∈ K iff fk(x0, ..., xn) = 0 for all k. This leads to thenotion of homogeneous ideal:
Definition 11.6. An ideal I ⊂ K[x0, ..., xn] is called homogeneous if for every f ∈ I we have fk ∈ I forevery k ∈ N, where fk is the homogeneous part in degree k of f .
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Example 11.7. A hypersurface of degree d in Pn is the solution set
Theorem 11.8. There is a one-to-one correspondence
{ projective varieties in Pn} ←→ { radical homogeneous ideals in K[x0, ..., xn]≥1 = ⊕k≥1K[x0, ..., xn]k}
Note that under this correspondence I(∅) = K[x0, ..., xn]≥1 since not points in Pn can have all homoge-neous coordinates equal to 0.
Definition 11.9. For a given projective variety V in Pn, its ring of homogeneous coordinates is defines as
S(V (I)) :=K[x0, x1, · · · , xn]
I(V ).
11.2 Maps between projective spaces
Definition 11.10. A map between projective spaces φ : Pm 99K Pn is given by
φ([x0 : · · · : xm]) = [φ0(x0, · · · , xm) : · · · : φn(x0, · · · , xm)]
where φ0(x0, · · · , xm), · · · , φn(x0, · · · , xm) are homogeneous polynomials of the same degree t. In this casewe say that the map φ has degree t.
Note that such a map may not be defined everywhere. We say that the map is a morphism on the setof points where it’s well defined.
Example 11.11. The projection π : Pn 99K Pn−1 given by π([x0 : · · · : xn]) = [x1 : · · · : xn] is well-definedeverywhere except at the point P = [1 : 0 : · · · : 0]. Hence it defines a morphism π : Pn \ {P} −→ Pn−1.We say that π is the projection centered at P onto the hyperplane V(x0) = Pn−1.
More generally:
Example 11.12. The projection π : Pn 99K Pn−k given by π([x0 : · · · : xn]) = [xk : · · · : xn] is well-defined everywhere except at the projective subspace Pk−1 = V(xk, . . . , xn). Hence it defines a morphismπ : Pn \ Pk−1 −→ Pn−k. We say that π = πP is the projection centered at the projective subspace Pk−1 =V(xk, . . . , xn) onto the projective subspace Pn−k = V(x0, . . . , xk−1).
Note that Pk−1 = V(xk, . . . , xn) is the smallest projective subspace which contains all the points P0 =[1 : 0 : · · · : 0], P1 = [0 : 1 : · · · : 0], . . . , Pk−1. We say that Pk−1 = V(xk, . . . , xn) is the projective subspacespanned by P0, P1, . . . , Pk−1. The projection π is a composition of k projections centered at points
PnπP0 // Pn−1
πP1 // Pn−2πP2 // · · · // Pn−k .
Example 11.13. More generally, given two projective subspaces L = Pk−1 and H = Pn−k such thatL⋂H = ∅ in Pn, we can define the projection πL : Pn \ L −→ H as follows: Given any point a ∈ Pn \ L,
we define πL(a) to be the unique point of intersection between H and the projective subspace spanned by Land a.
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Definition 11.14. A morphism of projective varieties V ⊆ Pm and W ⊂ Pn is a map φ : V → W suchthat for every a ∈ V , there exist φ0, · · · , φn are homogeneous polynomials of the same degree t such thatφ(a) = [φ0(a) : · · · : φn(a)].
Note that the choice of polynomials can change with a ∈ V :
Example 11.15. Conic parametrisation. Recall the parametrisation of the circle S1 = V(x2+y2−1) ⊂K2
φ : K1 → K2
t →(
1− t2
1 + t2,
2t
1 + t2
)which mapped K1 into S1 \ {(−1, 0)}. We can extend this map to projective spaces by substituting t = V
Uand x = X
Z , y = YZ and clearing denominators:
φ : P1 → P2
[U : V ] →[U2 − V 2 : 2UV : U2 + V 2
]Here P1 = K1
⋃{[0 : 1]} and φ([0 : 1]) = [−1 : 0 : 1] hence in this case Im(φ) = S1, the conic given by
X2 + Y 2 = Z2 in P2.
The inverse φ−1 : S1 → P1 is given by φ−1([X : Y : Z]) = [X + Z : Y ] = [Y : Z − X]. (note that[X + Z : Y ] is not well defined for [X : Y : Z] = [−1 : 0 : 1], but in that case we can use [Y : Z −X].
Other examples: Veronese and Segre embeddings.
12 Hilbert functions – Definitions and computations
Note that the splitting of a polynomial into homogeneous polynomials provides K[x0, x1, · · · , xn] with agraded ring structure
K[x0, x1, · · · , xn] = ⊕∞k=0 K[x0, x1, · · · , xn]k,
where K[x0, x1, · · · , xn]k is defined as the kth graded part of the ring. The product of polynomials alsodefines a map
K[x0, x1, · · · , xn]k ×K[x0, x1, · · · , xn]h −→ K[x0, x1, · · · , xn]k+h
This grading gives us an opportunity of measuring and comparing the sizes of coordinate rings, by measuringthe sizes of their graded parts. Indeed, while K[x0, x1, · · · , xn] is an infinite dimensional vector space overK, its kth graded part K[x0, x1, · · · , xn]k is finite dimensional, with basis
{xk00 xk11 · · ·x
knn ; k0 + · · · kn = k}.
Definition 12.1. Let X = V (I) be a projective variety in Pn with associated coordinate ring
S(X) =K[x0, x1, · · · , xn]
V(I).
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The Hilbert function of X, is the function
hX : N −→ Nk −→ hX(k) = dimS(X)k
which sends non-negative integers k to the dimension of S(X)k, the kth graded part of S(X), thought ofas a vector space over K.
Note that hX(0) = 1 as the only homogenous polynomials of degree 0 are the constants so S(X)0 = K.
Example 12.2. Let X = Pn. Then S(Pn) = K[x0, x1, · · · , xn] and hPn(k) is the number of monomials ofdegree k in n+ 1 variables. By the “stars and bars” method,
hPn(k) =
(n+ k
n
)=
(n+ k)!
n!k!=
(k + 1)(k + 2) · · · (k + n)
n!.
Note that this a polynomial expression in the variable k, of degree n = dimPn. Hence we can recovera geometric property of the space Pn (its dimension) from its Hilbert function. We will see that this is ageneral rule. Thus Hilbert polynomials will help us differentiate and classify projective varieties.
Example 12.3. Suppose X = {[1 : 0] [0 : 1]} ⊂ P1. Then I(X) = 〈x0x1〉, and so a suitable basis forS(X)k is {xk0, xk1} when k > 0.
=⇒ hX(k) =
{1, k = 0
2, k > 0.
Example 12.4. If X is a “double point” in P1, namely if we consider the non-radical ideal I(X) = 〈x20〉,
then a basis of S(X)k is {x0xk−11 , xk1} for k > 0. It follows that
hX(k) =
{1, k = 0
2, k > 0.
The Hilbert function regards the double point as though it is two separate points. Note though that the“double point” is more than just a projective variety – it is only defined as ”double point” by specifying anon-radical ideal I(X) = 〈x2
0〉. Such data (a set together with a possibly non-radical ideal) are examples ofmore general objects called algebraic schemes.
Example 12.5. Let X = {[1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1]}, a set of three non-collinear points in P2. ThenI(X) = 〈x0x1, x0x2, x1x2〉 in this case, indicating a suitable basis for S(X)k of {xk0, xk1, xk2} when k > 0.Therefore
hX(k) =
{1, k = 0
3, k > 0.
Example 12.6. Let X = {[1 : 0], [0 : 1], [1 : 1]} be a set of three collinear points in P1, with I(X) =〈x0x1(x0 − x1)〉. The number of x0’s in the monomial xk00 x
k11 can be reduced so long as k0 ≥ 2 and k1 ≥ 1
by repeatedly applying the identity [x20x1] = [x0x
21]. As a result, S(X)k has basis {x0, x1} if k = 1, and
{xk0, x0xk−11 , xk1} if k > 1, which leads to
hX(k) =
1, k = 0
2, k = 1
3, k > 1.
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We notice that in the last two examples, although the Hilbert functions differ for small values of k, forlarger values we have hX(k) = 3 in both cases. The same result is obtained if three collinear points in P2
are instead considered. Here for k > 2 we have hX(k) = 3 which can be thought of as a polynomial ofdegree 0 = dimX, while the actual value 3 corresponds to the number of points. Indeed, if X = {p1, p2, p3}then S(X) = S(p1)× S(p2)× S(p3) = K×K×K.
Example 12.7. Let Y be the image of the m-fold Veronese embedding
νm : Pn −→ P(n+mm )−1
[x0 : x1 : · · · : xn] 7−→[xi00 x
i11 · · ·x
inn
]ij≥0, i0+···+in=m
,
The morphism νm is injective and gives Pn ∼= Y . However, unlike with affine varieties, this isomorphismdoes not translate canonically into an isomorphisms of homogeneous coordinate rings, because of changes
in degrees. Indeed, if we let I = {(i0, i1, ..., in) ∈ Nn; i0 + · · ·+ in = m}, then a basis for S(P(n+mm )−1)1 can
be written as {y(i0,i1,...,in)}(i0,i1,...,in)∈I , the morphism νm gives a map ν∗m between the graded parts
S(Pn)m = K [x0, x1, · · · , xn]m ←− S(P(n+mm )−1)1
xi00 xi11 · · ·x
inn ←− y(i0,i1,...,in)
leading to isomorphismsK[x0, x1, · · · , xn]km ∼= S(Y )k.
By the definition of the Hilbert function,
hY (k) = dimK[x0, x1, · · · , xn]km)
=⇒ hY (k) =
(km+ n
n
)=
(km+ n)(km+ n− 1) · · · (km+ 1)
n!
Example 12.8. Let X = V (f) ⊂ Pn, the projective variety given as the zero locus of a homogeneouspolynomial f . Given d = deg(f), then for k ≥ d we have
hX(k) = dimS(X)k = dim
(K[x0, x1, · · · , xn]
〈f〉
)k
= dimK[x0, x1, · · · , xn]k
f ·K[x0, x1, · · · , xn]k−d.
⇒ hX(k) =
(k + n
n
)−(k + n− d
n
)=
(k + 1)(k + 2) · · · (k + n)− (k − d+ 1)(k − d+ 2) · · · (k − d+ n)
n!
which is a polynomial in the variable k with leading term d(n−1)!k
d.
The Hilbert functions in these examples have a number of similarities. They always agree with polynomialsof variable k for sufficiently large values of k. In this case, the degree n of the Hilbert polynomial agreeswith the dimensions of the associated variety, and the leading coefficients equals 1
n! times some positiveinteger, which we will find to always be the degree of the variety.
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13 Dimension and Degree of Projective Varieties
For affine spaces Kn and projective spaces Pn we have an obvious definition of dimension:
dim(Kn) = dim(Pn) = n.
For a hypersurface X = V(f) in Pn we also have natural notions of dimensions and degree:
dimV(f) = n− 1 and degV(f) = deg(f).
For varieties given by multiple equations in Pn it is not as straightforward to deduce the dimension anddegree directly from the number and degrees of their equations.
Example 13.1. The variety X = V (x21 − x0x2, x
22 − x1x3, x0x3 − x1x2) in P3 has dim(X) = 1 and
deg(X) = 3. (Based on the number and degrees of equations, we might have expected the dimension to be3-3=0 and the degree to be even).
Indeed, X is a degree 3 curve parametrised by
φ : P1 −→ P3
[u : v] −→ [u3 : u2v : uv2 : v3].
The 3 equations are not independent in the sense that they satisfy the relation
x0(x22 − x1x3) + x1(x0x3 − x1x2) + x2(x2
1 − x0x2) = 0.
Indeed, the 3 equations result from the 2× 2 minors of the matrix below (with ±):(x0 x1 x2
x1 x2 x3
)= φ∗
(u3 u2v uv2
u2v uv2 v3
)and the relation between the equations follows from the determinant
det
x0 x1 x2
x1 x2 x3
x0 x1 x2
= 0
From the relation above we note that
x2(x21 − x0x2) ∈ 〈x2
2 − x1x3, x0x3 − x1x2〉 but (x21 − x0x2) 6∈ 〈x2
2 − x1x3, x0x3 − x1x2〉
This the other similar relations show that all three equations x21−x0x2, x2
2−x1x3, x0x3−x1x2 are necessaryto define X: not one equation is in the ideal generated by the other two. This accounts for the unintuitiveresults for dimension and degree. If we think of X as being obtained by cut out by successive equations
X = V(x22 − x1x3)
⋂V(x0x3 − x1x2)
⋂V (x2
1 − x0x2),
we note that after using two first equations, the resulting variety is made of two components:
V(x22 − x1x3)
⋂V(x0x3 − x1x2) = V(x2
2 − x1x3, x0x3 − x1x2) = X⋃
P1.
In other words, the 3rd equation is useful only in eliminating the extra component P1 but it does notaffect the component X in any way, because X ⊂ V (x2
1 − x0x2).
Further examples from Grassmanians, Segre and Veronese embeddings...
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Example 13.2. A set of d points in Pn is said to have dimension 0 and degree d.
We can now define the dimension and degree of a general projective variety X by two methods: usingsections (cutting) and using projections (squashing). In both cases we will use linear projective spacesPn−1, Pn−2, ..., Pk.
Definitions based on Hyperplane Sections
Intuitively, whenever we cut an algebraic variety X by a hyperplane (1 linear equation) we expect to cutthe dimension by 1. Hence if we cut by k hyperplane and obtain a set of dimension 0 (a set of points),we expect X to have had dimension k to start with. However, this is not always true: it depends on theposition of the hyperplanes we chose.
Example 13.3. Let X = V(x0x3 − x1x2) in P3. This is a hypersurface hence it has dimension 2. Indeed,intersection by two hyperplanes
X⋂
V(x0)⋂
V(x3) = {[0 : 0 : 1 : 0], [0 : 1 : 0 : 0]}
a 0-dimensional set made of 2 points. However,
X⋂
V(x0)⋂
V(x1) = V(x0, x1) ∼= P1.
The reason for this anomaly is that
X⋂
V(x0) = V(x0, x1)⋃
V(x0, x2) = P1⋃
P1,
and one of the two components V(x0, x1) is already contained in V(x1).
Definition 13.4. Hyperplanes in General Position. For every projective variety X, we say that thehyperplanes H1, H2, ..., Hk are in general position with respect to X if at any intermediate stage l < k,the intersection X
⋂H1⋂· · ·⋂Hl has no component contained in Hl+1.
Definition 13.5. Dimension defined using Hyperplane Sections. If k hyperplanes H1, H2, ..., Hk
are in general position with respect to X and the intersection X⋂H1⋂· · ·⋂Hk is a non-empty set of
points, then we define say that dim(X) = k.
Definition 13.6. Degree defined using Hyperplane Sections. If k hyperplanes H1, H2, ..., Hk arein general position with respect to X and the intersection X
⋂H1⋂· · ·⋂Hl is a non-empty set of points,
then we define
deg(X) =∑
a∈X⋂H1
⋂···
⋂Hk
mX⋂H1
⋂···
⋂Hk
(a).
Here mX⋂H1
⋂···
⋂Hk
(a) denotes the multiplicity of intersection X⋂H1⋂· · ·⋂Hk at a.
Remark 13.7. If a = [a0 : a1 : ... : an] with ai 6= 0 is a point of intersection of some projective varietiesV1, . . . , Vh in Pn, then the multiplicity of the intersection V1
⋂· · ·⋂Vn at a can be calculated working in
the affine space Ui ∼= Kn given by the condition xi 6= 0 (see Definition ?? on Multiplicity of Intersection.)
Most hyperplanes H1, H2, ..., Hk in general position with respect to X intersect X with multiplicity 1at all points, so the definition above becomes deg(X) = |X
⋂H1⋂· · ·⋂Hk|.
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Remark 13.8. Intersection with Linear Projective space of Complementary Dimension. Sincethe intersection of k linearly independent hyperplanes in Pn is a Pn−k, we could rephrase the definitionabove as
dim(X) = k ⇐⇒ dim(X⋂
Pn−k) = 0, and in this case
deg(X) =∑
a∈X⋂
Pn−k
mX⋂
Pn−k(a)
for a Pn−k in general position with respect to X in Pn.
Reality Check: Check for curves in P2, curves and surfaces in P3.
Note: If dim(X) ≤ k one can always find k hyperplanes in general position wrt X, because one can alwaysfind a hyperplane which does not contain any component of a variety. (Exercise!)
It remains to prove that the Definitions above are consistent, namely that they do not depend onthe choice of k-tuples H1, ...,Hk of hyperplanes in general position. We will do that by use of Hilbertpolynomials.
Dimension and Degree based on Hilbert Polynomials
Theorem 13.9. Dimension and Degree from the Hilbert Polynomial. Consider X ⊂ Pn aprojective variety of dimension m. The Hilbert function of X, hX(k) := dim(S(X)k), is written as apolynomial pX(k)in the variable k when k � 0. Furthermore:
1. The degree of pX(k) is equal to m, the dimension of X.
2. The leading coefficient of pX(k) is 1m! deg(X).
Proof. Proof by induction on m. In the case when m = 0, we have seen how hX(k) is a constant polynomial= d if X is a finite set of d points.
For general m and X a general projective variety of dimension m, let f be the equation of a hyperplane,chosen so that no component of X lies in the hyperplane V(f). Now let Y = X
⋂V(f) = V(I, f), where
X = V(I) ⊂ Pn.We have that
S(Y )k ∼=S(X)k
f · S(X)k−1for all k ≥ 1.
Indeed, the restriction morphism S(X)k → S(Y )k is surjective with Kernel equal to f · S(X)k−1.
Next we note that the multiplication by f gives an injective map S(X)k−1 → f · S(X)k−1 ⊂ S(X)k.Indeed, assume that there exists some a ∈ S(X)k−1 such that
a · f = 0 but a 6= 0 in S(X)k−1
⇐⇒ a · f ∈ I
⇐⇒ V(a)⋃
V(f) = V(a · f) ⊇ V(I) = X.
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Since a 6= 0 in S(X), it follows that a /∈ I ⇐⇒ V(a) 6⊇ V(I), which would imply that V(f) contains somecomponent of X. However this contradicts the assumption that no component of X lies in the hyperplaneV(f). Hence means @ a such that a · f = 0 and a 6= 0 in S(X) and so ·f is injective. Now
·f injective ⇐⇒ dim(f · S(X)k−1) = dim(S(X)k−1,
and hence
S(Y )k ∼=S(X)k
f · S(X)k−1=⇒ hY (k) = hX(k)− hX(k − 1) for all k ≥ 1.
By applying the above relation repeatedly for k, k − 1, ..., 1 we get
hX(k) = hY (k) + hY (k − 1) + · · ·+ hY (1).
By construction we have Y = X⋂V(f), hence
dim(Y ) = dim(X)− 1 and deg(Y ) = deg(X).
We may now apply the lemma below to hY = Q and hX = P . Indeed, by induction
pY (k) =deg(Y )
(m− 1)!km−1 + · · · and hence by Lemma: pX(k) =
deg(X)
m!km + · · ·
Lemma 13.10. If P : N → N is a function such that P (k) − P (k − 1) = Q(k) for k � 1 for somepolynomial Q(k) = q0 + q1k + · · ·+ qm−1k
m−1, then P (k) is a polynomial expression with
deg(P ) = deg(Q) + 1 and P (k) =qm−1
mkm + · · ·
where · · · denote lower degree terms in k.
Proof. By ”telescoping” we get
P (k) = P (k)− P (k − 1) + P (k − 1)− P (k − 2) + · · ·+ P (1)− P (0)
= Q(k) +Q(k − 1) + · · ·+Q(0)
= q0S0(k) + q1S1(k) + · · ·+ qm−1Sm−1(k)
where Sn(k) =∑k
j=0 jn = 1n + 2n + · · ·+ kn. We can apply strong induction on n to prove that Sn(k) is a
polynomial expression in k and Sn(k) = 1n+1k
n+1 + · · · where · · · denote lower degree terms in k.
Indeed the binomial formula gives
(k − 1)n+1 = kn+1 −(n+ 1
1
)kn +
(n+ 1
2
)kn−1 + · · ·+ (−1)n+1
(n+ 1
n+ 1
)hence
kn+1 − (k − 1)n+1 =
(n+ 1
1
)kn −
(n+ 1
2
)kn−1 + · · ·+ (−1)n
(n+ 1
n+ 1
)By summing up the above equation for k, k − 1, ...., 1 and telescoping we get
kn+1 =
(n+ 1
1
)Sn(k)−
(n+ 1
2
)Sn−1(k) + · · ·+ (−1)nS0(k) =⇒
Sn(k) =1
n+ 1kn+1 +
1
n+ 1
(n+ 1
2
)Sn−1(k)− · · · − (−1)n
1
n+ 1S0(k)
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Hence by induction we obtain
Sn(k) =1
n+ 1kn+1 + · · · and hence P (k) = qm−1Sm−1(k) + · · · = qm−1
mkm + · · ·
Example 13.11. Let X = Pn.
=⇒ pX(t) =
(n+ t
n
)
=⇒ pX(t) =(n+ t)(n+ t− 1) . . . (t+ 1)
n!
=⇒ pX(t) =1
n!tn + . . .
The degree of pX is n, thus implying that the dimension of X = Pn is n as expected.
Example 13.12. Let Y be the image of the d-fold Veronese embedding vd : Pn → PN .
=⇒ pY (t) =
(dt+ n
n
)
=⇒ pY (t) =dn
n!tn + . . .
indicating that the dimension of Y is n.
Example 13.13. Suppose X = V (f, g) ⊂ Pn, where f and g are independent homogeneous polynomials ofdegrees c and d respectively. Firstly, X = V (f, g) ⊂ V (f) = Y . Then
S(X) =K[x0 : · · · : xn]
〈f, g〉
=⇒ S(X) =
(K[x0 : · · · : xn]
(f)
)/
(gK[x0 : · · · : xn]
(f)
)=⇒ S(X)k =
S(Y )tg · S(Y )k−.deg(g)
As a result,
pX(k) = pY (k)− pY (k − d)
=⇒ pX(k) =
(k + n
n
)−(k + n− c
n
)−(k + n− d
n
)+
(k + n− c− d
n
).
and direct calculations of the leading term show that dim(X) = n− 2 and deg(X) = cd.
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Bezout’s theorem
Notation . A short exact sequence is a sequence of the form
0→ ker(φ)→ V →W → 0.
where φ : V →W is a surjective homomorphism of vector spaces.
The Theorem of Dimension and Degrees based on Hilbert Polynomials and its proof can be generalisedto
Theorem 13.14. (Bezout’s theorem) If X ∈ Pn is a projective variety with dimension m, and if f isa homogeneous polynomial in K[x0, · · · , xn] chosen so that no component of X lies in V (f), then
deg(X ∩ V(f)) = deg(X) · deg(f).
Proof. Similar to previous cases, there exists a short exact sequence
0→ K[x0, · · · , xn]
I(X)
·f−→ K[x0, · · · , xn]
I(X)→ K[x0, · · · , xn]
I(X) + (f)→ 0
=⇒ hX∩V(f)(t) = pX(t)− pX(t− deg(f) where
pX(t) =deg(X)
m!tm + cm−1t
m−1 + · · ·
=⇒ pX(t− deg(f)) =deg(X)
m!(t− deg(f))m + cm−1(t− deg(f))m−1 + · · ·
=deg(X)
m!(tm − deg(f)tm−1m+ · · · ) + cm−1(tm−1 + · · · ) + · · ·
=⇒ hX∩V(f)(t) =deg(X) · deg(f)
(m− 1)!tm−1 + · · ·
=⇒ deg(X ∩ V(f)) = deg(X) · deg(f).
Dimension and Degree based on Projections
Let K be an algebrically closed field. The Proposition on Closed Projections 9.8 works similarly forprojective varieties:
Proposition 13.15. Closed Projections with Finite Fibres: For every projective variety V ⊂ Pn, wecan find a hyperplane Pn−1 ⊂ Pn and a projection map π : Pn 99K Pn−1 such that
• π(V ) is a projective variety in Pn−1, and moreover,
• for every b ∈ π(V ), the fibre π−1(b) = {a ∈ V ; π(a) = b} is a finite set of points.
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Proof. The proof is the same as for affine varieties: Given V = V(f1, . . . , fk), we can make a change ofvariables so that in the new coordinates each
fi = cixdi0 + . . .
with di = deg(fi) and ci = 0. Then π : Pn 99K Pn−1 is just the projection forgetting the coordinate x0 andπ(V ) is as described in the Proposition on Closed Projections 9.8. Moreover, for every b = [b1 : · · · : bn] ∈π(V ), the points a ∈ V with π(a) = b are
a = [x : b1 : · · · : bn] such that fi(x, b1, · · · , bn) = 0 for all i = 1, ..., k.
By construction these are non-trivial equations in the variable x and hence they will have at most finitelymany solutions.
By applying this Proposition successively we obtain:
Theorem 13.16. ”Noether Normalisation Theorem” (projective geometry formulation). Forevery projective variety X ⊂ Pn, we can find a linear projective subspace Pk ⊂ Pn and a projection mapπ : Pn 99K Pk which restricts to a morphism πX : X → Pk surjective with finite fibres.
Definition 13.17. Dimension defined using Projections. If the conditions in the theorem above aresatisfied, we say that dim(X) = k.
Definition 13.18. Degree defined using Projections. If the conditions in the theorem above aresatisfied, we define deg(X) to be the number of points in the largest fibre.
Note: Most fibres should have the same number of points, but there may be some fibres containing”points with multiplicity”.
Remark 13.19. The definitions of Dimension and Degree based on Projections are compatible with theDefinitions based on Hyperplane Sections and Hilbert Polynomials. In particular, they are consistent acrossall choices of suitable projections as above. Indeed, this can be checked by comparing with the Remark onIntersection with Linear Projective space of Complementary Dimension:
If there exists a map πX : X → Lk = Pk, surjective and with finite fibres, then as seen in the section onProjective Maps and Morphisms, the projection π : Pn 99K Lk = Pk is a composition of n− k projectionscentered at points
PnπP0 // Pn−1
πP1 // Pn−2πP2 // · · ·
πPn−k−1 // Pk = Lk
and the points P0, P1, . . . , Pn−k−1 span a linear projective subspace Ln−k−1 ⊂ Pn such that
π : Pn \ Ln−k−1 −→ Lk = Pk
is a projective morphism, i.e.π is the projection centered at Ln−k−1.
Note that Ln−k−1⋂Lk = ∅. Moreover, given b ∈ Lk, the points P0, P1, . . . , Pn−k−1, b span a linear
projective space Ln−k = 〈Ln−k−1, b〉 ∼= Pn−k and the subspace is in general position with respect to X (by
67
construction of the map π). Comparing with the Remark on Intersection with Linear Projective spaces ofComplementary Dimension, we get that
dim(X) = k ⇐⇒ dim(X⋂Ln−k) = 0, and in this case
deg(X) =∑
a∈X⋂Ln−k
mX⋂Ln−k
(a)
However, by construction of π we have that X⋂Ln−k = π−1(b) which is exactly as in the Definitions of
Dimension and Degree based on Projections.
Ln−k−1
Lk
14 Arithmetic genus, the Euler number, Geometric genus of a Curve
The arithmetic genus pa(X) of a curve X is defined by
pa(X) := 1− pX(0),
where pX is the Hilbert polynomial of X. For smooth complex curves, this quantity coincides with thegeometric genus of X, g(X) which is the number of “holes” that X has when represented as a two-dimensional real surface.
Example 14.1. The two-dimensional real sphere S2 satisfies g(S2) = 0, as S2 does not contain holes.
By Example 2.1, pPn(0) =(nn
)= 1, and so pa(P1) = 0. Indeed, S2 and P1K have the same real manifold
structure.
Example 14.2. Suppose X = V (f) is a curve in P2K, where f is a homogeneous polynomials of degree
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d. Then from Example 2.7,
pX(t) =
(t+ 2
2
)−(t+ 2− d
2
)
pX(t) =(t+ 2)(t+ 1)
2− (t+ 2− d)(t+ 1− d)
2
pa(X) = 1− 2− (d− 1)(d− 2)
2=
(d− 1)(d− 2)
2.
Example 14.3. Suppose X = V (f, g) is a curve in P3K, where f and g are homogeneous polynomials ofdegree d and e respectively. Then from Example 3.5,
pX(t) =
(t+ 3
3
)−(t+ 3− d
3
)−(t+ 3− e
3
)+
(t+ 3− d− e
3
)=
1
3![(t+ 3)(t+ 2)(t+ 1)− (t+ 3− d)(t+ 2− d)(t+ 1− d)− (t+ 3− e)(t+ 2− e)(t+ 1− e)
+ (t+ 3− d− e)(t+ 2− d− e)(t+ 1− d− e)]
=1
6[t3 + 6t2 + 11t+ 6− t3 − t2(6− 3d)− t(3d2 − 12d+ 11)− (6− 11d+ 6d2 − d3)− t3 − t2(6− 3e)
− t(3e2 − 12e+ 11)− (t− 11e+ 6e2 − e3) + t3 + t2(6− 3d− 3e) + t(11− 12d+ 3d2 − 12e+ 6de+ 3e2)
+ 6− 11d+ 6d2 − d3 − 11e+ 12de− 3d2e+ 6e2 − 3de2 − e3]
∴ pX(t) =12de− 3d2e− 3de2 + 6det
6
and so the arithmetic genus of X is
pa(X) = 1− 12de− 3d2e− 3de2
6= 1− 2de+
de(d+ e)
2.
Proving that the arithmetic genus and the geometric genus are equivalent relies on a number of concepts,which are set out below.
Theorem 14.4. The arithmetic genus of a smooth plane curve of degree d is equal to its geometric genus.
Proof. Let C ⊂ P2 be a smooth plane curve of degree d. Then its Hilbert polynomial is
hC(m) = md+ 1−(d− 1
2
)=⇒ pa(C) =
(d− 1
2
)
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15 Cech cohomology and the Hilbert function of a projective variety
Suppose a function π with domain Pn+1\{(0 : 0 : · · · : 1)} and codomain Pn removes the final coordinate ofits inputs – π(x0 : x1 : · · · : xn : xn+1) = (x0 : x1 : · · · : xn). π is then what is known as a one-dimensionalvector bundle over Pn.
Pn = (xn+1 = 0)
q = (0 : 0 : . . . : 1)
π
(x0 : x1 : . . . : xn+1)
(x0 : x1 : . . . : xn)
This concept can be extended to higher dimensions.
Definition 15.1. A k-dimensional vector bundle Eπ−→ X is a map π between manifolds E and X =
⋃i∈I Ui
(a union of open sets), such that for all i
π−1(Ui)φi∼= Ui ×Kk
and for each pair i, j, the transition functions
φij := φi|π−1(Ui⋂Uj) ◦ φ−1
j|π−1(Ui⋂Uj)
: (Ui⋂Uj)×Kk → (Ui
⋂Uj)×Kk
are linear on each fibre φij(x, v) = (x,Av), v being a k-dimensional vector and A a (k × k) square matrix.
This process may be represented by the following commutative diagram.
π−1(Ui ∩ Uj) (Ui ∩ Uj)×Kk 3 (x,Av)
(Ui ∩ Uj)×Kk 3 (x, v)
φi(π−1(Ui∩Uj))
φj(π−1(Ui∩Uj))φj ◦ φ−1
i =φij
If π is a continuous map, Eπ−→ X is called a topological vector bundle. If E and X are both algebraic
varieties and π is regular (can be written in polynomial form), Eπ−→ X is known as an algebraic vector
bundle.
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Example 15.2. Let L := Pn+1\{(0 : 0 : · · · : 1)}. The function π : L→ Pn is then a vector bundle of rank1.
To show this, first observe that Pn may be represented as the union of open sets⋃ni=0 Ui, where Ui =
{[x0 : x1 : · · · : xn]; xi 6= 0}. In fact, Ui ∼= Kn = {xi = 1} ⊆ Kn+1 by the relations
Ui 3 [x0 : x1 : · · · : xn] = [x0
xi: · · · : xn
xi]→ (
x0
xi, · · · , 1, · · · , xn
xi)
with inverse[u0 : u1 : · · · : 1 : · · · : un]← (u0, u1, · · · , ui, · · · , un)
Now observe that the function
π−1(Ui)π−→ Ui, π
−1(Ui) 3 [x0 : x1 : · · · : xn+1] 7−→ [x0 : · · · : xn] = [x0
xi: · · · : xn
xi]
can be completed to a natural isomorphism
π−1(Ui)φi−→ Ui ×K
[x0xi
: · · · : 1 : · · · : xn+1
xi
]([x0xi
: · · · : xnxi], xn+1
xi).
bijective
φi
Over the intersection π−1(Ui⋂Uj) this gives a commutative diagram
(x0xi : · · · : 1 : · · · : xn+1
xi) ((x0xi : · · · : xnxi ), xn+1
xi)
(x0xj : · · · : 1 : · · · : xn+1
xj) ((x0xj : · · · : xnxj ), xn+1
xj)
=
bijective
φi
φj◦φ−1i
φj
where φj ◦ φ−1i consists in multiplying the last coordinate by xi
xjto achieve a change of basis. π is thus a
vector bundle of rank 1. �
Definition 15.3. A section of an algebraic vector bundle Eπ−→ X is a regular map s : X → E such that
π ◦ s = idX .
The vector bundle π in Example 7.2 is commonly referred to as O(1). Its sections (or to be more accurate,its global sections) are x0, x1, . . . , xn, and the various linear combinations of these coordinates. Moreprecisely, if l =
∑i lixi, it can be identified with the section
s : Pn → L, with s([x0 : ... : xn]) = [x0 : ... : xn : l].
Similar bundles O(d) may also be defined for all d ∈ N, by
O(d) =
⋃ni=0(Ui ×K)
∼,
71
where ∼ is the equivalence relation
(x, v) ∼ (x,
(xixj
)dv) ∀x ∈ Ui ∩ Uj , x = [x0 : x1 : · · · : xn].
The global sections of O(d) are precisely the polynomials of degree d in x0, x1, . . . , xn. This is significantin that it allows for the areas of vector bundles and the Hilbert function to be linked. In fact – thoughwith the added condition that d must be sufficiently large – hX(d) is exactly the dimension of the space ofglobal sections of OX(d), the restriction to X of the vector bundle O(d).
Now suppose π : E → X is a general vector bundle. The set of global sections of π forms a vectorspace E(X). If U ⊆ X is open, a second vector bundle may be derived from E �U π−1(U)→ U . This newvector bundle has an associated vector space
E(U) = {s : U → E �U ; π ◦ s = idU},
where s is what is known as a local section of E on U .Letting X =
⋃i Ui as in Example 7.1 allows the following map to be defined:
E(X)δ0−→∏i
E(Ui)δ1−→∏i 6=jE(Ui ∩ Uj)
sδ0−→ (s�Ui)i; (si)i
δ1−→ si�(Ui∩Uj) − sj�(Ui∩Uj)
si is here defined on Ui and δ1 is the linear map of vector spaces with kernel
Ker(δ1) = {(si)i; si�(Ui∩Uj) = sj�(Ui∩Uj) on Ui ∩ Uj} = Im(δ0)
indicating that the above morphism sequence is exact. As well as this, it should be noted that Im(δ0) ∼=E(X).
This process may be extended like so:
0→∏i
E(Ui)δ1−→∏i,j
E(Ui ∩ Uj)δ2−→∏i,j,k
E(Ui ∩ Uj ∩ Uk)→ · · · ,
and δ2 is given by(sij)ij → (sij�(Ui∩Uj∩Uk) + sjk�(Ui∩Uj∩Uk) − sik�(Ui∩Uj∩Uk))ijk;
note that by the composition,
(si)iδ1−→ (si�(Ui∩Uj) − sj�(Ui∩Uj))ij
δ2−→ (si − sj + sj − sk + sk − si)�(Ui∩Uj∩Uk) = 0.
Im(δ1) ⊆ Ker(δ2), but equality is not guaranteed in this case, and so the sequence above is not necessarily
exact. With this in mind, define the vector space H1(X,E)Ker(δ2)Im(δ1) .
This process can be continued by defining
δk :∏
j1,j2,...,jk
E(Uj1 ∩ Uj2 ∩ ... ∩ Ujk)→∏
j1,j2,...,jk+1
E(Uj1 ∩ Uj2 ∩ ... ∩ Ujk+1),
δk(s) = (k+1∑l=1
(−1)k+1−lsj1j2···jl···jk+1|Uj1···jk+1
)j1···jk+1,
where s = (sj1j2...jk)j1,j2,...,jk and jl indicates that the jl component is absent from the calculations.
In the case of projective varieties a finite cover of open sets is considered, with the result that theprocess effectively concludes after a finite number of steps.
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References
• David Cox, John Little, Donal O’Shea: Ideals, Varieties and Algorithms chapters 1, 3, 4, 5, 7, 8, 9
• Miles Reid Undergraduate Algebraic Geometry (London Mathematical Society Student Texts)
• Frances Kirwan: Complex Algebraic Curves, London Mathematical Society Student Texts, 23, Cam-bridge University Press, 1992.
• William Fulton: Algebraic Curves. An Introduction to Algebraic Geometry, Reprint of 1969 original,Addison-Wesley, 1989.
• Igor R. Shafarevich: Basic Algebraic Geometry 1: Varieties in Projective Space
• Joe Harris: Algebraic Geometry – A first course (Graduate Texts in Mathematics) 1995
• Mumford The Red Book of Varieties and Schemes
• http://people.math.gatech.edu/ aleykin3/math4803spr13/BOOK/chapter4.pdf
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