Section 4-2

Radical Ideals and the Ideal-Variety Correspondence

by Pablo Spivakovsky-Gonzalez

-We ended Section 4-1 with Hilbert’s Nullstellensatz

-In this section we will look at Hilbert’s Nullstellensatz from

a different perspective

-If we have some variety V, can we identify those ideals that

consist of all polynomials which vanish on that variety?

Lemma 1:

Let V be a variety. If fm 2 I(V), then f 2 I(V).

Proof:

Let x 2 V. If fm 2 I(V), then (f(x))m = 0.

This can only be true if f(x) = 0.

This reasoning applies to any x 2 V, so we conclude that f 2 I(V).

-Therefore, I(V) has the property that if some power of a

polynomial is in the ideal, then that polynomial itself must

also belong to I(V).

-This leads to the definition of radical ideal.

Definition 2:An ideal I is radical if fm 2 I for some

integer

m ¸ 1 implies that f 2 I.

-We can now rephrase Lemma 1 using radical ideals.

Corollary 3:I(V) is a radical ideal.

Definition 4:

Let I ½ k[x1,…,xn] be an ideal. The radical of I,

denoted , is the set

{f : fm 2 I for some integer m ¸ 1 }.

Properties of :

1. I ½ , since f 2 I means f1 2 I and therefore f 2 by definition.

2. An ideal I is radical if and only if I = .

3. For any ideal I, is always an ideal.

Example:

Consider the ideal J = h x2, y3 i½ k[x, y].

Neither x nor y lie in J; but x 2 and y 2 .

Also, (x ¢ y)2 = x2y2 2 J, because x2 2 J.

Then x ¢ y 2 . Finally, x + y 2 . To see this, we note that

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 ,

by the Binomial Theorem.

Since each term above is a multiple of either x2 or y3, (x + y)4 2 J,

and therefore x + y 2 .

Lemma 5:

If I is an ideal in k[x1,…,xn] then is an ideal

in k[x1,…,xn] containing I. Furthermore, is a

radical ideal.

Proof:

I ½ has already been shown. We want to prove

is an ideal. Let f, g 2 ; then by definition there exist

m, l 2 Z+ so that fm, gl 2 I.Now consider the binomial expansion of (f + g)m+l-1 .

Every term in the expansion has a factor figj with

i + j = m + l – 1.

Therefore, either i ¸ m or j ¸ l, so either fi 2 I or

gj 2 I. This implies that figj 2 I, so every term of the

expansion lies in I.

Therefore, (f + g)m+l-1 2 I, so f + g 2 .

Finally, suppose f 2 and h 2 k[x1,…,xn].

This means that fm 2 I for some integer m ¸ l.

Therefore, hmfm 2 I, or (h¢f)m 2 I, so hf 2 .

This completes the proof that is an ideal. The book leaves the proof that is a radical ideal as an exercise at the end of the section.

Theorem 6 (The Strong Nullstellensatz):

Let k be an algebraically closed field. If I is an ideal in

k[x1,…,xn], then

I(V(I)) =

Proof:Consider any f 2 . Then by definition fm 2 I for some m ¸ l. Therefore fm vanishes on V(I), so clearly f must vanish on V(I) also.

It follows that f 2 I(V(I)) , so we have ½ I(V(I)) .

Conversely, suppose that f 2 I(V(I)). Then f vanishes on V(I). Now, by Hilbert’s Nullstellensatz, 9 m ¸ l such that fm 2 I.

This means that f 2 . And because f was arbitrary,

I(V(I)) ½ .

Before we had ½ I(V(I)), so clearly I(V(I)) =

This concludes our proof.

- Note: From now on, Theorem 6 will be referred to simply as “the Nullstellensatz”.

- The Nullstellensatz allows us to set up a “dictionary” between algebra and geometry => very important!

Theorem 7 (The Ideal-Variety Correspondence):

Let k be an arbitrary field.

1. The maps

I

affine varieties ===> ideals

V

ideals ===> affine varieties

are inclusion-reversing. If I1 ½ I2 are ideals,

then

V(I1) ¾ V(I2) and, similarly, if V1 ½ V2 are

varieties, then I(V1) ¾ I(V2).

In addition, for any variety V, we have V(I(V)) = V,

so that I is always one-to-one.

2. If k is algebraically closed, and we restrict ourselves to

radical ideals, then the maps

I

affine varieties ===> radical ideals

V

radical ideals ===> affine varieties

are inclusion-reversing bijections which are inverses

of each other.

Proof:

1.

The proof that I and V are inclusion reversing is given

as an exercise at the end of the section.

We will now prove that V(I(V)) = V, when

V = V(f1,…,fs) is a subvariety of kn.

By definition, every f 2 I(V) vanishes on V, so

V ½ V(I(V)).

On the other hand, we have f1,…,fs 2 I(V) from

definition of I. Therefore, h f1,…,fs i ½ I(V).

V is inclusion reversing, hence

V(I(V)) ½ V(h f1,…,fs i) = V.

Before we had V ½ V(I(V)), and now we

showed V(I(V)) ½ V.

Therefore V(I(V)) = V, and I is one-to-one

because it has a left inverse.

This completes the proof of Part 1 of Thm. 7.

2.

By Corollary 3, I(V) is a radical ideal.

We also know that V(I(V)) = V from Part 1.

The next step is to prove I(V(I)) = I whenever

I is a radical ideal.

The Nullstellensatz tells us I(V(I)) = . Also, if I is radical, I = (Exercise 4).

Therefore, I(V(I)) = I whenever I is a radical

ideal.

We see that V and I are inverses of each other.

V and I define bijections between the set of radical

ideals and affine varieties. This completes the proof.

Consequences of Theorem 7:

- Allows us to consider a question about varieties

(geometry) as an algebraic question about radical ideals,

and viceversa.

- We can move between algebra and geometry => powerful

tool for solving many problems!

- Note that the field we are working over must be

algebraically closed in order to apply Theorem 7.

Questions about Radical Ideals:

Consider an ideal I = h f1,…,fs i :

1. Radical Generators: Is there an algorithm to produce a set

{g1,…,gm} so that = h g1,…,gm i ?

2. Radical Ideal: Is there an algorithm to determine if I is radical?

3. Radical Membership: Given f 2 k[x1,…,xn], is

there an algorithm to determine if f 2 ?

Answer:-Yes, algorithms exist for all 3 problems.

-We will focus on the easiest question, #3, the Radical

Membership Problem.

Proposition 8 (Radical Membership):Let k be an arbitrary field and let

I = h f1,…,fs i ½ k[x1,…,xn]

be an ideal. Then f 2 if and only if the constant

polynomial 1 belongs to the ideal

= h f1,…,fs, 1 – yf i ½ k[x1,…,xn,y].

In other words, f 2 if and only if

= k[x1,…,xn,y].

Proof:

Suppose 1 2 . Then we can write 1 as:

s

pi (x1,…,xn, y) fi + q(x1,…,xn, y)(1 – yf),

i

for some pi , q 2 k[x1,…,xn, y].

We set y = 1 / f(x1,…,xn). Then our expression

becomes

s

pi (x1,…,xn, 1/f ) fi ,

i

Now we multiply both sides by fm :

s

fmAi fi ,i

for some polynomials Ai 2 k[x1,…,xn].

Therefore, fm 2 I, and so f 2 .

Going the other way, suppose that f 2 .

Then fm 2 I ½ for some m.

At the same time, 1 – yf 2 . Then,

1 = ymfm + (1 – ymfm ) =

= ymfm + (1 – yf)(1 + yf +…+ ym-1fm-1 ) 2

Hence, f 2 implies that 1 2 .

And before we had that 1 2 implies f 2 so the proof is complete.

Radical Membership Algorithm:

-To determine if

f 2 ½ k[x1,…,xn]

we first compute a reduced Groebner basis for:

h f1,…,fs, 1 – yf i ½ k[x1,…,xn,y].

-If the result is {1}, then f 2 .

-Example:

Consider the ideal I = h xy2 + 2y2, x4 – 2x2 + 1 i in k[x, y].

We want to determine if f = y – x2 + 1 lies in

Using lex order on k[x, y, z], we compute a

reduced Groebner basis of the following ideal:

= h xy2 + 2y2, x4 – 2x2 + 1, 1 – z (y – x2 + 1) i

The basis we obtain is {1}, so by Proposition 8

f 2 .

- In fact, (y – x2 + 1)3 2 I, but no lower power of f is in I.

Principal Ideals:

-If I = h f i, we can compute the radical of I as follows:

Proposition 9:

Let f 2 k[x1,…,xn] and I = h f i.

If is the factorization of f into a

product of distinct irreducible polynomials, then

= h f1f2 ··· fr i

Definition 10:

If f 2 k[x1,…,xn] is a polynomial, we define the

reduction of f, denoted fred , to be the polynomial such

that h fred i = , where I = h f i.

-A polynomial is said to be reduced, or square-free, if

f = fred .

-Section 4-2 ends with a formula for computing the radical

of a principal ideal => see pg. 181.

Sources Used- Ideals, Varieties, and Algorithms, by Cox, Little, O’Shea;

UTM Springer, 3rd Ed., 2007.

Thank You!

Stay tuned for the next lecture,

by ShinnYih Huang!