g=z g is cyclic then g is abelian.ccheng/courses/dummit_foote.pdf · 3.4.11 prove that if h is a...

3.1.36 Prove that if G/Z(G) is cyclic then G is abelian.

If G/Z(G) is cyclic with generator xZ(G). Every element in G/Z(G) can be written as xkzfor some k ∈ Z and z ∈ Z(G). Now, let g, h ∈ G, then g = xaz and h = xbw for z, w ∈ Z(G).We have gh = xazxbw = xa+bzw = xb+awz = xbwxaz = hg.

3.2.4 Show that if |G| = pq for some primes p, q (not necessarily distinct), theneither G is abelian or Z(G) = 1.

If |G| = pq and Z(G) 6= 1, we have |G/Z(G)| = p, q or 1. If |G/Z(G)| = 1, then G = Z(G), weare done. If |G/Z(G)| = p or q, G/Z(G) is cyclic. This implies G is abelian by Exercise 3.1.36.

3.2.8 Prove that if H and K are finite subgroups of G whose orders are relativelyprime then H ∩K = 1.

Since H ∩ K ≤ H and H ∩ K ≤ K and |HK| = |H||K||H∩K| . This implies that |H ∩ K| | |H|,

|H ∩K| | |K|. Since |H| and |K| are relatively prime, we have |H ∩K| = 1, hence H ∩K = 1.

3.2.10 Suppose H and K are subgroups of finite index in the (possible infinite)group G with |G : H| = m and |G : K| = n. Prove that l.c.m.(m,n) ≤ |G : H∩K| ≤ mn.Deduce that if m and n are relatively prime then |G : H ∩K| = |G : H| · |G : K|.

Note that H ∩ K is a subgroup of H and K respectively. So there are a, b ∈ Z such that

|H ∩K| divides a|H|+ b|K|. This implies that |H ∩K| divides a|G|mmn

+ b|G|mmn

. i.e. |H ∩K|

dividesa|G|n+ b|G|m

mn=an+ bm

mn|G| = g.c.d.(m,n)

mn|G| = 1

l.c.m.(m,n)|G|, which implies that

|H ∩K| ≤ |G|l.c.m.(m,n)

. Thus, we have l.c.m.(m,n) ≤ |G : H ∩K|.

On the other hand, since|H||K||H ∩K|

= |HK| ≤ G by Proposition 13. We have|H|

|H ∩K|≤ |G||K|

=

n. Thus, |G : H ∩K| ≤ |G||H|· |H||H ∩K|

= m · |H||H ∩K|

= mn. Therefore, we have

l.c.m(m,n) ≤ |G : H ∩K| ≤ mn

Now, if g.c.d.(m,n) = 1, l.c.m.(m,n) = mn. The above inequality becomes mn ≤ |G :H ∩K| ≤ mn. Hence, |G : H ∩K| = mn = |G : H||G : K|.

3.2.18 Let G be a finite group, let H be a subgroup of G and let N EG. Prove thatif |H| and |G : N | are relatively prime then H ≤ N .

Let f : G→ G/N with kernel ker f = N . Consider f |H : H → H/N , then |Imf |H | | |G/N | and|Imf |H | | |H| (image divides both domain and codomain). But |G/N and |H| are relativelyprime. So Im f |H = 1 which implies that H ≤ N = ker f .

3.2.19 Prove that if N is a normal subgroup of the finite group G and (|N |, |G :N |) = 1 then N is the unique subgroup of G of order |N |.

1

Let H ≤ G with |H| = |N |. Then (|H|, |G : N |) = (|N |, |G : N |) = 1. By 3.2.18 H ≤ N but|H| = |N |. So H = N .

3.2.20 If A is an abelian group with AEG and B is any subgroup of G prove thatA ∩B EAB.

For g ∈ AB, ∈ A ∩B. So x ∈ A and x ∈ B. Want to show that gxg−1 ∈ A ∩B. For g ∈ AB,write g = ab for a ∈ A and b ∈ B. Then (ab)x(ab)−1 = abxb−1a−1. Since x ∈ A, A E G andB ≤ G, we have bxb−1 ∈ A which implies that abxb−1a−1 ∈ A. Furthermore, since bxb−1 ∈ Aand A abelian, x ∈ B, we have a(bxb−1)a−1 = aa−1(bxb−1) = bxb−1 ∈ B. Thus, gxg−1 ∈ A∩Bfor all g ∈ AB, i.e. A ∩B EAB.

3.3.3 Prove that if H is a normal subgroup of G of prime index p then for all K ≤ Geither (i) K ≤ H or (ii) G = HK and |K : K ∩H| = 1.

Since H E G, we have H ≤ HK ≤ G. So p = |G : H| = |G : HK||HK : H|. Thus, either|HK : H| = 1 or p. In the first case, |HK : H| = 1 implies that HK = H, this gives K ≤ H.Otherwise, if |HK : H| = p, then |HK : H| = |K : K ∩H| = p.

3.3.7 Let M and N be normal subgroups of G such that G = MN . Prove thatG/M ∩N ∼= (G/M)× (G/N).

Consider

MN

M N

M ∩N

1

Define f : G → G/M × G/N by g 7→ (gM, gN). f is a homomorphism since f(g1g2) =(g1g2M, g1g2N) = (g1Mg2M, g1Ng2N) = (g1M, g1N)(g2M, g2N) = f(g1)f(g2). Also ker f =M ∩N as follows: for x ∈ ker f , f(x) = (xM, xN) = (M,N). Hence x ∈ M and x ∈ N . Onthe other hand, for x ∈M ∩N , x ∈M and x ∈ N , clearly x ∈ ker f since f(x) = (M,N). Theresult follows from the first isomorphism theorem.

3.4.1 Prove that G is an abelian simple group then G ∼= Zp for some prime p (donot assume G is a finite group).

Suppose that G is infinite. Let x 6= 1 ∈ G be an element of infinite order. Then 〈x〉 is asubgroup of G and actually it must be G since G is simple. So 〈x2〉 is a normal subgroup ofG = 〈x〉 (with index 2). This is a contradiction since G is normal.

Now, suppose that G is finite. If the order of G is prime, then G ∼= Zp. If |G| is not prime, say|G| = pm for some m. By Cauchy’s Theorem, G has an element of order p, but this implies

2

〈x〉 is a subgroup of G which is a contradiction. Hence, |G| ∼= Zp by Corollary 10.

In fact, every subgroup of an abelian group is normal. But G is simple, so G has no propersubgroup. So |G| must be prime, hence G ∼= Zp.

3.4.5 Prove that subgroups and quotient groups of a solvable groups are solvable.

Let G be a solvable group and H ≤ G a subgroup. Since G is solvable. There is a chain ofsubgroups

1 = G0 EG1 E · · ·EGn = G

such that Gi+1/Gi is abelian. Then it is clear that H ∩ Gi ≤ H ∩ Gi+1. And in factH ∩ Gi E H ∩ Gi+1. Since for x ∈ H ∩ Gi and g ∈ H ∩ Gi+1 we have gxg−1 ∈ H. Andgxg−1 ∈ Gi because Gi E Gi+1. Also it is clear that H ∩ Gi+1/H ∩ Gi is abelian. Thus H issolvable.

Similar way shows that G/H is solvable. Since Gi/H ≤ Gi+1/H and in fact Gi/H EGi+1/H.And (Gi+1/H)/(Gi/H) is abelian.

3.4.7 If G is a finite group and H E G prove that there is a composition series ofG, one of whose terms is H.

By Theorem 22 (Jordan-Holder Theorem), every finite group has a composition series. So wehave

1 = H0 EH1 E · · ·EHn = H

and

1 = K0/H EK1/H E · · ·EKm/H = G/H

By Lattice Isomorphism Theorem, this implies that

H = K0 EK1 E · · ·EKm = G

Therefore,

1 = H0 EH1 E · · ·EH EK1 E · · ·EKm = G

3.4.11 Prove that if H is a nontrivial normal subgroup of the solvable group Gthen there is a nontrivial subgroup A of H with AEG and A is abelian.

3.4.12 Prove (without using the Feit-Thompson Theorem) that the following areequivalent:(i) every group of odd order is solvable.(ii) the only simple groups of odd order are those of prime order.

(i) ⇒ (ii)

Let G be a simple group of odd order. Suppose that |G| = n = pm for some m 6= 1. If G isabelian, there exists an element x 6= 1 ∈ G of order p, i.e. 〈x〉 E G, a contradiction. If G isnon-abelian, since G is solvable and every finite group has a composition series. So we have

1 = H0 EH1 E · · ·EHn = G

By Exercise 1, the length of the composition series must be at least 2. So Hn−1 6= 1. But thisimplies that Hi−1 EG, a contradiction since G is simple. Therefore, the only simple groups of

3

odd order are those of prime order.

(ii) ⇒ (i)

3.5.9. Prove that the (unique) subgroup of order 4 in A4 is normal and is isomor-phic to V4.

As shown in the Figure 8 on page 111, the unique subgroup of order 4 in A4 is H =〈(12)(34), (13)(24)〉. Let σ = (12)(34) and τ = (13)(24). Since στ = (12)(34)(13)(24) =(14)(23) and all three elements σ, τ, στ have order 2. So H ∼= V4. Now, show that H is nor-mal. Only need to check on the generators σ, τ and only need to check that gσg−1 ∈ V4 andgτg−1 ∈ V4 for g all three cycles containing 3. i.e.

(123)(12)(34)(132) = (14)(23) ∈ V4

(123)(13)(24)(132) = (12)(34) ∈ V4

and

(134)(12)(34)(143) = (14)(23) ∈ V4

(134)(13)(24)(143) = (12)(34) ∈ V4

Thus, H EG.

3.5.10. Find a composition series for A4. Deduce that A4 is solvable.

Consider the chain of subgroups

1 ≤ 〈(12)(34)〉 ≤ V4 ≤ A4

Note that 〈(12)(34)〉 E V4 because of index 2. So each subgroup is a normal subgroup ofthe group on the right-hand side. And we just proved in previous exercise that V4 E A4.Furthermore, the composition factors are

S4/A4∼= Z2 A4/V4

∼= Z3 V4/〈(12)(34)〉 ∼= Z2 〈(12)(34)〉/1 ∼= Z2

are all simple. So this is a composition series, hence A4 is solvable.

4.1.1 Let G act on the set A. Prove that if a, b ∈ A and b = g · a for some g ∈ G,then Gb = gGag

−1 (Ga is the stabilizer of a). Deduce that if G acts transitively onA then the kernel of the action is ∩g∈GgGag−1.

For x ∈ Gb, x · b = b. So g−1xg · a = g−1xb = g−1b = a. Hence g−1xg ∈ Ga which implies thatx ∈ gGag−1. On the other hand, for x ∈ Ga we have gxg−1 · b = gxg−1ga = gx · a = g · a = b.Thus, Gb = gGag

−1.

Let K be the kernel of the action, i.e. K = {g ∈ G | g · a = a for all a ∈ A}. Show thatK = ∩g∈GgGag−1.

For x ∈ K, x · a = a. Since g−1xg · a = g−1g · a = a for all g ∈ G. We have g−1xg ∈ Ga for allg ∈ G. Hence, x ∈ gGag−1 for all g ∈ G. i.e. x ∈ ∩g∈GgGag−1.

On the other hand, let x ∈ ∩g∈GgGag−1. Then x ∈ gGag−1 for all g ∈ G. So for any

b ∈ A we have x · b = ghg−1 · b for some h ∈ Ga (i.e. h · a = a). But G acts transitively on A,4

i.e. g ·a = b. Thus, x·b = ghg−1 ·b = gh·a = g ·a = b. So x ∈ K. Therefore, K = ∩g∈GgGag−1.

4.1.2 Let G be a permutation group on the set A (i.e., G ≤ SA), let σ ∈ G and leta ∈ A. Prove that σGaσ

−1 = Gσ(a). Deduce that if G acts transitively on A then

∩σ∈GσGaσ−1 = 1.

Let x ∈ σGaσ−1. And let x = σgσ−1 for g ∈ Ga. Then x ·σ(a) = σgσ−1 ·σ(a) = σg · a = σ(a).Hence x ∈ Gσ(a). On the other hand, let x ∈ Gσ(a). So x · σ(a) = σ(a). We have

σ−1xσ · a = σ−1(σ(a)) = a. Hence, σ−1xσ ∈ Ga, i.e., x ∈ σGaσ−1 Therefore σGaσ−1 = Gσ(a).

Now, if G acts transitively on A, by previous exercise the kernel is ∩g∈GgGg−1. Let ϕ : G→ SAbe the homomorphism associated to the action. Since G ≤ SA, ϕ is injective, the kernel is 1.

4.1.3 Assume that G is an abelian, transitive subgroup of SA. Show that σ(a) 6= afor all σ ∈ G−{1} and all a ∈ A. Deduce that |G| = |A|. [use the preceding exercise.]

Suppose that σ(a) = a for some a ∈ A and some σ ∈ G. This implies that Ga 6= 1 hence⋂σ∈G σGaσ

−1 6= 1, a contradiction to the result in preceding exercise.

For a fixed a ∈ A, and any b ∈ A, there is σ ∈ G such that σ(a) = b. To show this σ is unique,suppose there is τ ∈ G such that σ(a) = τ(a) = b. Then τ−1σ(a) = a, so τ−1σ ∈ Ga = 1.Thus, σ = τ . Hence we have an injective map ϕa : A → G sending a to σ. Hence |A| ≤ |G|.Similarly, if we define ψ : G→ A by ψ(σ) = σ ·a. Then ψ is injective since σ(a) = τ(a) impliesthat τ−1σ ∈ Ga = 1 which again implies thatσ = τ . Therefore, |G| = |A|.

4.2.8 Prove that if H has finite set index n then there is a normal subgroup K ofG with K ≤ H and |G : K| ≤ n!.

Let G act by left multiplication on the set A of left cosets (i.e. G/H). And let πH be theassociated permutation representation afforded by this action

πH : G→ SA

By Theorem 3, K = kerπH =⋂x∈G xHx

−1 and K is the normal subgroup of G containedin H. By first isomorphism theorem, G/K ∼= ImπH ≤ SA. Since H has index n, |SA| = n!.

Hence, we have [G : K] =

∣∣∣∣GK∣∣∣∣ ≤ |SA| = n!.

4.2.9 Prove that if p is a prime and G is a group of order pα for some α ∈ Z+, thenevery subgroup of index p is normal in G. Deduce that every group of order p2

has a normal subgroup of order p.

Since |G| = pα. Since p is the smallest prime dividing |G|. By Corollary 5, any subgroup ofindex p is normal. In particular, let α = 2, then |G| = p2. By Cauchy’s Theorem, G has anelement x of order p. So H = 〈x〉 and [G : H] = p. H has index p, hence H is normal.

4.2.11 Let G be a finite group and let π : G→ SG be the left regular representation.Prove that if x is an element of G of order n and |G| = mn, then π(x) is a productof m n-cycles. Deduce that π(x) is an odd permutation if and only if |x| is even

5

and |G||x| is odd.

The permutation representation π : G → SG is defined by π(x) = σx and for each g ∈ G, thepermutation of G is σx(g) = x · g.

Let H = 〈x〉. Since H is cyclic. For each g ∈ G, the stabilizer of g in H is 1 (GH(x) = 1), sothe action is faithful (i.e. the kernel of the action is 1). So each orbit of H in G has order nand is the cycle containing g in the decomposition of π(x). i.e. for each g,

Hg = (g xg x2g · · · xn−1g)

Since |G| = mn, G = Hg1 ∪ · · · ∪Hgm. Hence the image of x of left regular representation is a

product of m of n-cycles. Now, if |x| is even and |G||x| is odd, by Proposition 25 π(x) is an odd

permutation. Conversely, if π(x) is odd and |x| is odd, the number of cycles of even length inthe decomposition is zero, which is even. By Proposition 25 again, π(x) is even, a contradiction.

4.2.12 Let G and π be as in the preceding exercise. Prove that if π(G) contains anodd permutation then G has a subgroup of index 2. [use Exercise 3 in Section 3.3]

4.2.13 Prove that if |G| = 2k where k is odd then G has a subgroup of index 2. [UseCauchy’s Theorem to produce an element of order 2 and then use the precedingtwo exercises.]

By Cauchy’s Theorem, there is an element x ∈ G of order 2, so |x| = 2. By Exercise 11, π(x)is an odd permutation, then by Exercise 12 G has a subgroup of index 2.

4.2.14 Let G be a finite group of composite order n with the property that G hasa subgroup of order k for each positive integer k dividing n. Prove that G is notsimple.

Let p be the smallest prime dividing n. So n = pm for some m ∈ Z. By assumption, G has asubgroup H of order m. Then H has index p, hence is normal by Corollary 5. Therefore G isnot simple.

4.3.4 Prove that if S ⊆ G and g ∈ G then gNG(S)g−1 = NG(gSg−1) and gCG(S)g−1 =CG(gSg−1).

(1) gNG(S)g−1 = NG(gSg−1)

(⊆) Let gag−1 ∈ NG(gSg−1), for a ∈ NG(S) = {x ∈ G | xgSg−1x−1 = gSg−1}. We have

(gag−1)gSg−1(gag−1)−1 = gag−1gSg−1ga−1g−1 = gaSa−1g−1 = gSg−1

since aSa−1 = S. So gag−1 ∈ NG(gSg−1).

(⊇) Let x ∈ NG(gSg−1). Then xgSg−1x−1 = gSg−1. This implies that g−1xgSg−1x−1g = S,i.e. (g−1xg)S(g−1xg)−1 = S. Thus g−1xg ∈ NG(S), so x ∈ gNG(S)g−1.

(2) gCG(S)g−1 = CG(gSg−1)

6

(⊆) Let gag−1 ∈ gCG(S)g−1, for some a ∈ CG(S) = {x ∈ G | xsx−1 = s for all x ∈ S}. Thenfor any s ∈ S we have

(gag−1)gsg−1(gag−1)−1 = gag−1gsg−1ga−1g−1 = gasa−1g−1 = gsg−1

since asa−1 = s. Thus, gag−1 ∈ NG(gSg−1).

(⊇) Let x ∈ CG(gSg−1). Then xgsg−1x−1 = gsg−1 for all s ∈ S. This implies thatg−1xgsg−1x−1g = s, i.e. (g−1xg)s(g−1xg)−1 = s for all s ∈ S. Thus g−1xg ∈ CG(S), sox ∈ gCG(S)g−1.

4.3.5 If the center of G is of index n, prove that every conjugacy class has at mostn elements.

By Orbit-Stabilizer Theorem [G : Gg][Gg : Z(G)] = [G : Z(G)] = n, we see that [G : Gg] di-vides n. When the action is conjugation, the number of conjugates of g is precisely |G : CG(g)|.Thus, |G : CG(g)| divides n, which implies that every conjugacy class has at most n elements.

4.3.6 Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use thefact that 〈g〉 ≤ CG(g) for all g ∈ G to show that there is at most one possible classequation for G. [Use Exercise 36, Section 3.1]

Center Z(G) is a subgroup of G. Since G is non-abelian, |Z(G)| can only be 1, 3, 5. Suppose|Z(G)| = 5. By Orbit- Stabilizer Theorem [G : Gg][Gg : Z(G)] = [G : Z(G)] = 15/5 = 3. Sothe number of conjugates of any non-center element g ∈ G, i.e. |G : CG(g)|, must divides 3.And it must be 3, otherwise g would be in the center. However, by Class Equation,

|G| = |Z(G)|+r∑i=1

|G : CG(gi)|

we have∑r

i=1 |G : CG(gi) = 15− 5 = 10 which is not divisible by 3, a contradiction.

A quicker way: suppose |Z(G)| is prime, by Exercise 36 in Section 3.1, G/Z(G) is cyclic, so Gis abelian, a contradiction. Hence |Z(G)| = 1.

Now, since 〈g〉 ≤ CG(g) for all g ∈ G (Exercise 6 in Section 2.2). Let g ∈ G be any non-identityelement in G. Since Z(G) = 1 � 〈g〉 ≤ CG(g) � G. |CG(g)| can only be 3 or 5. Since |G| = 15,by Lagrange Theorem, we must have 〈g〉 = CG(g). So the number of conjugates in everyconjugacy class is 3 or 5. By Class Equation,

∑ri=1 |G : CG(gi)| = 15− 1 = 14. Thus, we have

the relation 3a + 5b = 14 where a denotes the number of conjugacy classes of order 3 and bdenotes the number of conjugacy classes of order 5. The only one solution is a = 3 and b = 1.Therefore, there is at most one possible class equation for G. i.e. |G| = 15 = 1 + 3 + 3 + 3 + 5.

4.3.13 Find all finite groups which have exactly two conjugacy classes.

Every element in an abelian group is a conjugacy of its own. So abelian groups having exactlytwo conjugacy classes must have two element hence is isomorphic to Z2. Now, let G be non-abelian group. Note that the identity in G forms a conjugacy class of its own. So |Z(G)| = 1.Since we assume that G has exactly two conjugacy classes, there is an element g 6∈ Z(G) suchthat

|G| = 1 + |G : CG(g)|7

by class equation. And |G : CG(g)| is the number of elements in the other conjugacy class. Sowe have

|G| − 1 = |G : CG(g)|

this implies that the number of conjugates in the other conjugacy class divides |G| − 1. Butthis happens only when |G| = 2. Hence G is isomorphic to Z2.

4.3.23 Recall that a proper subgroup M of G is called maximal if wheneverM ≤ H ≤ G, either H = M or H = G. Prove that if M is a maximal subgroupof G then either NG(M) = M or NG(M) = G. Deduce that if M is a maximal sub-group of G that is not normal in G then the number of non-identity elements of Gthat are contained in conjugates of M is at most (|M | − 1)|G : M |.

Since M ≤ NG(M) ≤ G by Exercise 6 in Section 2.2 and the fact that NG(M) is a subgroupof G. But M is a maximal subgroup of G, so it is clear that either NG(M) = M or NG(M) = G.

If M is not normal, i.e. NG(M) � G, then M = NG(M). So we have∣∣∣∣ G

NG(M)

∣∣∣∣ =

∣∣∣∣ GM∣∣∣∣

By Proposition 6, the number of conjugates of M , gMg−1, is [G : NG(M)]. And each conju-gate of M has cardinality |gMg−1| = |M |. The largest number of non-identity elements in theconjugates of M occurs when these conjugates intersect trivially. Hence, the number of non-identity elements of G that are contained in the conjugates of M is at most (|M | − 1)|G : M |.

4.3.24 Assume H is a proper subgroup of the finite group G. Prove that G 6=∪g∈GgHg−1, i.e. G is not the union of the conjugates of any proper subgroup. [PutH in some maximal subgroup and use the preceding exercise.]

Let M be a maximal subgroup of G containing H. If M is normal in G, we have ∪g∈GgHg−1 ⊆∪g∈GgMg−1 = M 6= G. If M is not normal in G, we still have ∪g∈GgHg−1 ⊆ ∪g∈GgMg−1. Byprevious exercise, we know that ∪g∈GgMg−1 contains at most (|M | − 1)|G : M | non-identityelements of G. Hence,

|⋃g∈G

gHg−1| ≤ (|M | − 1)|G : M | = |G| − |G||M |

+ 1 < |G| <∞

since|G||M |

> 1.

4.3.27 Let g1, g2, · · · , gr be representatives of conjugacy classes of the finite groupG and assume these elements pairwise commute. Prove that G is abelian.

If g1, g2, · · · , gr are pairwise commute, then gigj = gjgi which implies that gjgig−1j = gi. Thus,

for each i, we have gj ∈ CG(gi) for all j = 1, · · · , r, so |C(gi)| ≥ r. By Class Equation, |G| isthe sum of all distinct conjugacy classes. Each conjugacy class has |G : CG(gi)| elements, andthere are r of them. Hence

|G| =r∑i=1

|G : CG(gi)| =r∑i=1

|G||CG(gi)|

≤r∑r=1

|G|r

= |G|

8

This implies that |CG(gi)| = r. i.e. CG(gi) = {g1, g2, · · · , gr}. We claim that G = CG(gi).G ⊇ CG(gi) is clear. And every element x ∈ G is in the centralizer of itself, hence G ⊆ CG(gi).Therefore, G = CG(gi), hence G is abelian.

4.3.29 Let p be a prime and let G be a group of order pα. Prove that G has asubgroup of order pβ, for every β with 0 ≤ β ≤ α. [Use Theorem 8 and inductionon α.]

Suppose |G| = p, then G has a subgroup order 1 and p, so this is clear.

Suppose any group of order pa has a subgroup of order pb for every b with 0 ≤ b ≤ a. Let Gbe a group of order |G| = pa+1, want to prove that G has a subgroup of order pb for every0 ≤ b ≤ a + 1. By Theorem 8, Z(G) 6= 1. Then H = Z(G) is a normal subgroup of G. Forany subgroup A of G containing H, we have the following correspondence:

G/H //

��

G

��A/H //

��

A

��1 // H

Since |G/H| = pc for some c with 0 ≤ c ≤ b. By induction hypothesis, G/H has subgroup oforder pb for every b with 0 ≤ b ≤ c. Hence by Lattice Isomorphism Theorem, G has a subgroupof order pb for every 0 ≤ b ≤ n+ 1.

4.3.30 If G is a group of odd order, prove for any nonidentity element x ∈ G thatx and x−1 are not conjugate in G.

First, note that x 6= x−1. since if x = x−1, then x2 = 1 and |x| = 2, a contradiction since |G|is even and cannot have an element of even order.

Suppose x and x−1 are conjugate, want to show that CG(x) has even order, a contradiction.By definition, CG(x) = {g ∈ G | gxg−1 = x}.

4.4.13 Let G be a group of order 203. Prove that if H is a normal subgroup oforder 7 in G then H ≤ Z(G). Deduce that G is abelian in this case.

|G| = 203 = 7 · 29. Since H is normal, by Corollary 13 |G/CG(H) is isomorphic to a subgroupof Aut(H). And |Aut(H)| = 6, so |G/CG(H)| must divide 6. Since |G| = 7 · 29, the onlypossibility is |G/CG(H)| = 1. Hence G = CG(H). i.e. H ≤ Z(G). So |H| divides |Z(G)|.Since |P | = 7, we have |Z(G)| = 7 or 7 · 29. This implies that |G/Z(G)| = 1 or 29, henceG/Z(G) is cyclic. Therefore, G is abelian.

4.5.13 Prove that a group of order 56 has a normal Sylow p-subgroup for someprime p dividing its order.

9

|G| = 56 = 23 · 7. Since

n2 ≡ 1 (mod 2), n2 | 7 so n2 = 1, 7

n7 ≡ 1 (mod 7), n2 | 8 so n7 = 1, 8

If n7 = 1, the Sylow 7-subgroup is normal in G. Then we are done. If n7 = 8, G has 8 · 6 = 48non-identity elements of order 8 in these Sylow 7-subgroups. The rest 56−48 = 8 are preciselyelements in Sylow 2-subgroup. This implies that G has a unique Sylow 2-subgroup.


|G| = 56 = 23 · 3 · 13. Since

n3 ≡ 1 (mod 3), n3 | 8 · 13 so n3 = 1, 4, 13, 52

n13 ≡ 1 (mod 13), n13 | 8 · 13 so n13 = 1

So Syl13 is normal in G.


|G| = 351 = 33 · 13. Since

n3 ≡ 1 (mod 3), n3 | 13 so n3 = 1, 13

n13 ≡ 1 (mod 13), n13 | 33 so n13 = 1, 27

If n13 = 1, the Sylow 13-subgroup is normal in G. Then we are done. If n13 = 27, Ghas 27 · 12 = 324 non-identity elements of order 13 in these Sylow 13-subgroups. The rest351− 324 = 27 are precisely elements in Sylow 3-subgroup. This implies that G has a uniqueSylow 3-subgroup.

4.5.17 Prove that if |G| = 105 then G has a normal Sylow 5-subgroup and a normalSylow 7-subgroup.

(The solution basically follow the proof of the example on page 143 where |G| = 30).|G| = 105 = 3 · 5 · 7. Since

n5 ≡ 1 (mod 5), n5 | 3 · 7 so n5 = 1, 21

n7 ≡ 1 (mod 7), n7 | 3 · 5 so n7 = 1, 15

Let P = Syl5 and Q = Syl7. If n5 = 1 and n7 = 1, then we are done. If either n5 = 1 or n7 = 1,i.e., either P or Q is normal. Let H = PQ. Then H ≤ G by Corollary 15, and both P and Qare characteristic subgroups of H by statement (2) on page 135 that every subgroup of a cyclicgroup is characteristic (H is cyclic by Example on page 143), or it is easy to see that n5 = 1 andn7 = 1, hence PEH and QEH. So P and Q are characteristic in H. Since HEG by Corollary5. By exercise 4.4.8 or the statement (3) on page 135, both P and Q are normal subgroup of G.

Now, assume that neither Sylow subgroup is normal, i.e., n5 = 21 and n7 = 15. Then G has21 · 4 = 84 elements of order 5 and 15 · 6 = 90 elements of order 7. So total elements would be84 + 90 = 174 > 105, a contradiction. Therefore, one of P or Q, hence both, must be normalin G.

10

n7 ≡ 1 (mod 7), n7 | 32 · 5 so n7 = 1, 15

If Sylow 3-subgroup, P = Syl3, is normal, G/CG(P ) is isomorphic to a subgroup of Aut(P ) byCorollary 13. Since |Aut(P )| = 6 or 48. And P is a square of a prime, P is an abelian group,hence P ≤ CG(P ) by Exercise 6(b) in Section 2.2. It follows that |CG(P )| is divisible by 9,which implies that |G/CG(P )| = 1, 5, 7, 35. Together these imply |G/CG(P )| = 1. Hence,G = CG(P ). i.e. P ≤ Z(G). So |G/Z(G)| =1, 5, 7 or 35. If |G/Z(G)| = 1, then G is abelian.If |G/Z(G)| = 5 or 7, G/Z(G) is cyclic hence G is abelian. If |G/Z(G)| = 35, it is easy to seethat n5 = 1 and n7 = 1 then G/Z(G) is cyclic by Example on page 143. Therefore, G is abelian.

4.5.32 Let P be a Sylow p-subgroup of H and let H be a subgroup of K. Let P EHand HEK, prove that P is normal in K. Deduce that if P ∈ Sylp(G) and H = NG(P ),then NG(H) = H.

If P is a Sylow p subgroup of H and assume that P E H, then P is characteristic in H byCorollary 20. Since H is normal in K, by the statement (3) P is normal in K. Hence, ifP ∈ Sylp(G) and H = NG(P ) (i.e., P EH), then by above argument P EK. This implies that

K = NK(P ) ≤ NG(P ) = H

Thus, H = K. Since H EK (i.e. K = NG(H)), we have H = NG(P )EK which implies thatNG(H) = NG(NG(P )) = K = NG(P ) = H, as desired. (the solutions was not correct, willcome back later!)

4.5.33 Let P be a normal Sylow p-subgroup of G and let H be any subgroup of G.Prove that P ∩H is the unique Sylow p-subgroup of H.

It suffices to show that P ∩H is Sylow p subgroup of H. Then by Exercise 24 in Section 3.1,P ∩H EH.

Let |G| = pαm with p - m. Then |H| = pβm where 0 ≤ β ≤ α and |P | = pα. Since P E G,HP ≤ G. By Proposition 13 in Section 3.2, we have

|H||H ∩ P |

=|HP ||P |

divides|G||P |

=pαm

pα= m

Hence |P ∩H| = pβ. Thus, P ∩H is the unique Sylow p-subgroup of H.

Alternatively, since P ∩H EH and P ∩H is a p-subgroup of H. Suppose that P ∩H is not aSylow p-subgroup of H. There is an element x ∈ H\P of order p. But this is not possible sinceP is the unique Sylow p-subgroup of G, that contains all elements of order p, a contradiction!

4.5.34 Let P ∈ Sylp(G) and assume N EG. Use the conjugacy part of Sylow’s Theo-rem to prove that P ∩N is a Sylow p-subgroup of N . Deduce that PN/N is a Sylowp-subgroup of G/N

Suppose Q is a Sylow p-subgroup of N . Q is also a p-subgroup of G, by Theorem 18(2), thereexists g ∈ G such that Q ⊆ gPg−1. Since Q ⊆ N , we have Q ⊆ gPg−1 ∩N . But N is normal.so Q ⊆ gPg−1 ∩ gNg−1 = g(P ∩ N)g−1. This implies that g−1Qg ⊆ P ∩ N . Now, g−1Qgis a Sylow p-subgroup of N since |g−1Qg| = |Q|. The group P ∩ N is a p-subgroup of N , sog−1Qg = P ∩N . This shows that P ∩N is a Sylow p-subgroup of N .

12

Let |P | = pa and |Q| = pb, so a Sylow p-subgroup of G/N has order pa−b. By 2nd IsomorphismTheorem,

PN

N=

P

P ∩NSo |PN/N | = |P |/|P ∩N | = pa/pb = pa−b. Thus, PN/N is a Sylow p-subgroup of G/N .

G/N G

PN/N PN

1 N

10.1.5 For any left ideal I of R define

IM = {∑

finite

aimi | ai ∈ I,mi ∈M}

to be the collection of all finite sums of elements of the form am where a ∈ I andm ∈M . Prove that IM is a submodule of M .First, prove that IM is a subgroup. Note that the identity exists since 0 ∈M and take m = 0,then 0m ∈ IM . For x, y ∈ IM , x =

∑aimi, y =

∑bimi, we have x+y =

∑(ai+bi)mi ∈ IM .

Finally, let y =∑−aimi, then x+ y =

∑(ai − ai)mi = 0, so y is the inverse of x. Thus, IM

is a subgroup.

IM is closed under the action of ring elements: let r ∈ R, rx = r∑aimi =

∑raimi ∈ IM

since rai ∈ I. Hence, IM is a submodule.

10.1.7 Let N1 ⊆ N2 ⊆ · · · be an ascending chain of submodules of M . Prove that∪∞i=1Ni is a submodule of M .

We prove by Proposition 1. First, ∪∞i=1Ni 6= φ since Ni 6= φ for each i. Furthermore, for r ∈ Rand x, y ∈ ∪∞i=1Ni, we have x ∈ Nm and y ∈ Nk for some m, k. Without loss of generality,assume m ≤ k. Then x ∈ Nk and x+ ry ∈ Nk hence x ∈ ∪∞i=1Ni.

10.1.9 If N is a submodule of M , the annihilator of N in R is defined to be{r ∈ R | rn = 0 for all n ∈ N}. Prove that the annihilator of N in R is a 2-sidedideal of R.

(1) Prove that Ann(N) is a subring.Note that 0 ∈ N , so Ann(N) is nonempty. For x, y ∈ Ann(N), xn = 0 and yn = 0 forall n ∈ N . So (x − y)n = xn − yn = 0 for all n ∈ N . Hence, x − y ∈ Ann(N). Now,(xy)n = x(yn) = x · 0 = 0. Hence Ann(N) is a subring.

(2) Prove that Ann(N) is closed under right and left multiplication by element of r ∈ R.Let x ∈ Ann(N), xn = 0 for all n ∈ N . (rx)n = r(xn) = r · 0 = 0, hence rx ∈ Ann(N).Similarly, (xr)n = x(rn) = x · 0 = 0.

Therefore, Ann(N) is a 2-sided ideal of R.

13

10.1.19 Let F = R, let V = R2 and let T be the linear transformation from V to Vwhich is projection onto the y-axis. Show that V , 0, the x-axis and the y-axis arethe only F [x]-submodules for this T .

Since the F [x]-submodules of V are precisely the T -invariant subspace of V . We see thatT (0) = 0 ⊆ V , T (V ) ⊆ V , T (x-axis) = 0 ⊆ x-axis and T (y-axis) = y-axis ⊆ y-axis. Hence,these are F [x]-modules.

In general, for L be any 1-dimensional subspace on R2, neither x-axis nor y-axis. It is easy tosee that T (L) 6⊆ L. Hence the only F [x]-submodules for T are 0, V , x-axis and y-axis.

10.2.4 Let A be any Z-module, let a be any element of A and let n be a positiveinteger. Prove that the map ϕa : Z/nZ → A given by ϕa(k) = ka is a well-defineZ-module homomorphism if and only if na = 0. Prove that HomZ(Z/nZ, A) ∼= An,where An = {a ∈ A | na = 0} (so An is the annihilator in A of the ideal (n) of Z).

Since ka = ma if and only if (k − m)a = 0 if and only if a ∈ nZ if and only if na = 0.This proves that ϕa is well-defined if and only if na = 0. For x, y ∈ Z/nZ, r ∈ Z, we haveϕa(x + y) = (x + y)a = xa+ ya = ϕa(x) + ϕa(y). And ϕa(rx) = rxa = rϕa(x). Hence, ϕa isZ-module homomorphism.

Define f : Hom(Z/nZ, A)→ An by f(ϕa) = ϕa(k) = ka for k ∈ Z/nZ. First, f is well-definedsince ϕa is. For ϕa, ϕb ∈ Hom (Z/nZ, A), we have f(ϕa +ϕb)(k) = ϕa(k) +ϕb(k) = ka+ kb =ϕa(k) + ϕb(k) = f(ϕa) + f(ϕb). Furthermore, let r ∈ R, then we have f(rϕa) = rϕa(k) =rka = r(ka) = rϕa(k) = rf(ϕa). So f is a group homomorphism. To show that f is surjective,for a ∈ An with na = 0, let k = 1, then we have ϕa(1) = 1 · a = a. Suppose ϕa(k) = ϕb(k),then ka = kb which implies that a = b (by cancellation property since A is an abelian group).i.e. ϕa = ϕb. Therefore, f is an isomorphism.

10.2.6 Prove that HomZ(Z/nZ,Z/mZ) ∼= Z/(n,m)Z.

The homomorphism between them will be determined by where the generator in Z/nZ goes.Let ϕ : Z/nZ→ Z/mZ be a Z-module homomorphism, then ϕ(0) = 0. Hence,

0 = ϕ(0) = ϕ(1 + 1 + · · ·+ 1) = ϕ(1) + · · ·+ ϕ(1) = nϕ(1)

To have a homomorphism, we must have nϕ(1) ∈ mZ and this is true if and only ϕ(1) =m

gcd(m,n), or its multiples. These elements form a group generated by

m

gcd(m,n)and is cyclic

of order gcd(m,n). Therefore, HomZ(Z/nZ,Z/mZ) ∼= Z/(m,n)Z.

10.2.7 Let z be a fixed element of the center of R. Prove that the map m 7→ zm isan R-module homomorphism from M to itself. Show that for a commutative ringR the map from R to EndR(M) given by r 7→ rI is a ring homomorphism (where Iis the identity endomorphism).

Let ϕ : R × M → M be defined by ϕ(m) = zm. For x, y ∈ M and r ∈ R, we haveϕ(rx + y) = z(rx + y) = z(rx + y) = zrx + zy = rzx + zy = rϕ(x) + ϕ(y). Hence, ϕ isan R-module homomorphism.

14

Recall that EndR(M) =HomR(M,M). Let ψ : R →EndR(M) be defined by ψ(r) = rI. Forr, s ∈ R, we see that ψ(r+s) = (r+s)I = rI+sI = ψ(r)+ψ(s). Also, ψ(rs) = rsI = rI ·sI =ψ(r)ψ(s). Therefore, ψ is a R homomorphism.

10.2.11 Let A1, A2, · · · , An be R-modules and let Bi be submodule of Ai for eachi = 1, 2, · · · , n. Prove that (A1 × · · · ×An)/B1 × · · · ×Bn) ∼= (A1/B1)× · · · × (An/Bn).

Define a map ϕ : A1 × · · · × An −→ (A1/B1) × · · · × (An/Bn) by (a1, · · · , an) 7→ (a1 +B1, · · · , an + Bn) for ai ∈ Ai, i = 1, 2, · · ·n. Since on each component Ai −→ Ai/Bi is anatural projection hence is an R-module homomorphism by Proposition 3. Therefore, ϕ is anR-module homomorphism.

Prove that ϕ is a bijection.

First, note that ϕ is surjective since each component ϕi : Ai → Ai/Bi is a natural projectionhence surjective (well, I thought it is clear that the product of a surjective map is surjective).It suffices to prove that kerϕ = B1 × · · · ×Bn.

(1) kerϕ ⊆ B1 × · · · ×BnIf a = (a1, · · · , an) ∈ kerϕ, then ϕ(a) = (a1 + B1, · · · , an + Bn) = (0, · · · , 0) ∈ (A1/B1) ×· · · × (An/Bn). So we have ai + Bi = 0 hence ai ∈ Bi for each i. Thus, a = (a1, · · · , an) ∈B1 × · · · ×Bn.

(2) kerϕ ⊇ B1 × · · · ×BnFor b = (b1, · · · , bn) ∈ B1× · · · ×Bn, we have ϕ(b) = ϕ(b1, · · · , bn) = (b1 +B1, · · · , bn +Bn) =(B1, · · · , Bn) = (0, · · · , 0) in (A1/B1)× · · · × (An/Bn). Hence, b ∈ kerϕ.

Therefore, kerϕ = B1 × · · · × Bn. By first isomorphism theorem for modules (Theorem 4 inSection 2), (A1 × · · · ×An)/B1 × · · · ×Bn) ∼= (A1/B1)× · · · × (An/Bn).

10.2.13 Let I be a nilpotent ideal in a commutative ring R, let M and N be R-modules and let ϕ : M → N be an R-module homomorphism. Show that if theinduced map ϕ : M/IM → N/IN is surjective, then ϕ is surjective.

Mϕ //

p

��

N

q

��M/IM

ϕ // N/IN

Given n ∈ N , want to show that there is m ∈ M such that ϕ(m) = n by showing thatq(ϕ(m)− n) = 0 ∈ N/IN . Indeed, by commutativity of the diagram we have

q(ϕ(m)− n) = q · ϕ(m)− q(n)

= ϕ · p(m)− q(n)

= ϕ(m+ IM)− q(n)

= n+ IN − (n+ IN)

= 015

Thus, ϕ(m)−n ∈ IN . This is true for all n ∈ N , hence ϕ(M)−N = IN . i.e. N = ϕ(M)+IN .This implies that

N = ϕ(M) + I(ϕ(M) + IN)

= ϕ(M) + Iϕ(M) + I2N

= ϕ(M) + I2N

Inductively, we obtain N = ϕ(M) + IrN . Since I is nilpotent, Ik = 0 for some k ≥ 1. There-fore, N = ϕ(M) and ϕ is surjective.

10.3.2 Assume R is commutative. Prove that Rn ∼= Rm if and only if n = m, i.e.two free R-modules of finite rank are isomorphic if and only if they have the samerank. [Apply Exercise 12 of Section 2 with I a maximal ideal of R. You mayassume that F is a field, then Fn ∼= Fm if and only if n = m.]

By Exercise 12 in Section 10.2, we have Rn/IRn ∼= I/IR× · · · × I/IR. If I ⊂ R is a maximalideal, then R/I is a field. Also, IR = I. Hence we have

Rn/IRn ∼= R/I × · · · ×R/ILet F = R/I we obtain

Rn/IRn ∼= F × · · · × F ∼= Fn

Similarly, we have

Rm/IRm ∼= R/I × · · · ×R/I ∼= Fm

Therefore,

m = n⇔ Fm = Fn ⇔ Rm/IRm = Rn/IRn ⇔ Rm = Rn

10.3.7 Let N be a submodule of M . Prove that if both M/N and N are finitelygenerated then so is M .

Let f : M → M/N be a projection homomorphism defined by m 7→ m+N . Let {y1, · · · , yn}be a basis for N and {x1, · · · , xm} be a basis for M/N . For m ∈M , its image m in M/N canbe written as m =

∑i aixi. This implies that m+N =

∑i ai(xi +N) =

∑i aixi +N . Thus,

m −∑

i aixi ∈ N , which implies that m −∑

i aixi =∑

j bjyj . Thus, m =∑

i aixi +∑

j bjyj .

Note that M ∼= N ⊕M/N , therefore M is finitely generated.

10.3.9 (a) Show that M is irreducible if and only if M 6= 0 and M is cyclic modulewith any nonzero element as generator.

(⇒)Suppose M is not cyclic, let M = 〈a1, · · · , an〉 for n > 1. i.e. M is finitely generated bya1, · · · , an with ai 6= 0 for all i. Then M has a submodule N 6= 0, and N (M . This is becausewe can let A = {a1, · · · , ak} for k < n, then N = RA = {r1a1, · · · , rkak | ri ∈ R, ai ∈ A} is asubmodule. This proved that M is cyclic.

Now, wuppose M is irreducible, pick a nonzero element m ∈ M , then 0 6= Rm ⊂ M . ClearlyRm is a submodule. Hence Rm = M since M is irreducible. Therefore, M is a cyclic module.

(⇐)Suppose M is a cyclic module with any nonzero element as generator. i.e. M = Ra for anya ∈M . Want to prove that M is irreducible. Assume that M has a proper submodule N , then

16

0 6= N ⊂M . Pick a nonzero element n in N , so we have 0 6= Rn ⊂ N ⊂M . But Rn = M , soN = M hence M is irreducible.

Now, determine all the irreducible Z-modules.Let M be an irreducible Z-module. Note that Z-module can be seen as an abelian group, soby the structure of an abelian group, M has torsion part and free part. Since M is irreduciblehence does not have any proper submodule. But Tor(M) is a submodule of M . So Tor(M) iseither 0 or Tor(M) = M .

Suppose Tor(M)=0, take m ∈ M and m 6= 0. We have am 6= bm for a, b ∈ Z, a 6= b. So wehave 0 6= 2Zm ( Zm ⊂M . Contradiction to the fact that M is irreducible.

On the other hand, suppose Tor(M) = M , take m ∈M and m 6= 0. By what we just proved in(⇒), M = Zm. Now, since Z is a PID, so Ann(m) = (k) for some k ∈ Z. Then M ∼= Z/kZ. Weclaim that k is a prime. If not, k = qr for some q, r ∈ Z, then 0 6= Z/qZ ( Z/kZ ⊂ M . Con-tradiction to the fact that M is irreducible. So k is a prime. Hence we proved that M ∼= Z/pZsome some prime p.

Finally, if p is a prime, by what we proved in (⇐), Z/pZ is irreducible.

10.3.10 Assume R is commutative. Show that an R-module M is irreducible if andonly if M is isomorphic (as a R-module) to R/I where I is a maximal ideal of R.

(⇒)Suppose M is irreducible. Define ϕ : R → M by r 7→ rm for m 6= 0, by previous exercise, Mis cylic, i.e. M = Rm for some m 6= 0. So ϕ is actually surjective. Prove that kerϕ =Ann(m).For r ∈ kerϕ, ϕ(r) = rm = 0 implies that r ∈Ann(m). The converse is also clear. Thus wehave R/I ∼= M where I = kerϕ =Ann(M).

Now, prove that I is a maximal ideal. First, I is an ideal since for any s ∈ R, s(rm) = s ·0 = 0.So SI ⊆ I. To show that I is maximal, suppose I is not a maximal ideal, there is an ideal Jsuch that I ( J ( R.

R/I ∼= M R

J/I J

1 I

Then M ∼= R/I has a submodule J/I which contradicts to the irreducibility of M . Hence, Imust be a maximal ideal.

(⇐)Suppose M ∼= R/I for a maximal ideal I of R. If M is not irreducible, there is a submoduleN such that N ( M . By Lattice Theorem again, there is an ideal J with I ( J ( R. Thisimplies that J is not maximal.

17

10.3.11 If M1 and M2 irreducible R-modules, prove that any nonzero R-modulehomomorphism from M1 to M2 is an isomorphism.

Let ϕ : M1 −→M2 be a nonzero R-module homomorphism. Want to prove that ϕ is surectiveand injective hence is a bijection. First, note that kerϕ is a submodule of M1 and im ϕ is asubmodule of M2 by 10.2.1 in last homework assignment. But M1 and M2 are irreducible, sokerϕ = 0 (note that kerϕ can not be M , or it would imply that ϕ is a zero map which contra-dicts to the assumption), so ϕ is injective. On the other hand, we obtained that im ϕ = M2

(note that im ϕ 6= 0, if it were, ϕ is a zero map which again contradicts to the assumption).This implies that ϕ is surjective. Therefore, ϕ is a bijection, hence is an isomorphism.

Now, if M is irreducible, prove that EndR(M) is a division ring.

Let ϕ : M −→M be a nonzero R-module homomorphism. So ϕ ∈ EndR(M). But we have justproved that ϕ is a bijection in this case, which implies that ϕ has an inverse. In other words,every nonzero element in EndR(M) has a multiplicative inverse. Note that the “multiplication”in EndR(M) is the composition. i.e. there exists ψ ∈ EndR(M) such that ϕ ◦ψ = ψ ◦ϕ = 1M .

Therefore, EndR(M) is a division ring.

10.4.4 Show that Q ⊗Z Q and Q ⊗Q Q are isomorphic left Q-modules. [Show theyare both 1-dimensional vector spaces over Q.]

By definition on page 367, since Q is a left Z-module and a (Q,Z) bimodule, hence Q⊗Z Q isa left Q-module. Also, since Q is a left Q-module and a (Q,Q) bimodule, thus Q⊗Q Q.

Now, show that they are isomorphic to 1-dimensional vector spaces over Q.

For a⊗ b ∈ Q⊗QQ for a, b ∈ Q, we have a⊗ b = a ·1⊗ b = a(1⊗ b) = a(1⊗ b ·1) = a(1 · b⊗1) =a(b · 1⊗ 1) = ab(1⊗ 1). Thus, Q⊗QQ is SpanQ{1⊗ 1} ∼= Q. Hence, Q⊗QQ is a 1-dimensionalvector space over Q.

On the other hand, since Q⊗Z Q is a left Q-module defined by

s(∑

a⊗ b) =∑

sa⊗ b for s ∈ Q and a, b ∈ Q

So

a⊗ b = a⊗ p

q= ap⊗ 1

q=ap

qq ⊗ 1

q= a

p

q⊗ 1 = ab⊗ 1

Thus, Q⊗Z Q is a group generated by α ⊗ 1 for α ∈ Q. And α ⊗Z 1 = α(1⊗ 1) in Q-moduleQ⊗Z Q. Hence Q⊗Z Q is a 1-dimensional vector space over Q.

(Example: Q⊗Z Q:2

3⊗Z

1

2=

1

3· 2⊗Z

1

2=

1

3⊗Z 2 · 1

2=

1

3⊗Z 1.

However, Q⊗Q Q:2

3⊗Q

1

2= 1 · 2

3⊗Q

1

2= 1⊗Q

2

3· 1

2= 1⊗Q

1

3.)

10.4.5 Let A be a finite abelian group of order n and let pk be the largest powerof the prime p dividing n. Prove that Z/pkZ ⊗Z A is isomorphic to the Sylow p-subgroup of A.

18

Let P be a Sylow p-subgroup of A with |P | = pk. Define ϕ : Z/pkZ×A→ P by (xmodpkZ, a) 7→xa.

ϕ is bilinear: for r1, r2 ∈ Z, x1, x2 ∈ Z/pkZ and a1, a2 ∈ Aϕ(r1x1+r2x2, a) = ϕ(r1x1+r2x2, a) = (r1x1+r2x2)a = r1x1a+r2x2a = r1ϕ(x1, a)+r2ϕ(x2, a).

ϕ(x1, r1a1 + r2a2) = x1(r1a1 + r2a2) = x1r1a1 + x1r2a2 = r1x1a1 + r2x1a2 = r1ϕ(x1, a1) +r2ϕ(x1, a2).

Thus, ϕ is bilinear. Note that A is subgroup of finite abelian group, hence is abelian. So P isa Z-module. Since Z/pkZ and A are Z-modules and let Z/pkZ⊗Z A be the tensor product ofZ/pkZ and A over Z. By Corollary 12 there is a Z-module homomorphism Φ : Z/pkZ⊗ZA→ P .

Z/pkZ×A ι //

ϕ''

Z/pkZ⊗Z A

Φ��P

Since Φ maps x⊗ a to xa ∈ P which is an element of order pk. Finally, Z/pkZ⊗Z A has orderat least pk because pk(x⊗ a) = pkx⊗ a = 0⊗ a = 0. Hence, Φ is an isomorphism.

10.4.6 If R is any integral domain with quotient field Q, prove that (Q/R)⊗R(Q/R) =0.

R is an integral domain and Q is the quotient field. So Q = {mn | m,n ∈ R}. Form1,m2, n1, n1 ∈ R with ni 6= 0 we have

(m1

n1mod R)⊗ (

m2

n2mod R) = n2(

m1

n1n2mod R)⊗ (

m2

n2mod R)

= (m1

n1n2mod R)⊗ n2(

m2

n2mod R)

= (m1

n1n2mod R)⊗ (m2 mod R)

= 0

since (m2 mod R) = 0.

10.4.11 Let {e1, e2} be a basis of V = R2. Show that the element e1 ⊗ e2 + e2 ⊗ e1 inV ⊗R V cannot be written as a simple tensor v ⊗ w for any v, w ∈ R2.

Suppose v = ae1 + be2, w = ce1 + de2 for a, b, c, d ∈ R. Then

v ⊗ w = (ae1 + be2)⊗ (ce1 + de2) = ac(e1 ⊗ e1)⊗ ad(e1 ⊗ e2) + bc(e2 ⊗ e1) + bd(e2 ⊗ e2)

If v ⊗ w = e1 ⊗ e2 + e2 ⊗ e1, then

ac(e1 ⊗ e1)⊗ (ad− 1)(e1 ⊗ e2) + (bc− 1)(e2 ⊗ e1) + bd(e2 ⊗ e2) = 0

Since {e1 ⊗ e1, e1⊗2, e2 ⊗ e1, e2 ⊗ e2} is a basis, we have ac = 0, ad − 1 = 0, bc − 1 = 0 andbd = 0. Thus, a = 0 and c = 0, but this implies that −1 = 0, a contradiction!

19

10.4.12 Let V be a vector space over the field F and let v, v′ be nonzero elementsof V . Prove that v ⊗ v′ = v′ ⊗ v in V ⊗F V if and only if v = av′ for some a ∈ F .

Let {v1, v2, · · · , vn} be a basis for V . (Here we assume that V is finite dimensional so that forany element in V , its expression of a FINITE linear combination of the basis makes sense. IfV is infinite, then let’s choose a finite dimensional subspace W ⊂ V containing v and v′ toproceed the proof). So V ⊗F V = 〈vi ⊗ vj | i, j = 1, 2, · · · , n〉. Let v, v′ ∈ V , then v = Σaiviand v′ = Σa′ivi. So we have v ⊗ v′ = (Σaivi)⊗ (Σa′ivi) = ΣiΣj(aia

′i)vi ⊗ vj . Similarly, we have

v′ ⊗ v = ΣiΣj(a′iai)vi ⊗ vj . Hence we have v ⊗ v′ = v′ ⊗ v if and only if aia

′j = a′iaj for all i, j.

Now, for fixed j, we have aia′j = a′iaj for all i if and only if ai = a′iaj(a

′j)−1 if and only

if ai = (aj(a′j)−1)a′i since F is commutative and every element in F has a multiplica-

tive inverse. Since this is is true for some fixed j, let’s denote the term a = (aj(a′j)−1),

note that a ∈ F . So for this fixed j, we have ai = a · a′i, i = 1 · · · , n. Hence we havev = Σaivi = Σ(aa′i)vi = aΣa′ivi = av′.

Therefore, we proved that v ⊗ v′ = v′ ⊗ v if and only if v = av′ for some a ∈ F .

10.4.15 Show that tensor products do not commute with direct products in general.

Consider Q⊗Z(∞⊕i=1

Z/2iZ), this implies

Q⊗Z (⊕Z/2iZ) ∼= (Q⊗Z Z/2Z)⊕ (Q⊗Z Z/22Z)⊕ (Q⊗Z Z/23Z)⊕ · · ·

Since for each i, Z/2iZ is a torsion abelian group (i.e. every element has a finite order), byexercise 10.4.8(d) (prove later) each term on the right hand side is 0, hence the direct productis 0. However, the left hand side is not zero since

⊕Z/2iZ is not a torsion abelian. This is

because there is no integer m such that m(⊕Z/2iZ) = 0 when i goes arbitrarily large. Hence,

we conclude that tensor products do not commute with direct products in general.

Now we prove 10.4.8(d): If A is an abelian group group, show that Q⊗Z A = 0 if and only ifA is a torsion abelian group (i.e. every element of A has finite order).

(⇒)Suppose Q ⊗Z A = 0. For a ∈ A and q ∈ Q, q = m

n for m,n in Z and n 6= 0, we have

0 = q⊗ a = mn ⊗ a = m · 1

n ⊗ a = 1n ⊗m · a. Since 1

n 6= 0, so m · a must be zero. So a has finiteorder m. Since a ∈ A is arbitrary, so every element in A has finite order hence A is a torsiongroup.

(⇐)Suppose a ∈ A has finite order, say m. i.e. ma = 0 with some m ∈ Z+. Let q ∈ Q. Then wehave q ⊗ a = m · qm ⊗ a = q

m ⊗m · a = qm ⊗ 0 = 0. Hence Q⊗Z A = 0.

10.5.3 Let P1 and P2 be R-modules. Prove that P1 ⊕ P2 is a projective R-module ifand only if both P1 and P2 are projective.

20

(⇐)Suppose P1 and P2 are projective modules, by Proposition 30(4), P1 and P2 are direct sum-mand of free R-modules. i.e. P1 ⊕Q1 and P2 ⊕Q2 are free for some modules Q1 and Q2. Butfree module is isomorphic to a direct sum of regular modules, so P1 ⊕ Q1 ⊕ P2 ⊕ Q2 is alsoa direct sum of modules, hence is free. Since P1 ⊕ Q1 ⊕ P2 ⊕ Q2

∼= (P1 ⊕ P2) ⊕ (Q1 ⊕ Q2),therefore, P1 ⊕ P2 is a direct summand of a free module hence is projective.

(⇒)

Suppose P1⊕P2 is projective. For R-modules M,N , let Mϕ−→ N −→ 0 be an exact sequence,

and f : P1 −→ N an R-module homomorphism. Want to prove that f lifts to an R-modulehomomorphism into M . i.e. f lifts to F : P1 −→M .

Let π : P1 ⊕ P2 −→ P1 be the natural projection (x, y) 7→ x for x ∈ P1 and y ∈ P2. Thenthe composition f ◦ π is an R-module homomorphism. Let nx be the image of x in N . i.e.f(x) = nx. Since ϕ is surjective (by exactness), so there exists an element mx ∈M such thatϕ(mx) = nx. Now, since P1 ⊕ P2 is a projective R-module, f ◦ π lifts to F ′ : P1 ⊕ P2 −→ Msuch that ϕ ◦ F ′ = f ◦ π. i.e. the big diagram commutes and F ′ : (x, y) 7→ mx.

Now, want to prove that f lifts to an R-module homomorphism F : P1 −→ M and the smalldiagram commutes. i.e. f = ϕ ◦ F .

Let i1 : P1 → P1 ⊕ P2 be an inclusion and define F = F ′ ◦ i1 : P1 −→ M by sending x tomx. Certainly F is an R-module homomorphism since F ′ is hence the composition with theinclusion is. Also we claim that f = ϕ ◦ F . Indeed, for x ∈ P1, f(x) = nx ∈ N . On the otherhand, ϕ ◦ F (x) = ϕ(F (x)) = ϕ(mx) = nx. Hence the small diagram commutes and thereforeP1 is a projective R-module.

Similarly we can prove that P2 is a projective R-module by replacing P1 by P2 in the proof.

10.5.7 Let A be a nonzero finite abelian group.(a) Prove that A is not a projective Z-module.(b) Prove that A is not a injective Z-module.

(a) By Example (3) on page 391, free Z-modules have no nonzero elements of finite order sono nonzero finite abelian group can be isomorphic to a submodule of free module. Since A isabelian, A is not isomorphic to a submodule of a free module. So A is not a direct summandof a free module. Hence A is not projective.

(b) Since A is a finite abelian group, A ∼= Zpα11×· · ·×Zpαkk . Since each Zpαii is not divisible: for

a ∈ Zpαii and for n = pαi1 , there is no x such that xn = a since xpαii = 0. Hence Zpα11

×· · ·×Zpαkkis not divisible. Therefore A is not injective.

10.5.8 Let Q be a nonzero divisible Z-module. Prove that Q is not a projectiveZ-module. Deduce that the rational numbers Q is not a projective Z-module.

Lemma If F is a any free module then ∩∞n=1nF = 0.

21

Let {x1, · · · , xk} be a basis for F . Suppose x ∈ ∩∞n=1nF . Note that {nx1, · · · , nxk} be a basisfor nF . So if x ∈ ∩∞n=1, x ∈ F , x ∈ 2F , x ∈ 3F, · · · , hence

x = a1x1 + a2x2 + · · ·+ akxk

= a21(2x1) + · · ·+ a2

2(2x2) + · · ·+ a2k(2xk)

...

= an1 (nx1) + · · ·+ an2 (nx2) + · · ·+ ank(nxk)

(Note: the notation aji , here j is an index, not an exponent!)

Thus, 0 = (a1 − nan1 )x1 + · · · + (ak − nank)xk for all k. Hence a1 = nan1 , i.e. a1 = 2a21 =

3a31 = 4a4

1 = · · · . So n | a1 for all n. Similarly, ak = nank which implies that n | ak for all n.Therefore, a1, · · · , ak must be zero, hence x = 0. �

Now, since Q is a divisible Z-module, nQ = Q for all n. Suppose Q is a projective Z-module.Then Q is a direct summand of a free module F , i.e. Q ⊕ K ∼= F . So Q is isomorphic to asubmodule of a free module. But ∩∞n=1nF = 0 while ∩∞n=1nQ = ∩∞n=1Q = Q a contradiction!Therefore, Q is not projective.

10.5.9 Assume R is commutative with 1.(a) Prove that the tensor product of two free R-modules is free. [Use the fact thattensor products commute with direct sums.](b) Use (a) to prove that the tensor product of two projective R-modules is pro-jective.

(a) By Corollary 19 in Sec 10.4.(b) Suppose M,N are projective. Then M ⊕ P = F1 and N ⊕Q = F2 for some P,Q and freemodules F1 and F2. Thus,

F1 ⊗R F2 = (M ⊕ P )⊗ (N ⊕Q)

= (M ⊕ P )⊗N ⊕ (M ⊕ P )⊗Q= (M ⊗R N)⊕ (P ⊗N)⊕ (M ⊗Q)⊕ (P ⊗Q)

By (a) F1 ⊗R F2 is free. Thus, M ⊗R N is the direct summand of a free module, hence isprojective.

10.5.15 Let M be a left Z-module and let R be a ring with 1.(a) Show that HomZ(R,M) is a left R-module under the action (rϕ)(r′) = ϕ(r′r).

(b) Suppose that 0 −→ Aψ−→ B is an exact sequence of R-modules. Prove that if

every Z-module homomorphism f from A to M lifts to a Z-module homomorphismF from B to M with f = F ◦ ψ, then every R-module homomorphism f ′ from A toHomZ(R,M) lifts to an R-module homomorphism F ′ from B to HomZ(R,M) withf ′ = F ′ ◦ ψ.(c) Prove that if Q is an injective Z-module then HomZ(R,Q) is an injective R-module.

(a) For ϕ ∈ HomZ(R,M), r, r′ ∈ R, define rϕ : R −→M by rϕ(r′) = ϕ(r′r). First prove thatrϕ ∈ HomZ(R,M).

22

For r1, r2 ∈ R, and a ∈ R, we have- (rϕ)(r1 + r2) = ϕ((r1 + r2)r) = ϕ(r1r + r2r) = ϕ(r1r) + ϕ(r2r) = rϕ(r1) + rϕ(r2)- (rϕ)(ar1) = ϕ((ar1)r) = ϕ(a(r1r)) = aϕ(r1r) = arϕ(r1)by the definition of rϕ and that ϕ is a homomorphism.

Hence rϕ ∈ HomZ(R,M).

Now, prove that this action of R on HomZ(R,M) makes it into a left R-module.Let ϕ,ψ ∈ HomZ(R,M). First, by Proposition 2 in 10.2, define ϕ + ψ by (ϕ + ψ)(r) =ϕ(r) + ψ(r), then ϕ + ψ ∈ HomZ(R,M) and with this operation HomZ(R,M) is an abeliangroup.

Prove that HomZ(R,M) satisfies the R-module criteria: let r, r′, s ∈ R,(1) (r + s)ϕ = rϕ+ sϕ(r + s)ϕ(r′) = ϕ(r′(r + s)) = ϕ(r′r + r′s) = ϕ(r′r) + ϕ(r′s) = rϕ(r′) + sϕ(r′) = (rϕ+ sϕ)(r′).

(2) (rs)ϕ = r(sϕ)(rs)ϕ(r′) = ϕ(r′(rs)) = ϕ((r′r)s) = sϕ(r′r) = r(sϕ(r′)) since sϕ ∈ HomZ(R,M).

(3) r(ϕ+ ψ) = rϕ+ rψr(ϕ+ ψ)(r′) = (ϕ+ ψ)(r′r) = ϕ(r′r) + ψ(r′r) = rϕ(r′) + rψ(r′) = (rϕ+ rψ)(r′)since ϕ+ ψ ∈ HomZ(R,M).

(4) 1 · ϕ = ϕ1 · ϕ(r′) = ϕ(r′ · 1) = ϕ(r′).

Hence, HomZ(R,M) is a left R-module.

(b) Given an R-module homomorphism f ′ : A −→ HomZ(R,M) and a ∈ A, define f : A −→M by a 7→ f ′(a)(1R). f is well-defined since f ′ is. Also, f is a Z-module homomorphism sincefor a, b ∈ A and r ∈ Z, we have- f(a+ b) = f ′(a+ b)(1R) = ((f ′(a) + f ′(b))(1R) = f ′(a)(1R) + f ′(b)(1R) = f(a) + f(b).- f(ra) = f ′(ra)(1R) = rf ′(a)(1R) = rf(a).

Now, since ψ is injective, let b ∈ B be the image of a. Since by assumption f lifts toF : B −→ M and f = F ◦ ψ. So F : b 7→ f(a). Now, define F ′ : B −→ HomZ(R,M)by F ′(b)(r) = F (rb). Show that F ′ is an R-module homomorphism.

For b1, b2 ∈ B, r, s,∈ R, we have- F ′(b1 + b2)(r) = F (r(b1 + b2)) = F (rb1 + rb2) = F (rb1) + F (rb2) = F ′(b1)(r) + F ′(b2)(r)- F ′(sb)(r) = F (r(sb)) = F ((rs)b) = F ′(b)(rs) = sF ′(b)(r)by the definition of F ′ and the fact that B is an R-module, F is an R-module. The last equalityholds because F ′(b) ∈ HomZ(R,M) hence by (a) we have F ′(b)(rs) = sF ′(b)(r).

Hence F ′ is an R-module homomorphism.

Finally, prove that the diagram commutes, i.e. f ′ = F ′ ◦ ψ.Indeed, f ′(a)(r) = f ′(a)(r · 1R) = rf ′(a)(1R) = rf(a) since f ′(a) is a homomorphism. On theother hand, (F ′ ◦ ψ)(a)(r) = F ′(ψ(a))(r) = F ′(b)(r) = F (rb) = rF (b) = rf(a). Hence, the

23

diagram commutes.

(c) If Q is an injective Z-module, let M = Q, then by (b) any R-module homomorphismf ′ : A −→ HomZ(R,Q) lifts to an R-module homomorphism F ′ : B −→ HomZ(R,Q) and thediagram commutes. So by Proposition 34 HomZ(R,Q) is an injective R-module.

10.5.16 Prove that every left R-module M is contained in an injective left R-module.(a) Show that M is contained in an injective Z-module Q.(b) Show that HomR(R,M) ⊆ HomZ(R,M) ⊆ HomZ(R,Q).(c) Use the R-module isomorphism M ∼= HomR(R,M) and the previous exerciseto conclude that M is contained in an injective R-module.

(a) If M is a Z-module, by Corollary 37, M is contained in an injective Z-module Q.

(b) By (a), we know that M is contained in an injective Z-module Q, let i : M ↪→ Q be an

inclusion. So we have an exact sequence 0 → Mi↪→ Q. Apply HomZ(R,−) on this sequence,

by Proposition 27 we have

0 −→ HomZ(R,M)i∗−→ HomZ(R,Q)

which is also exact. Hence, we have HomZ(R,M) ⊆ HomZ(R,Q).

On the other hand, for ϕ ∈ HomR(R,M), ϕ : R −→ M is an R-module homomorphism. Butas modules, a fortiori R and M are abelian groups. Note that Z-module homomorphisms arethe same as abelian group homomorphisms, hence ϕ is also an abelian group homomorphism,hence an Z-module homomorphism. This implies that ϕ ∈ HomZ(R,M). Hence we haveHomR(R,M) ⊆ HomZ(R,M).

Therefore, we proved HomR(R,M) ⊆ HomZ(R,M) ⊆ HomZ(R,Q).

(c) If M is an R-module, by exercise 10.5.10(b) we have M ∼= HomR(R,M). By (b), we haveM ⊆ HomZ(R,Q). But 10.5.15(c) says that if Q in an injective Z-module, HomZ(R,Q) is aninjective R-module. Hence, we proved that M is contained in an injective R-module.

10.5.25 Prove that A is a flat R-module if and only if for every finitely generatedideal I of R, the map A⊗R I → A⊗R R ∼= A induced by the inclusion I ⊆ R is againinjective. (or equivalently, A⊗R I ∼= AI ⊆ A).

10.5.26 Suppose R is a PID. This exercise proves that A is a flat R-module if andonly if A is torsion free R-module (i.e. if a ∈ A is nonzero and r ∈ R, then ra = 0implies r = 0).(a) Suppose that A is flat and for fixed r ∈ R consider the map ψr : R→ R definedby multiplication by r : ψr(x) = rx. If r is nonzero show that ψr is an injection.Conclude from the flatness of A that the map from A to A defined by mapping ato ra is injective and that A is torsion free.(b) Suppose that A is torsion free. If I is a nonzero ideal of R, then I = rR forsome nonzero r ∈ R. Show that the map ψr in (a) induces an isomorphism R ∼= I

of R-modules and that composite Rψ−→ I

η−→ R of ψr with the inclusion : I ⊆ R is

multiplication by r. Prove that the composite A ⊗R R1⊗ψr−−−→ A ⊗R I

1⊗η−−→ A ⊗R R24

corresponds to the map a 7→ ra under the identification A ⊗R R = A and that thiscomposite is injective since A is torsion free. Show that 1⊗ ψr is an isomorphismand deduce that 1 ⊗ η is injective. Use the preceding exercise to conclude that Ais flat.

Proof

12.1.5 Let R = Z[x] and let M = (2, x) be the ideal generated by 2 and x, consideredas a submodule of R. Show that {2, x} is not a basis of M . [Find a nontrivialR-linear dependence between these two elements.] Show that the rank of M is 1but not free rank of 1.

Since x · 2 + (−2) · x = 0 for x,−2 6= 0. Thus {2, x} is not linearly independent, hence is not abasis of M . Now, prove that the rank of M is 1. Since

M = {2p(x) + xq(x) | p(x), q(x) ∈ Z[x]}∼= 2Z[x] + xZ[x]∼= 2Z⊕ xZ[x]∼=ϕ Z[x]/(x)⊕ Z[x]

where ϕ : (2, 0) 7→ (1, 0) and (0, x) 7→ (0, 1). Note that Z[x]/(x) is a torsion submodule sincefor any 0 6= m ∈ Z[x]/(x), m ∈ Z, we have x ·m = 0 because (x) is an ideal. And Z[x] is torsionfree. By exercise 12.1.1(b), rank(M) = rank(Z[x]) = 1. Hence the rank of M is 1. Finally,suppose M is free of rank 1, R can be seen as a R-module. Then R/M = Z[x]/(2, x) ∼= Z2 isa torsion R-module. By exercise 12.1.1(b), R has rank 1, a contradiction!

12.1.8 Let R be a PID, let B be a torsion R-module and let p be a prime in R.Prove that if pb = 0 for some nonzero b ∈ B, then Ann(B) ⊆ (p).

For r ∈ Ann(B), then rb = 0 for all b ∈ B. Let (b) be a cyclic submodule generated by b ∈ B,Ann((b)) ⊆ R is an ideal. Since R is a PID, Ann((b)) = (s) for some s ∈ R. By assumption,p ∈ Ann((b)) = (s), so p = sr for some r ∈ R. But p is a prime. In PID, prime p is irreducible.Since s cannot be a unit, thus r is a unit. Hence, s = pr−1 ∈ (p). This implies that (s) ⊆ (p).Thus, Ann(B) ⊆ Ann((b)) = (s) = (p).

12.1.9 Give an example of an integral domain R and a nonzero torsion R-moduleM such that Ann(M) = 0. Prove that if N is finitely generated torsion R-modulethen Ann(N) 6= 0.

Let R = Z, M = Z/Z⊕Z/2Z⊕Z/3Z⊕· · · . If m ∈M is nonzero, then m = (a1, a2, · · · , an) forai ∈ Z/iZ, i = 1, 2, · · · , n. We see that m ∈ Tor(M): let x = l.c.m.(1, 2, · · · , n) then xm = 0.i.e. m is annihilated by l.c.m(1, 2, · · · , n). Now prove Ann(M) = 0. Let 0 6= m ∈ Ann(M)with m ∈ Z. But mx 6= 0 for x ∈ Z/(m + 1)Z, a contradiction! Therefore, Ann(M) must bezero.

Suppose now N is finitely generated torsion R-module. N = {r1a1 + · · ·+ rmam | r1, · · · , rm ∈R, a1, · · · , am ∈ A} for {a1, · · · , am} ∈ A and for each ai, there exists ri ∈ R such thatriai = 0. Let r =

∏ri. Since R is an integral domain, r 6= 0. Hence rN = 0, this implies that

25

Ann(N) 6= 0.

12.2.3 Prove that two 2 × 2 matrices over F which are not scalar matrices aresimilar if and only if they have the same characteristic polynomial.

Let A and B be two 2× 2 matrices. Suppose A and B are similar, then A = PBP−1 for someinvertible 2× 2 matrix P . Let λ be an eigenvalue of A. We have

A− λI = PBP−1 − λI = P (B − λI)P−1

Hence

det(A− λI) = det(P (B − λI)P−1)

= det(P )det(B − λI)det(P−1)

= det(P )det(P−1)det(B − λI)

= det(PP−1)det(B − λI)

= det(B − λI)

Hence A and B have the same characteristic polynomials.

Conversely, if A and B have the same characteristic polynomial p(x). If p(x) is irreducible, thenp(x) is the invariant factor for both A and B. So A,B are similar. Suppoe p(x) = f(x)g(x)with f, g linear. If f, g have different roots, then p(x) is the minimal polynomial for bothA and B. Then A and B have the same invariant factor, hence they are similar. Now, ifp(x) = (x−a)2, then the minimal polynomial is f(x) = (x−a). Then A,B are scalar matrices,a contradiction. Hence, A,B are similar.

12.2.4 Prove that two 3× 3 matrices are similar if and only if they have the samecharacteristic and same minimal polynomials. Give an explicit counterexample tothis assertion for 4× 4 matrices.

(⇒)Let A and B be two 3× 3 matrices. Suppose A and B are similar, then A = PBP−1 for someinvertible 3× 3 matrix P . Let λ be an eigenvalue of A. We have

A− λI = PBP−1 − λI = P (B − λI)P−1

Hence

det(A− λI) = det(P (B − λI)P−1)

= det(P )det(B − λI)det(P−1)

= det(P )det(P−1)det(B − λI)

= det(PP−1)det(B − λI)

= det(B − λI)

Hence A and B have the same characteristic polynomials.

(⇐)Suppose A and B have the same characteristic polynomials p(x). Consider the following cases:

(1) p(x) is irreducible.Since by Cayley-Hamilton Theorem, the minimal polynomial m(x) divides p(x), so m(x) is

26

irreducible. So p(x) is the only one invariant factor. Then by Theorem, A and B have thesame invariant factors hence are similar.

(2) p(x) = f(x)g(x) = (x− a)(x− b)2 with a, b ∈ F a field and a 6= b.Since characteristic polynomial p(x) and minimal polynomial m(x) have the same roots, wehave m(x) = (x − a)(x − b) or m(x) = (x − a)(x − b)2. If m(x) = (x − a)(x − b), since thecharacteristic polynomial of A is the product of all the invariant factors of A. So the invariantfactors are

(x− b), (x− a)(x− b)The rational canonical form is a 0 0

0 0 −ab0 1 a+ b

A and B have the same invariant factors and rational canonical form, hence they are similar.

(3) p(x) = f(x)g(x) = (x− a)(x2 + bx+ c) where g(x) is irreducible.The possible minimal polynomials are m(x) = x − a or m(x) = x2 + bx + c. Supposem(x) = x2 + bx + c, but x − a does not divide m(x) so this is not possible. On the otherhand, if m(x) = x−a, but g(x) does not divide m(x). Hence m(x) must be equal to p(x). Butm(x) is the largest invariant factor, so A and B have only one and the same invariant factors.So they have the same rational canonical form, hence are similar.

(4) p(x) = (x− a)(x− b)(x− c) where a, b, c are distinct.In this case, m(x) = p(x) since they must have the same roots. As in (3), A and B have thesame invariant factors and rational canonical form hence they are similar.

(5) p(x) = (x− a)3

m(x) can be (x− a)3, (x− a)2 or (x− a).If m(x) = (x− a)3, then p(x) = m(x), so as before A and B are similar.If m(x) = (x− a)2, then they have invariant factors

(x− a), (x− a)2

In this case, they have the same rational canonical form:a 0 00 0 −a2

0 1 2a

Hence A and B are similar.

If m(x) = (x− a), A and B are scalar matrices since the invariant factors

(x− a), (x− a), (x− a)

and the rational canonical form: a 0 00 a 00 0 a

27

Hence A and B are similar.

Example of an 4× 4 matrix that have the same characteristic polynomial but are not similar.

Let

A =

1 0 0 00 1 0 00 0 0 −10 0 1 2

, B =

0 −1 0 01 2 0 00 0 0 −10 0 1 2

pA(x) = pB(x) = (x− 1)4 and mA(x) = mB(x) = (x− 1)2. But A and B are not similar since

they do not have the same rational canonical form.

12.2.8 Verify that the characteristic polynomial of the companion matrix0 0 0 · · · 0 −a0

1 0 0 · · · 0 −a1

0 1 0 · · · 0 −a2...

......

......

0 0 0 · · · 1 −an−1

is

xn + an−1xn−1 + · · ·+ a1x+ a0.

We prove by induction. When n = 2, the companion matrix is

[0 −a0

1 −a1

]so we have

Ix−A =

[x −a0

1 +− a1

]So the characteristic polynomial p(x) = det(Ix−A) = x(x+ a1) + a0 = x2 + a1x+ a0.

Suppose it is true for n− 1, we have

x 0 · · · 0 a0

−1 x · · · 0 a1

0 −1. . .

... a2...

... x an−3

0 0 · · · −1 x+ an−2

And p(x) = det(Ix−A) = xk−1 + ak−2x

k−2 + · · ·+ a1x+ a0.When n = k,

28

xI −A =

x 0 · · · 0 a0

−1 x · · · 0 a1

0 −1. . .

... a2...

... x an−2

0 0 · · · −1 x+ an−1

n×n

= x

x 0 · · · 0 a0

−1 x · · · 0 a1

0 −1. . .

... a2...

... x an−3

0 0 · · · −1 x+ an−2

(n−1)×(n−1)

+ a0

x 0 · · · 0−1 x · · · 0

0 −1. . .

......

... x0 0 · · · −1

(n−1)×(n−1)

So

p(x) = det(xI −A)

= x(xn−1 + ak−1xk−2 + ak−2x

k−3 + · · ·+ a2x+ a1) + a0(−1)k−1 · (−1)k+1

= xk + ak−1xk−1 + · · ·+ a2x

2 + a1x+ a0

as desired.

12.2.10 Find all similarity classes of 6×6 matrices over Q with minimal polynomial(x+ 2)2(x− 1).

If m(x) = (x+ 2)2(x− 1), consider the following cases of invariant factors:1. (x+ 2), (x+ 2), (x+ 2), (x+ 2)2(x− 1).2. (x− 1), (x− 1), (x− 1), (x+ 2)2(x− 1).3. (x− 1), (x− 1)(x+ 2), (x+ 2)2(x− 1).4. (x+ 2), (x− 1)(x+ 2), (x+ 2)2(x− 1).5. (x+ 2), (x+ 2)2, (x+ 2)2(x− 1).6. (x+ 2)2(x− 1), (x+ 2)2(x− 1).

This is because the matrix is 6 × 6, the product of all invariant factors (which is the char-acteristic polynomial) must have degree 6. Also each invariant factor divides the next. i.e.a1(x) | a2(x) | · · · | an(x). Hence, the rational canonical form are below:

1.

−2

−2−2

0 0 41 0 00 1 −3

2.

1

11

0 0 41 0 00 1 −3

3.

1

0 21 −1

0 0 41 0 00 1 −3

4.

−2

0 21 −1

0 0 41 0 00 1 −3

29

5.

−2

0 −41 −4

0 0 41 0 00 1 −3

6.

0 0 41 0 00 1 −3

0 0 41 0 00 1 −3

12.2.11 Find all similarity classes of 6× 6 matrices over C with characteristic poly-nomial (x4 − 1)(x2 − 1).

The characteristic polynomials is p(x) = (x4−1)(x2−1) = (x2+1)(x2−1)2 = (x2+1)(x−1)2(x+1)2. The possible m(x) are (x2 +1)(x−1)(x+1), (x2 +1)(x−1)2(x+1), (x2 +1)(x−1)(x+1)2,(x2 + 1)(x− 1)2(x+ 1)2. Since p(x) and m(x) have the same roots and m(x) | p(x). Thus, thecorresponding invariant factors are

1. (x− 1)(x+ 1), (x2 + 1)(x− 1)(x+ 1) = x4 − 1.2. (x+ 1), (x2 + 1)(x− 1)2(x+ 1) = x5 − x4 − x+ 1.3. (x− 1), (x2 + 1)(x− 1)(x+ 1)2 = x5 + x4 − x− 1.4. (x2 + 1)(x− 1)2(x+ 1)2 = x6 − x4 − x2 + 1.

So the rational canonical forms are

1.

0 11 0

0 0 0 11 0 0 00 1 0 00 0 1 0

2.

−1

0 0 0 0 −11 0 0 0 10 1 0 0 00 0 1 0 00 0 0 1 1

3.

1

0 0 0 0 11 0 0 0 10 1 0 0 00 0 1 0 00 0 0 1 −1

4.

0 0 0 0 0 −11 0 0 0 0 00 1 0 0 0 10 0 1 0 0 00 0 0 1 0 10 0 0 0 1 0

12.2.15 Determine up to similarity all 2 × 2 rational matrices (i.e. ∈ M2(Q)) ofprecise order 4 (multiplicatively, of course). Do the same if the matrix has entriesfrom C.

Since the matrix has order 4, i.e. A4 = 1, the minimal polynomial divides x4 − 1 = (x2 +1)(x+ 1)(x− 1). The possibilities for m(x) are

x+ 1, x− 1, (x+ 1)(x− 1), (x2 + 1), (x2 + 1)(x+ 1), (x2 + 1)(x− 1), (x2 + 1)(x+ 1)(x− 1).

But the order is precisely 4, so we exclude x+ 1, x− 1 and (x+ 1)(x− 1). Furthermore, A isa 2 matrix, degree m(x) ≤ 2. So the only possibilities are (x2 + 1), hence

A =

(0 −11 0

)Over C, x2 + 1 = (x− i)(x+ i). So the possibilities for m(x) are

x+ i, x− i, (x+ i)(x− i), (x± i)(x± 1).30

For example, by computation we see that

A4 =

(i 00 i

)= I

So the 2× 2 matrix with m(x) = x+ i has order 4. Same for other three cases. Therefore, thecorresponding rational canonical forms are(

i 00 i

),

(−i 00 −i

),

(0 −11 0

),

(±1 00 ±i

).

Note that the second and the third are similar.

12.3.9 Prove that the matrices

A =

−8 −10 −17 9 13 2 0

, B =

−3 2 −44 −1 44 −2 5

both have (x−1)2(x+1) as characteristic polynomial but that one can be diagonal-

ized and the other cannot. Determine the Jordan canonical form for both matrices.

By computation, we see that they both have characteristic polynomial p(x) = (x− 1)2(x+ 1).Since A− I 6= 0 and A+ I 6= 0, the minimal polynomial mA(x) must have degree at least two.But by computation, (A − I)2 6= 0 and (A − I)(A + I) 6= 0. Thus, mA(x) = (x − 1)2(x + 1)has no distinct roots, hence A is not diagonalized.

On the other hand, B − I 6= 0 and B + I 6= 0, but (B − I)(B + I) = 0. Thus, mB(x) =(x− 1)(x+ 1) which is linear and has distinct root. Hence B is diagonalizable. Their JordanCanonical forms are as follows:

JA =

1 1 00 1 00 0 −1

, JB =

1 0 00 1 00 0 −1

12.3.22 Prove that an n× n matrix A with entries from C satisfying A3 = A can bediagonalized. Is the same statement true over any field F?

Since A3 = A, the minimal polynomial m(x) divides x3− x = x(x+ 1)(x− 1). Since m(x) hasno repeated roots, A can be diagonalized. However, if charF = 2, take

A =

(1 10 1

)It is easy to see that A3 = A but A is not diagonalized.

12.3.23 Suppose A is a 2 × 2 matrix with entries from Q for which A3 = I butA 6= I. Write A in rational canonical form and in Jordan canonical form viewed asa matrix over C.

Since A3 = I, so minimal polynomial m(x) divides x3−1 which can be factored as (x−1)(x2 +

x + 1) over Q. But over C, x2 + x + 1 = (x + 1−√

3i2 )(x + 1+

√3i

2 ). Since A is a 2 × 2 ma-trix, so characteristic polynomial is of degree 2 hence minimal polynomial must have a degree

≤ 2. So m(x) can be (x + 1−√

3i2 ), (x + 1+

√3i

2 ), (x − 1)(x + 1−√

3i2 ), (x − 1)(x + 1+

√3i

2 ) or31

(x + 1−√

3i2 )(x + 1+

√3i

2 ) because A is a 2 × 2 matrix and A 6= I. Hence the possible invariantfactors will be

(1) (x+ 1−√

3i2 ), (x+ 1−

√3i

2 )

(2) (x+ 1+√

3i2 ), (x+ 1+

√3i

2 )

(3) (x− 1)(x+ 1−√

3i2 )

(4) (x− 1)(x+ 1+√

3i2 )

(5) (x+ 1−√

3i2 )(x+ 1+

√3i

2 )

and the corresponding rational canonical forms are

(1) [−1−

√3i

2 0

0 −1−√

3i2

]

(2) [−1+

√3i

2 0

0 −1+√

3i2

]

(3) Since (x− 1)(x+ 1−√

3i2 ) = x2 − (1+

√3i

2 )x− 1−√

3i2 , we have[

0 1−√

3i2

1 1+√

3i2

]

(4) Since (x− 1)(x+ 1+√

3i2 ) = x2 − (1−

√3i

2 )x− 1+√

3i2 , we have[

0 1+√

3i2

1 1−√

3i2

]

(5) Since (x− 1+√

3i2 )(x− 1−

√3i

2 ) = x2 + x+ 1, we have[0 −11 −1

]Now write A in Jordan form.

(1) If m(x) = (x + 1−√

3i2 ), since the characteristic polynomial p(x) and minimal polynomial

m(x) have the same roots. And p(x) has degree 2 (since A is a 2 × 2 matrix), so p(x) =

(x+ 1−√

3i2 )2. Hence we have

J =

[−1−

√3i

2 0

0 −1−√

3i2

](2) If m(x) = (x+ 1+

√3i

2 ), then p(x) = (x+ 1+√

3i2 )2. So we have

J =

[−1+

√3i

2 0

0 −1+√

3i2

]32

(3) If m(x) = (x− 1)(x+ 1−√

3i2 ), since minimal polynomial divides characteristic polynomial

and p(x) has degree 2. Hence p(x) = m(x). We have

J =

[1 0

0 −1−√

3i2

](4) If m(x) = (x− 1)(x+ 1+

√3i

2 ), then p(x) = m(x). We have

J =

[1 0

0 −1+√

3i2

](5) If m(x) = (x+ 1−

√3i

2 )(x+ 1+√

3i2 ), then p(x) = m(x). We have

J =

[−1−

√3i

2 0

0 −1+√

3i2

]

12.3.24 Prove there are no 3× 3 matrices A over Q with A8 = A but A4 6= I.

Suppose there is a 3 × 3 matrix A over Q with A8 = A. The minimal polynomial dividesx8− x = (x4 + 1)(x2 + 1)(x+ 1)(x− 1). Since A is a 3× 3 matrix, degree m(x) ≤ 3. Thus, thepossibilities for m(x) are

(x− 1), (x+ 1), (x2 + 1), (x− 1)(x+ 1), (x2 + 1)(x− 1), (x2 + 1)(x+ 1), (x2 + 1)(x− 1)(x+ 1)

But by assumption A4 6= I, this implies that m(x) does not divide x4 − 1. Therefore, none ofthe above works, hence there are no such matrices exist.

13.1.1 Show that p(x) = x3 + 9x + 6 is irreducible in Q[x]. Let θ be a root of p(x).Find the inverse of 1 + θ in Q(θ).

Proof

p(x) = x3 + 9x + 6 is irreducible in Q[x] by Eisenstein at p = 3. Let θ be a root of p(x). Tofind the inverse of 1 + θ, we can proceed by Euclidean Algorithm in Q[x] as follows:

Note that x3 + 9x+ 6 = (x+ 1)(x2 − x+ 10)− 4 by long division. Rearranging the equationwe have 4 = −(x3 + 9x+ 6) + (x+ 1)(x2 − x+ 10). Now, dividing each term by 4 gives

1 = −1

4(x3 + 9x+ 6) +

1

4(x2 − x+ 10)(x+ 1)

In the quotient field Q[x]/(x3 + 9x+ 6) this equation becomes

1 =1

4(x2 − x+ 10)(x+ 1)

so that the inverse of 1 + θ in Q(θ) is precisely1

4(θ2 − θ + 10).

13.1.5 Suppose α is a rational root of a monic polynomial in Z[x]. Prove that α isan integer.

Let f(x) = xn + an−1xn−1 + · · ·+ a1x+ a0 ∈ Z[x] be a monic polynomial having α as a root.

Want to prove that α is an integer.

33

Let α =p

qin lowest term, q 6= 0 and (p, q) = 1. So we have f(α) = 0 which is

αn + an−1αn−1 + · · ·+ a1α+ a0 = 0

Subtracting a0 from both sides we obtain

αn + an−1αn−1 + · · ·+ a1α = −a0

Since α =p

q, so we have

pn

qn+ an−1

pn−1

qn−1+ · · ·+ a1

p

q= −a0

Factoring out p we get

p

(pn−1

qn+ an−1

pn−2

qn−1+ · · ·+ a1

1

q

)= −a0

Since there’s no factor 1p in the term

(pn−1

qn + an−1pn−2

qn−1 + · · ·+ a11q

)and a0 is an integer. Hence

p must divide a0.

On the other hand, rearrange f(p

q) = 0 in different way we have

pn

qn= −

(an−1

pn−1

qn−1+ · · ·+ a1

p

q+ a0

)Multiply qn−1 on both sides, we obtain

pn

q= −

(an−1 + an−2p

n−1q + · · ·+ a1pqn−2 + a0q

n−1)

since the there’s no factor q on the right hand side. Also the right side is actually an integerssince all p, q, ai are. Hence q must be 1.

This proves that α =p

q=p

1= p which is an integer.

13.1.8 Prove that x5 − ax − 1 ∈ Z[x] is irreducible unless a = 0, 2 or −1. Thefirst two correspond to linear factors, the third corresponds to the factorization(x2 − x+ 1)(x3 + x2 − 1).

Let f(x) = x5 − ax − 1. First, note that when a = 0, we have f(x) = x5 − 1 = (x − 1)(x4 +x3 + x2 + x + 1); when a = 2, we have f(x) = x5 − 2x − 1 = (x + 1)(x4 − x3 + x2 − x + 1);when a = −1, we have f(x) = x5 + x− 1 = (x2 − x+ 1)(x3 + x2 − 1).

Now, consider the case where a 6= 0, 2,−1.

Case 1 : If f(x) has at least one linear factor, say x− α, where α ∈ Z.

In this case, we have α5− aα− 1 = 0, so α5− aα = 1. Factoring out α we have α(α4− a) = 1.So α must divide 1 and since α and a are integers, so α must be ±1. If α = 1, we haveα4 − a = 1 − a = 1 so a = 0. Contradiction! If α = −1, then α4 − a = 1 + a = −1 so a = 2.Again contradiction!

Case 2 : If f(x) = (x2 + bx + 1)(x3 + dx2 + ex − 1) = 0, i.e. f(x) can be factored into theproduct of two irreducible polynomials in Z[x] of degree 2 and 3. (Remark, the product of the

34

constant terms in these two factors is the constant term in f(x) which is −1, so they must beeither 1 or −1).

In this case, we will get a contradiction by doing the following computation.

f(x) = (x2+bx+1)(x3+dx2+ex−1) = x5+(b+d)x4+(1+e+bd)x3+(be+d−1)x2+(e−b)x−1 =0, we obtain

b+ d = 01 + e+ bd = 0be+ d− 1 = 0

From the first one we have b = −d, then plug into the second and the third we get 1+e−d2 = 0and −de+ d− 1 = 0. The former one gives e = d2 − 1, plugging into the latter one we obtaind3 − 2d + 1 = 0 which gives (d− 1)(d2 + d− 1) = 0. So d = 1 since d is an integer. Then wealso get e = 0. Now, since a = e− b, so we have a = 0− 1 = −1. Contradiction!

Case 3 : If f(x) = (x2 + bx − 1)(x3 + dx2 + ex + 1) = 0, similar to Case 2 except swap thenegative sign on the constant terms of the two irreducible factors.

We will do similar computation. So f(x) = (x2 + bx− 1)(x3 + dx2 + ex+ 1) = x5 + (b+ d)x4 +(+e+ bd− 1)x3 + (be− d+ 1)x2 + (b− e)x− 1 = 0, we obtain

b+ d = 0e+ bd− 1 = 0be− d+ 1 = 0

Similarly, we get b = −d from the first equation, the plug into the second and the third. Butby solving them simultaneously we get d3 + 2d− 1 = 0 which has no integer root since neither1 nor −1 is a root of this equation (by plugging ±1 into it).

Therefore we proved that this f(x) = x5−ax−1 is irreducible over Z[x] except at x = 0, 2,−1.

13.2.7 Prove that Q(√

2 +√

3) = Q(√

2,√

3). Conclude that [Q(√

2 +√

3) : Q] = 4. Findan irreducible polynomial satisfied by

√2 +√

3.

First, prove that Q(√

2 +√

3) = Q(√

2,√

3). (⊆) is clear, so we prove (⊇).

Denote√

2+√

3 by α. So α(√

2−√

3) = (√

2+√

3)(√

2−√

3) = −1. This gives√

2−√

3 = − 1α .

Combining this equation with α =√

2 +√

3 and solving for√

2 we get 2√

2 = α − 1α , hence√

2 = 12(α − 1

α) so√

2 ∈ Q(α). Similarly, solving for√

3 we get√

3 = 12(α + 1

α), hence√3 is also in Q(α). So we proved that Q(

√2,√

3) ⊆ Q(α) = Q(√

2 +√

3). Therefore,Q(√

2 +√

3) = Q(√

2,√

3).

Now prove that [Q(√

2 +√

3) : Q] = 4. Consider the extensions of fields Q ⊆ Q(√

2) ⊆Q(√

2,√

3).

35

Q(√

2)/Q is degree 2 since the minimal polynomial for√

2 over Q is x2− 2 (monic, irreducibleby Eisenstein at p = 2 and having

√2 as a root) so [Q(

√2) : Q] = 2. On the other hand,

Q(√

2,√

3)/Q(√

2) has degree at most 2 and is 2 if and only if x2−3 is irreducible over Q(√

2).It remains to prove that

√3 6∈ Q(

√2). If it were, then

√3 = a+ b

√2 for a, b ∈ Q. So we have

3 = a2 + 2b2 + 2ab√

2. If a, b 6= 0, solve for√

2 we get

√2 =

3− a2 − 2b2

2ab

a contradiction since this implies that√

2 ∈ Q which is not true. Now, if a = 0 we have√3 = b

√2, multiplying both side by

√2 we found that

√6 is a rational, a contradiction.

If b = 0, we would have that√

3 = a a rational, again a contradiction. This proved that√3 6∈ Q(

√2).

Hence Q(√

2,√

3)/Q(√

2) has degree 2, then by Theorem 14, the multiplication rule for fieldsextensions, we have [Q(

√2,√

3) : Q] = 4. But we just proved that Q(√

2 +√

3) = Q(√

2,√

3).Hence [Q(

√2 +√

3) : Q] = 4, as desired.

Finally, find an irreducible polynomial satisfied√

2 +√

3.

Let α =√

2 +√

3. Square both sides we get α2 = 5 + 2√

6, so α2 − 5 = 2√

6. Square bothsides again we have α4 − 10α2 + 25 = 24 which is α4 − 10α2 + 1 = 0. Hence the polynomialx4 − 10x2 + 1 satisfies α =

√2 +√

3. This polynomial is monic and has α as a root. Since wejust proved that [Q(

√2,√

3) : Q] = 4 and by Proposition 11, [Q(√

2,√

3) : Q] = deg mα(x).This implies that the minimal polynomial for α has degree 4 so it is actually the polynomialx4 − 10x2 + 1. This implies that x4 − 10x2 + 1 is irreducible.

13.2.8 Let F be a field of characteristic 6= 2. Let D1 and D2 be elements of F ,neither of which is a square in F . Prove that F (

√D1,√D2) is of degree 4 over F if

D1D2 is not a square in F and is of degree 2 over F otherwise.

The proof is basically similar to 13.2.7. Consider the field extensions F ⊆ F (√D1) ⊆

F (√D1,√D2).

First, show that [F (√D1) : F ] = 2.

Since D1 is not a square, so√D1 6∈ F . And D1 is a root of x2 −D1. Since the other root of

x2 −D1 is −√D1 and neither of them is in F , so x2 −D1 is irreducible hence is the minimal

polynomial of√D1 over F . Hence [F (

√D1) : F ] = deg m√D1

(x) = 2.

Note that [F (√D1,√D2) : F (

√D1)] is at most 2 and is 2 if and only if x2 −D2 is irreducible

over F (√D1), and this is true if and only if

√D2 6∈ F (

√D1). If

√D2 ∈ F (

√D1), then√

D2 = a + b√D1 for a, b ∈ F . Squaring both sides we get D2 = a2 + b2D1 + 2ab

√D1. If

a, b 6= 0, solve for√D1 we get √

D1 =a2 +D1b

2

2ab∈ F

which is a contradiction since D1 is not a square in F .

If b = 0, we have√D2 = a ∈ F , a contradiction.

If a = 0, then√D2 = b

√D1. Multiplying both sides by

√D2 we get D2 = b

√D1D2. This

36

implies that√D2 ∈ F (

√D1) if D1D2 is a square in F . Hence [F (

√D1,√D2) : F ] = [F (

√D1) :

F (√D1)] = 2 if D1D2 is a square in F .

If D1D2 is not a square in F , from above we know that√D2 6∈ F (

√D1). So [F (

√D1,√D2) :

F (√D1)] = 2. Hence [F (

√D1,√D2) : F ] = [F (

√D1,√D2) : F (

√D1)][F (

√D1) : F ] = 2·2 = 4.

13.2.13 Suppose F = Q(α1, α2, · · · , αn) where α2i ∈ Q for i = 1, 2, ·, n. Prove that

3√

2 6∈ F .

Note that each αi satisfies the polynomial x2 − α2i ∈ Q[x]. So

[Q(α1, α2, · · · , αi+1) : Q(α1, α2, · · · , αi)] ≤ 2.

Thus, [F : Q] = 2r for some r ≤ n. Suppose 3√

2 ∈ F , then Q( 3√

2) ⊆ F . But [Q( 3√

2) : Q] = 3.The degrees of the fields extensions Q ⊆ Q( 3

√2) ⊆ F imply that 3 | 2r, a contradiction!

13.2.14 Prove that if [F (α) : F ] is odd then F (α) = F (α2).

Consider the fields extension F ⊆ F (α2) ⊆ F (α). Since α satisfies the polynomial x2−α2 overF (α2) which has degree 2. This implies that [F (α) : F (α2)] ≤ 2. Assume that [F (α) : F ] isodd, we must have [F (α) : F (α2)] = 1. i.e. F (α) = F (α2).

13.2.17 Let f(x) be an irreducible polynomial of degree n over a field F . Let g(x)be any polynomial in F [x]. Prove that every irreducible factor of the compositepolynomial f(g(x)) has degree divisible by n.

Without loss of generality we assume that f(x) is monic (if not, we could divide f(x) by itsleading coefficient to obtain a monic polynomial). Let p(x) be an irreducible factor of f(g(x)),so p(x) divides f(g(x)). So p(x) is monic and has degree k ≤ n. Let α be a root of p(x), wehave [F (α) : F ] = deg minα(x) = k.

Since α is a root of p(x), so α is also a root of f(g(x)). So f(g(α)) = 0 and this implies that g(α)is a root of f(x). So f(x) is the minimal polynomial for g(α). Hence we have [F (g(α)) : F ] =deg ming(α)(x) = n. Now consider the field extensions

F ⊆ F (g(α)) ⊆ F (α)

By the multiplication rule of fields extension, we see that [F (g(α)) : F ] divides [F (α) : F ]. i.e.n | k.

13.2.18 Let k be a field and let k(x) be the field of rational functions in x with

coefficients from k. Let t ∈ k(x) be the rational functionP (x)

Q(x)with relatively prime

polynomials P (x), Q(x) ∈ k[x], with Q(x) 6= 0.(a) Show that the polynomial P (X) − tQ(X) in the variable X and coefficients ink(t) is irreducible over k(t) and has x as a root.(b) Show that the degree of P (X) − tQ(X) as a polynomial in X with coefficientsin k(t) is the maximum of the degrees of P (x) and Q(x).

(c) Show that [k[x] : k(t)] = [k(x) : k(P (x)

Q(x))] = max (deg P (x), deg Q(x)).

37

(a) First, P (X) − tQ(X) is irreducible over (k[X])[t] since it is nothing but a polynomial ofdegree one in t. But because (k[t])[X] = k[t,X] = k[X, t] = (k[X])[t], and P (X), Q(X) arerelatively prime. So P (X) − tQ(X) is irreducible over (k[t])[X]. (If not, P (X) − tQ(X) =F (x, t)G(X), note that F and G can’t be both in t since their product has degree one in t.But this implies that G(X) divides both P (X) and Q(X), contradiction!). Now, since k[t] is aUFD, by Gauss Lemma P (X)−tQ(X) is irreducible over its field of fractions which is (k(t))[x].

Furthermore, if we plug x into P (X)− tQ(X) we get P (x)− tQ(x). But t = P (X)Q(X) , so we have

P (x)− tQ(x) = P (x)− P (X)Q(X)Q(x) = 0 hence P (X)− tQ(X) has x as a root.

(b) Let P (x) = xn + an−1xn−1 + · · · + a1x + a0 and Q(x) = xm + bm−1x

m−1 + · · · + b1x + b0(without loss of generality, we assume P (X) and Q(X) are monic, if not, we could divide eachof them by its leading coefficient to obtain a monic polynomial.) We assume that n > m > 0,then we have

P (X)− tQ(X) = (xn + an−1xn−1 + · · ·+ a1x+ a0)− t(xm + bm−1x

m−1 + · · ·+ b1x+ b0)= (xn + an−1x

n−1 + · · ·+ a1x+ a0)− txm − tbm−1xm−1 − · · · − tb1x− tb0

= xn + · · ·+ (am − tbm)xm + · · ·+ (a1 − tb1)x+ (a0 − tb0)

Hence degree of P (X)− tQ(X) is n which is the maximum of the degree of P (x) and Q(x).

(c) The first equality is clear since t = P (x)Q(x) by assumption. On the other hand, in part (a) we

assumed that the polynomial P (X) − tQ(X) is monic, and proved that it is irreducible andhas x as a root hence is the minimal polynomial of x over k(t). Also part (b) proved that ithas degree n. Hence we have [k(x) : k(t)] = deg minx(X) = max (deg P (x), deg Q(x)) = n.

13.2.22 Let K1 and K2 be two finite extensions of a field F contained in the field K.Prove that the F -algebra K1⊗FK2 is a field if and only if [K1K2 : F ] = [K1 : F ][K2 : F ].

Let {αi} be a basis for K1 over F and {βj} be a basis for K2 over F . Then {αi ⊗ βj} is abasis for K1 ⊗F K2 over F . Define a map ϕ : K1 ⊗F K2 → K1K2 by ϕ(αi ⊗ βj) = αiβj , andextend it by linearity. It is easy to check that ϕ is an F -algebra homomorphism. The map ϕis surjective because the elements αiβj span K1K2 as an F -vector space.

(⇐) By assumption the F -vectors spaces K1 ⊗F K2 and K1K2 have the same dimension overF , namely [K1 : F ][K2 : F ]. Thus by linear algebra ϕ is injective and hence an isomorphism.Therefore, K1 ⊗F K2 is isomorphic to the field K1K2.

(⇒) Conversely, if K1 ⊗F K2 is a field then it has no non-zero ideals so ker(ϕ) = 0. Thereforeϕ is an isomorphism and [K1K2 : F ] = [K1 ⊗F K2 : F ] = [K1 : F ][K2 : F ].

13.4.1 Determine the splitting filed and its degree over Q for x4 − 2.

Since f(x) = x4 − 2 can be factored as (x − i 4√

2)(x + i 4√

2)(x − 4√

2)(x + 4√

2), the roots are± 4√

2, ±i 4√

2.

First, prove that the splitting field for f(x) is Q( 4√

2, i).Let K be the splitting field for f(x). By definition of the splitting field, K contains all roots of

38

f(x) hence contains 4√

2 and the ratio of two roots i 4√

2, 4√

2 which is i. So K ⊇ Q( 4√

2, i). Onthe other hand, all roots ± 4

√2, ±i 4

√2 clearly lie in the field Q( 4

√2, i), so we have K ⊆ Q( 4

√2, i).

Hence, K = Q( 4√

2, i).

Now, prove that [Q( 4√

2, i) : Q] = 8

Consider the extensions of fields: Q ⊆ Q( 4√

2) ⊆ Q( 4√

2, i).

(1) The extension Q( 4√

2)/Q has degree 4 since it satisfies the polynomial x4 − 2 which is ir-reducible over Q by Eisenstein at p = 2. So x4 − 2 is the minimal polynomial for 4

√2 over Q.

And it is monic. Hence [Q( 4√

2) : Q] = deg min 4√2(x) = 4.


2, i)/Q( 4√

2) has degree at most 2 since it satisfies the polynomial x2 +1and is precisely 2 if and only if x2 + 1 is irreducible. Indeed, x2 + 1 is irreducible over Q (noneof its roots i, −i are in Q( 4

√2). So we have [Q( 4

√2, i) : Q( 4

√2)] = 2.

Hence, by multiplication rule for field extensions, we have

[Q(4√

2, i) : Q)] = [Q(4√

2, i) : Q(4√

2)][Q(4√

2) : Q)] = 4 · 2 = 8

i.e. [Q( 4√

2, i) : Q)] has degree 8.

13.4.2 Determine the splitting filed and its degree over Q for x4 + 2.

First, find the roots of this polynomial. If α is a root of this equation, then α4 = −2, then(ξα)4 = −2 where ξ is any 4th root of −1. Hence the solutions of this equation are

ξ4√

2, ξ a 4th root of − 1

To compute ξ explicitly, since x4 + 1 = (x2 + i)(x2 − i), so x2 = i = eπ2i hence x =

±eπ4i = ±(

√2

2 +√

22 i). On the other hand, we have x2 = −i = e−

π2i which implies that

x = ±e−π4i = ±(

√2

2 −√

22 i). Hence ξ = ±

√2

2 ±√

22 i. The roots of the polynomial x4 + 2 are

therefore (±√

22 ±

√2

2 i)4√

2.

Prove that the splitting field for the polynomial x4 + 2 is Q( 4√

2, i).

Let K be the splitting field for x4 + 2 over Q. K must contains all roots of this polynomial.

Let α1 = (√

22 +

√2

2 i)4√

2, α2 = (√

22 −

√2

2 i)4√

2, α3 = (−√

22 +

√2

2 i)4√

2 and α4 = (−√

22 −

√2

2 i)4√

2.

Hence 4√

2 = (α1 + α4)/(α1α4) and i = (α1 + α3)(α1α4

α1 + α4)3. So K contains 4

√2 and i. i.e.

K ⊇ Q( 4√

2, i).

On the other hand, all roots lie in the field Q( 4√

2, i). For example, 4√

2(√

22 +

√2

2 i) is a product

of 2√

4 and√

22 +

√2

2 i, but the latter is just a linear combination of√

2 and i, where√

2 = ( 4√

2)2.

Hence the root 4√

2(√

22 +

√2

2 i) lies in the field Q( 4√

2, i). Similarly all other roots also lie in the

field Q( 4√

2, i), so K ⊆ Q( 4√

2, i). Hence, the splitting field for x4 + 2 over Q is Q( 4√

2, i). i.e.K = Q( 4

√2, i).

Now, compute the degree extension. The degree is actually the same as the exercise 13.4.1since the polynomials x4 − 2 and x4 + 2 have the same splitting field over Q. So I will just do

39

the copy and paste:

Consider the extensions of fields: Q ⊆ Q( 4√

2) ⊆ Q( 4√

2, i).


2)/Q has degree 4 since it satisfies the polynomial x4 − 2 which isirreducible over Q by Eisenstein at p = 2 (Remark: although we just proved that Q( 4

√2, i) is

the splitting field for the polynomial x4 + 2. To compute the degree of extension, we shouldknow that the minimal polynomial would be x4 − 2 over Q since 4

√2 is a root of x4 − 2 but

not of x4 + 2). So x4 − 2 is the minimal polynomial for 4√

2 over Q. And it is monic. Hence[Q( 4√

2) : Q] = deg min 4√2(x) = 4.


2, i)/Q( 4√

2) has degree at most 2 since it satisfies the polynomial x2 +1and is precisely 2 if and only if x2 + 1 is irreducible. Indeed, x2 + 1 is irreducible over Q (noneof its roots i, −i are in Q). So we have [Q( 4

√2, i) : Q( 4

√2)] = 2.

Hence, by multiplication rule for field extensions, we have

[Q(4√

2, i) : Q)] = [Q(4√

2, i) : Q(4√

2)][Q(4√

2) : Q)] = 4 · 2 = 8

i.e. [Q( 4√

2, i) : Q)] has degree 8.

13.5.3 Prove that d divides n if and only if xd − 1 divides xn − 1.

If d divides n, then n = dq for some integer q. So we have xn − 1 = xdq − 1 = (xd − 1)(xq−1 +xq−2 + · · ·+ x+ 1). Hence xd − 1 divides xn − 1.

On the other hand, let n = dq + r we have

xn − 1 = (xdq+r − xr) + (xr − 1) = xr(xdq − 1) + (xr − 1)

So if xd − 1 divides xn − 1, since xd − 1 also divides xdq − 1 = (xd − 1)(xq−1 + xq−2 + · + 1).Thus xd−1 must also divides xr−1. But r < d, r must be 0. Therefore, n = dq. i.e. d divides n.

13.5.4 Let a > 1 be an integer. Prove for any positive integers n, d that d dividesn if and only if ad − 1 divides an − 1. Conclude in particular that Fpd ⊆ Fpn if andonly if d divides n.

By plugging a into previous exercise we get d divides n if and only if ad − 1 divides an − 1.

To prove that Fpd ⊆ Fpn if and only if d divides n.

Note that Fpd is the field consisting of the roots of xpd − x over Fp. Similarly Fpn is the

field consisting of the roots of xpn − x over Fp. If d divides n, by previous exercise we have

pd − 1 divides pn − 1, this implies that xpd−1 − 1 divides xp

n−1 − 1. Hence xpd − x divides

xpn−x which means that the roots of xp

d−x is contained in the roots of xpn−x. i.e. Fpd ⊆ Fpn .

On the other hand, if Fpd ⊆ Fpn . Then the roots of xpd−x is contained in the roots of xp

n−x.

Hence we have xpd − x divides xp

n − x which implies that xpd−1 − 1 divides xp

n−1 − 1. Byprevious exercise, pd − 1 divides pn − 1, hence d divides n.

40

13.5.7 Suppose K is field of characteristic p which is not a perfect field: K 6= Kp.Prove there exist irreducible inseparable polynomials over K. Conclude that thereexist inseparable finite extensions of K.

Suppose K 6= Kp, then there is an element α in K with α 6= βp for every β ∈ K. Sof(x) = xp − α has no root in K. Hence f(x) is irreducible over K. Moreover, f(x) is insep-arable since its derivative f ′(x) = pxp−1 = 0 is relatively prime to f(x). Therefore, f(x) isinseparable by Corollary 36 in 13.5.

13.5.8 Prove that f(x)p = f(xp) for any polynomial f(x) ∈ Fp[x].

By Proposition 35, for any a1, a2 ∈ Fp, (a1 + a2)p = ap1 + ap2. Suppose (a1 + · · · + an−1)p =ap1+· · ·+apn−1, by induction we have (a1+· · ·+an−1+an)p = (a1+· · ·+an−1)p+apn = ap1+· · ·+apn.

Moreover, by Fermat’s Little Theorem ap ≡ a (mod p) for a ∈ Z we have

f(x)p = (xn + an−1xn−1 + · · ·+ a1x+ a0)p

= xnp + apn−1xp(n−1) + · · ·+ ap1x

p + ap0

= (xp)n + an−1(xp)n−1 + · · ·+ a1(xp) + a0

= f(xp)

Hence f(x)p = f(xp) for any polynomial f(x) ∈ Fp[x].

13.6.15 Let p be an odd prime not dividing m and let Φm(x) be the mth cyclotomicpolynomial. Suppose a ∈ Z satisfies Φm(a) ≡ 0 (mod p). Prove that a is relativelyprime to p and that the order of a in (Z/pZ)× is precisely m.

Note that xm − 1 =∏d|m

Φd(x) and can also be written as Φm(x) ·∏

d|m,d<m

Φd(x). Let p be an

odd prime and p - n. Suppose a ∈ Z satisfies Φm(a) ≡ 0 (mod p), then p divides Φm(a). Hencep divides am − 1. i.e am − 1 ≡ 0 (mod p) or am ≡ 1 (mod p). Thus a is relatively prime top (if not, then a = pb for some integer b, then we would have bmpm ≡ 0 (mod p), contradiction.).

To compute the order of a ∈ Z/pZ, note that the order of a (mod p) must be less than orequal to m from what we just proved. Suppose a has order less than m, then al ≡ 1 (mod p)for some integer l with l < m and l | m. This implies that al − 1 ≡ 0 (mod p). So we haveal − 1 =

∏d|l Φd(a) ≡ 0 (mod p). This implies that Φk(a) ≡ 0 (mod p) for some integer k ≤ l

and k | l. Now, Φk(x) has a as a root (mod p) with k ≤ l < m and k | m. This implies thatxm − 1 would have a as a multiple root mod p. Since we know that xm − 1 is separable whichmeans it has distinct roots (example (2) on page 547). Contradiction! Hence the order of a(mod p) is preciesly m.

13.6.16 Let a ∈ Z. Show that if p is an odd prime dividing Φm(a) then either pdivides m or p ≡ 1 (mod m).

Suppose p - m. Since p divides Φm(a), by 13.6.15 the order of a (mod p) is m. i.e. am ≡ 1(mod p). But a ∈ (Z/pZ)×, so m divides |Z/pZ| = p− 1 by Lagrange. Hence p− 1 ≡ 0 (modm) or p ≡ 1 (mod m).

41

Hence we proved that either p divides m or p ≡ 1 (mod m).

13.6.17 Prove there are infinitely many primes p with p ≡ 1 (mod p).

We will prove by assuming 13.6.14. Consider Φm(1),Φm(2), · · · ,Φm(n), · · · ∈ Z[x]. These aremonic polynomial in Z[x] of degree at least one, so by 13.6.14 there are infinitely many distinctprimes dividing Φm(i) for i = 1, 2, · · · . But in 13.6.16 we just proved that either p divides m orp ≡ 1 (mod m). Since only finitely many of them are divisors of m because m is finite, hencethere are infinitely many primes p with p ≡ 1 (mod m).

14.1.4 Prove that Q(√

2) and Q(√

3) are not isomorphic.

Suppose Q(√

2) and Q(√

3) are isomorphic. Want to draw a contradiction.

Let σ : Q(√

2)→ Q(√

3) be an isomorphism, since σ is a homomorphism, we have σ(√

2 ·√

2) =σ(√

2)σ(√

2) =√

3√

3 = 3. On the other hand, σ(√

2 ·√

2) = σ(2) = σ(1) + σ(1) = 1 + 1 = 2by the fact that an homorphism maps the identity to identity. Hence we obtain 2 = 3, acontradiction!

Here is alternative less-technical proof:Let σ : Q(

√2) −→ Q(

√3) be an isomorphism. Then σ takes 1 to 1 i.e. σ(1) = 1, it follows

that σ(a) = a for a in the prime subfield Q. Hence σ fixes the prime subfield Q.

Now, since Q(√

2) is the field generated by√

2 over Q. And the isomorphism σ fixes Q, so σis completely determined by its action on

√2, i.e. by σ(

√2). So let σ(

√2) = a+ b

√3 for some

a, b ∈ Q. The minimal polynomial for√

2 over Q is x2 − 2 (since it is monic, irreducible byEisenstein Criterion at p = 2 and has

√2 as a root). Then σ(

√2) must also be a root of this

polynomial (any automorphism permutes the roots of an irreducible polynomial). However,σ(√

2) of the form a+ b√

3 cannot be its root.

If it were, then (a + b√

3)2 − 2 = a2 + 2ab√

3 + 3b2 − 2 = 0. If a = b = 0, we have −2 = 0,contradiction (or, in this case, σ maps

√2 to zero which means that σ is a zero map, contra-

diction). If a = 0, b 6= 0, 3b2 = 2, then b =√

23 6∈ Q, contradiction. If a 6= 0, b = 0, then we

have a =√

2, a contradiction. Finally if a 6= 0, b 6= 0, then we have√

3 = 2−a2−3b2

2ab , again a

contradiction since√

3 is not rational.

Hence σ(√

2) cannot be of the form of a+ b√

3 which implies that Q(√

2) is not isomorphic toQ(√

3).

14.1.5 Determine the automorphisms of the extension Q( 4√

2)/Q(√

2) explicitly.

Note that 4√

2 is satisfies the polynomial x2−√

2 over Q(√

2) so the [Q( 4√

2) : Q(√

2) has degreeat most 2 but is precisely 2 since the root 4

√2 is not contained in Q(

√2). So Q( 4

√2)/Q(

√2) is

a quadratic extension. Any automorphism σ ∈ Aut(Q( 4√

2)/Q(√

2) is completely determinedby its action on 4

√2. i.e. σ sends 4

√2 to − 4

√2 fixing Q(

√2).

42

14.1.7 Determine Aut(R/Q).(a) Prove that any σ ∈ Aut (R/Q) takes squares to squares and takes positive realsto positive reals. Conclude that a < b implies σa < σb for every a, b ∈ R.(b) Prove that − 1

m < a− b < 1m implies − 1

m < σa− σb < 1m for every positive integer

m. Conclude that σ is a continuous map on R.(c) Prove that any continuous map on R which is the identity on Q is the identitymap, hence Aut(R/Q) = 1.

(a) For α2 ∈ R, we have σ(α2) = σ(α · α) = σ(α)σ(α) = (σ(α))2 ∈ R since σ is an auto-morphism. On the other hand, for α ∈ R and α > 0, since α is a real number, α = β2

for some β ∈ R and β 6= 0. So we have σ(α) = α(β2) = (σ(β))2 > 0 (since β 6= 0, soσ(β) 6= 0 because σ is an automorphism hence injective). Thus σ ∈ Aut (R/Q) takes squaresto squares and takes positive reals to positive reals. Now if a < b for a, b ∈ R, then b−a > 0, wehave σ(b−a) = σ(b)−σ(a) > 0 since σ takes positive reals to positive reals. Hence σ(b) > σ(a).

(b) Since any automorphism σ ∈ Aut(R/Q) fixes Q. For every positive integer m, we have

− 1

m= σ(− 1

m) < σ(a− b) = σ(a)− σ(b) < σ(

1

m) =

1

m

Now, we conclude that σ is continuous on R. Given any ε > 0, for any positive integer m andlet δ = 1

m < ε. If | a− b |< δ, we have

| σ(a)− σ(b) |< 1

m< ε

since ε is arbitrary, hence we conclude that σ is a continuous map on R for every a, b ∈ R.

(c) In real analysis we know that for any real number a ∈ R, there exists a sequence of rationalnumbers {qn} such that {qn} → a as n→∞ (Q is dense in R). Now, let σ be any continuousmap on R which is the identity on Q. We have

a = limn→∞

qn = limn→∞

σ(qn) = σ( limn→∞

qn) = σ(a)

since σ fixes Q and note that limit can exchanges with the continuous function σ.

Hence σ(a) = a. i.e. Any automorphism from R to R fixing Q must be the identity.

14.1.8 Prove that the automorphisms of the rational function field k(t) which fix

k are precisely the fractional linear transformations determined by t 7→ at+ b

ct+ dfor

a, b, c, d ∈ k, ad− bc 6= 0.

Let σ : k(t) −→ k(t) be an automorphism fixing the field k. Since σ is completely deter-

mined by its action on t, let σ : t 7→ p(t)

q(t)for polynomials p(t), q(t) in k[t] with q(t) 6= 0. i.e.

σ(t) =p(t)

q(t). Want to prove that p(t) and q(t) are both linear.

Now, want to use exercise the result of 13.2.18 in previous homework, by adapting suitablevariables in this exercise. We have

[k(t) : k(σ(t))] = [k(t) : k(p(t)

q(t))] = max (deg p(t),deg q(t))

43

As proved in exercise 13.2.18, k(t) is an extension of k(p(t)

q(t)). So k(

p(t)

q(t)) ⊆ k(t). But σ is

an automorphism mapping t top(t)

q(t), hence [k(t) : k(

p(t)

q(t))] must be 1. This implies that max

(deg p(t), deg q(t)) is 1 which means that both p(t) and q(t) has degree no greater than one.That is, p(t) = at + b, q(t) = ct + d for a, b, c, d ∈ k and p(t), q(t) are relatively prime, i.e.

ad 6= bc (If ad−bc = 0, then ad = bc, rearrange we have a =bc

d, plugging into

at+ b

ct+ dwe obtain

bcd t+ b

ct+ d=

bd(ct+ d)

ct+ d=b

dwhich means at + b and ct + d are not relatively prime). Therefore,

the automorphism σ is determined by t 7→ at+ b

ct+ dfor a, b, c, d ∈ k and ad− bc 6= 0.

On the other hand, if ad − bc 6= 0, then σ(t) =at+ b

ct+ ddefines an automorphism. Clearly σ

is well-defined since it is independent of choice of t. And it maps a rational functionp(t)

q(t)to

a rational functionp(at+bct+d)

q(at+bct+d)∈ k(t) by just simplifying by doing multiplication on the fraction

and use common denominator ct+ d. The inverse is given by σ−1(t) =dt− b−ct+ a

, by taking the

inverse on the matrix (a bc d

)An easy verification can be done as below:

σ−1σ(t) = σ−1(at+ b

ct+ d) =

d(at+bct+d)− b−c(at+bct+d) + a

=adt+bd−bct−bd

ct+d−act−bc+act+ad

ct+d

=t(ad− bc)ad− bc

= t

Similarly, we have

σσ−1(t) = σ(dt− b−ct+ a

) =a( dt−b−ct+a) + b

c( dt−b−ct+a) + d

=adt−ab−bct+ab−ct+a

cdt−bc−cdt+ad−ct−a

=t(ad− bc)ad− bc

= t

Hence σ is a bijection. Also, it is a homomorphism since for f(t), g(t) ∈ k(t), we have

σ(f + g)(t) = (f + g)(at+ b

ct+ d) = f(

at+ b

ct+ d) + g(

at+ b

ct+ d) = σ(f(t)) + σ(g(t))

σ(fg)(t) = (fg)(at+ b

ct+ d) = f(

at+ b

ct+ d)g(

at+ b

ct+ d) = σ(f(t))σ(g(t))

Therefore σ(t) =at+ b

ct+ dfor ad− bc 6= 0 defines an automorphism on k(t).

14.2.3 Determine the Galois group of (x2 − 2)(x2 − 3)(x2 − 5). Determine all thesubfields of the splitting field of this polynomial.

The extension K = Q(√

2,√

3,√

5) is Galois over Q since it is the splitting field of the polyno-mial (x2 − 2)(x2 − 3)(x5 − 2) since all roots are in K. On the other hand, any field containingthese roots contains K. Any automorphism σ is completely determined by its action on thegenerators

√2,√

3 and√

5 which must be mapped to ±√

2, ±√

3 and ±√

5 respectively. Hence44

the only possibilities for automorphisms are the maps

σ:

√

2→ −√

2√3→

√3√

5→√

5

τ :

√

2→√

2√3→ −

√3√

5→√

5

δ:

√

2→√

2√3→

√3√

5→ −√

5

It’s clearly to see that σ2 = τ2 = δ2 = 1. Moreover,

στ = τσ:

√

2→ −√

2√3→ −

√3√

5→√

5

σδ = δσ:

√

2→ −√

2√3→

√3√

5→ −√

5

τδ = δτ :

√

2→√

2√3→ −

√3√

5→ −√

5

That is, three maps commute with each other. Furthermore,

στδ:

√

2→ −√

2√3→ −

√3√

5→ −√

5

So the Galois group G is isomorphic to Z2 × Z2 × Z2.

Now, find all fixed fields corresponding to the subgroups of G.

From above we know that the subgroup 〈σ〉 fixes the field Q(√

3,√

5), 〈τ〉 fixes the fieldQ(√

2,√

5), also 〈δ〉 fixesQ(√

2,√

3). Furthermore, 〈στ〉 fixesQ(√

5,√

6), 〈σδ〉 fixesQ(√

3,√

10)and 〈τσ〉 fixes Q(

√2,√

15)....etc. In summary we have

subgroups fixed fields

{1} Q(√

2,√

3,√

5)

{1, σ} Q(√

3,√

5)

{1, τ} Q(√

2,√

5)

{1, δ} Q(√

2,√

3)

{1, στ} Q(√

5,√

6)

{1, τδ} Q(√

2,√

15)

{1, σδ} Q(√

3,√

10)

{1, στδ} Q(√

6,√

10,√

15) ∼= Q(√

6,√

10)

{1, σ, τ, στ} Q(√

5)

{1, σ, δ, σδ} Q(√

3)

{1, τ, δ, τδ} Q(√

2)

{1, σ, τδ, στδ} Q(√

15)

{1, τ, σδ, τσδ} Q(√

10)

{1, δ, στ, δστ} Q(√

6)

{1, στ, τδ, σδ} Q(√

30)〈σ, τ, δ〉 Q

which is the correspondence between the subgroups of the Galois group Z2 × Z2 × Z2 and thefixed fields. And |Gal(K/Q)| = [Q(

√2,√

3,√

5) : Q] = 8.

14.2.4 Let p be a prime. Determine the elements of the Galois group of xp − 2.

45

By example in 13.4 on page 541, the roots of the polynomial xp− 2 are ξ p√

2 where ξ is the pth

root of unity. The splitting field for xp − 2 is Q( p√

2, ξp) and [Q( p√

2, ξp) : Q] = p(p− 1).

Since Q( p√

2, ξ) is the splitting field for the separable polynomial xp − 2 (all roots ξ p√

2 aredistinct), so it is Galois over Q. Hence [Q( p

√2, ξp) : Q] =| Gal(Q( p

√2, ξp)/Q) |= p(p− 1).

Now, determine the elements of the Galois group. Any automorphism fixes the prime field Qand is completely determined by its action on the generators p

√2 and ξ which must be mapped

to one of the roots of xp − 2 (since xp − 2 is irreducible over Q by Eisenstein at p = 2) byProposition 2 in 14.1.

Define σ:

{p√

2→ ξ p√

2ξ → ξ

and τ :

{p√

2→ p√

2ξ → ξa

where ξa is the generator of the multiplicative group F∗p with |ξa| = p − 1. Thus σp = 1 and

τp−1 = 1.

Also σaτ :

{p√

2→ p√

2→ ξa p√

2ξ → ξa → ξa

and τσ:

{p√

2→ ξ p√

2→ ξa p√

2ξ → ξ → ξa

So σaτ = τσ.

Hence, the Galois group G = 〈σ, τ | σp = τp−1 = 1, σaτ = τσ〉.

14.2.5 Prove that the Galois group of xp − 2 for p a prime is isomorphic to thegroup of matrices (

a b0 1

)where a, b ∈ Fp, a 6= 0.

The roots of xp − 2 are p√

2, ξ p√

2, ξ2 p√

2, · · · , ξp−1 p√

2 where ξ is a primitive pth root of unity.Hence the splitting field can be written as Q( p

√2, ξ p√

2) (example in 13.4, p541). Any automor-phism σ ∈ G maps each of these two elements to one the roots of xp − 2 hence can be definedby

σa,b :

{p√

2→ ξb p√

2ξ → ξa

Since every element in G must map these generator ξ and p√

2 to their conjugate so there areat most p(p− 1) choices. And p(p− 1) = |Gal(Q(ξp,

p√

2)/Q)| = [Q(ξp,p√

2) : Q]. So σa,b wherea, b ∈ Zp, a 6= 0 are precisely the elements of G.

Now let A be the group of matrices above. Define a map f : G→ A by

f : σa,b 7→(a b0 1

)- f is well-defined since if σa,b = σc,d then we have(

a b0 1

)=

(c d0 1

)- f is injective: if σ ∈ ker(f), f(σ) is the identity matrix (note that here the operation of thematrix group is a multiplication). This implies that a = 1, and b = 0. Hence σ1,0 is an identity

46

in G.- f is clearly surjective by the way it is defined.- f is a group homomorphism: for σa,b, σc,d ∈ G, since

σc,d · σa,b =

{p√

2→ ξb p√

2→ ξbc+d p√

2ξ → ξa → ξac

So we have σc,d · σa,b = σbc+d,ac.

Hence f(σc,d · σa,b) = f(σac,bc+d) =

(ac bc+ d0 1

)=

(c d0 1

)(a b0 1

)= f(σc,d)f(σa,b)

Therefore we proved that f is an isomorphism.

14.2.10 Determine the Galois group of the splitting field over Q of x8 − 3.

(1) Prove that the splitting field for x8−3 is Q( 8√

3, ξ), where ξ is a primitive 8th root of unity.

First, find the roots of this polynomial. If α is a root of this equation, then α8 = 3, then(ξα)8 = 3 where ξ is any primitive 8th root of 1. Hence the solutions of this equation are

ξ8√

3, ξ a 8th root of 1

Compute ξ explicitly, first note that x8− 1 = (x4− 1)(x4 + 1) = (x− 1)(x+ 1)(x2 + 1)(x4 + 1).

And exercise 13.4.2 proved that the roots for x4 + 1 are ±√

22 ±

√2

2 i. In addition, ±i are the

roots of x2 + 1. Hence ξ is precisely (±√

22 ±

√2

2 i)i which is again ±√

22 ±

√2

2 i.

Let K be the splitting field for x8 − 3 over Q. By definition, K contains all roots of x8 − 3hence contains 8

√3 and ξ which is the ratio of two roots 8

√3 and ξ 8

√3. So K ⊇ Q( 8

√3, ξ). On

the other hand, all roots are clearly contained in Q( 8√

3, ξ). Therefore, K = Q( 8√

3, ξ).

Now prove that Q( 8√

3, ξ) = Q( 8√

3,√

2, i) where ξ is a primitive 8th root of 1, i.e.√

22 +

√2

2 i.

⊆ is clear. The other containment ⊇ is also true because√

2 = ξ + ξ7 and i = ξ2. Hence weproved that the splitting field field is actually Q( 8

√3,√

2, i).

(2) Prove that [Q( 8√

3,√

2, i) : Q] = 32.

Consider the extension of fields Q ⊆ Q( 8√

3) ⊆ Q( 8√

3,√

2) ⊆ Q( 8√

3,√

2, i)

For the first containment, Q( 8√

3) has degree 8 over Q since it satisfies x8 − 3 which is monic,irreducible by Eisenstein at p = 3 and has 8

√3 as a root. Also, [Q( 8

√3,√

2) : Q( 8√

3)] has degreeat most 2 and is precisely 2 if and only if x2 − 2 is irreducible over Q( 8

√3). And this is true if

and only if√

2 6∈ Q( 8√

3).

So we want to prove that√

2 6∈ Q( 8√

3). First, prove that√

2 6∈ Q( 4√

3). We know alreadythat

√2 6∈ Q(

√3) (use the same argument in example(2) on page 526). So if

√2 ∈ Q( 4

√3) =

Q(√

3)( 4√

3), then√

2 = a+b 4√

3, for a, b ∈ Q(√

3). But Q( 4√

3)/Q(√

3) is a quadratic extensionhence is a degree 2 Galois extension. So there are two automorphisms: an identity and σ,say. Hence σ :

√2 7→ −

√2. Since an automorphism will maps a + b 4

√3 to its conjugate, so

we have σ : a + b 4√

3 7→ a − b 4√

3. But we just assumed that√

2 = a + b 4√

3, so this implies47

σ :√

2 7→ a− b 4√

3. In addition, automorphism σ fixes Q. Combine all these we have

σ(2) = σ(√

2√

2) = σ(√

2)σ(√

2) = (a− b 4√

3)(a− b 4√

3) = (a− b 4√

3)2

On the other hand, we have

1(2) = 1(√

2√

2) =√

2√

2 = (a+ b4√

3)(a+ b4√

3) = (a+ b4√

3)2

So these two should be equal. Hence we have (a + b√

2)2 = (a − b√

2)2. After simplifying weget 2ab 4

√3 = 0. So either a or b must be 0.

If b = 0, a 6= 0, we have√

2 = a, a contradiction. If a = 0, b 6= 0, we get√

2 = b 4√

3. Raise

to both sides, we have b =

√2

4√

3, a contradiction since b ∈ Q(

√3). Hence we proved that

√2 6∈ Q( 4

√3).

Now, proceed to the next stage. Want to prove that√

2 6∈ Q( 8√

3). Assuming that√

2 ∈Q( 8√

3) = Q( 4√

3)( 8√

3), then√

2 = a + b 8√

3 for a, b ∈ Q( 4√

3). Use the same method above,we can get a contradiction, hence prove that

√2 6∈ Q( 8

√3). Therefore, we proved that

[Q( 8√

3,√

2) : Q( 8√

3)] = 2.

Finally, Q( 8√

3,√

2, i) has degree 2 extension over Q( 8√

3,√

2) since i 6∈ Q( 8√

3,√

2) and satisfiesthe degree 2 irreducible polynomial x2 + 1.Thus, by the multiplication rule of field extensions, we have

[Q(8√

3,√

2, i) : Q] = [Q(8√

3,√

2, i) : Q(8√

3,√

2)][Q(8√

3,√

2) : Q(8√

3)][Q(8√

3) : Q] = 2 ·2 ·8 = 32

(3) Determine the Galois group

Any automorphism fixes the prime field Q and is completely determined by its action on thegenerators 8

√3,√

2 and i.

Define σ:

8√

3→ ξ 8√

3i→ i√

2→√

2ξ → ξ

τ :

8√

3→ 8√

3i→ −i√

2→√

2ξ → ξ7

δ:

8√

3→ 8√

3i→ i√

2→ −√

2ξ → ξ5

So we have στσ:

8√

3→ ξ 8√

3→ ξ7 8√

3→ 8√

3i→ i→ −i→ −i√

2→√

2→√

2→√

2ξ → ξ → ξ7 → ξ7

which is actually τ . Hence στσ = τ .

Also, an easy computation gives us σ8 = τ2 = 1. So we found G contains a subgroupH = 〈σ, τ | σ8 = τ2 = 1, στσ = τ〉 which is isomorphic to D16 and is a normal subgroup(with index 2). On the other hand, we see that δ2 = 1, so G also contains a cyclic subgroup Kof order 2. Note that K is also a normal subgroup of G since σδ = δσ and τδ = δτ as below:

48

σδ :

8√

3→ 8√

3→ ξ 8√

3i→ i→ i√

2→ −√

2→ −√

2ξ → ξ5 → ξ5

, δσ :

8√

3→ ξ 8√

3→ ξ 8√

3i→ i→ −i√

2→√

2→ −√

2ξ → ξ → ξ5

and

τδ :

8√

3→ 8√

3→ 8√

3i→ i→ −i√

2→ −√

2→ −√

2ξ → ξ5 → ξ35 = ξ3

, δτ :

8√

3→ 8√

3→ 8√

3i→ −i→ −i√

2→ −√

2→ −√

2ξ → ξ7 → ξ35 = ξ3

so δ commutes with 〈δ〉 and 〈τ〉 hence commutes with all elements in G.

Since H and K are normal subgroups of G, so HK is a subgroup of G by Corollary 15 insection 3.2. But HK has order 32 by Proposition 13 in section 3.2 below

| HK |= |H||K||H ∩K|

if H and K are finite subgroups of G (note that here H ∩K = 1). But G has order 32, hencewe must have HK = G.

Now since both H and K are normal subgroup and H ∩K = 1, by Theorem 9 in section 5,4we have G = HK ∼= H ×K. But H ∼= D16 and K ∼= Z2, hence we have G ∼= D16 × Z2.

14.2.13 Prove that if the Galois group of the splitting field of a cubic over Q is thecyclic group of order 3 then all the roots of the cubic are reals.

Let K be the splitting field of a cubic polynomial f over Q. Let G denote the Galois groupGal(K/Q). Suppose G is a cyclic group of order 3. We claim that in this case all roots arereal. Suppose not, since complex roots are conjugate so must be in pair. Let β1 and β2 be theconjugate complex roots and α the real root. The conjugation induces a transposition thatpermutes these two roots. And it is an automorphism of K fixing Q. So the Galois group Gcontains an element of order two. But this is a contradiction since by assumption G is cyclicof order 3 that can’t contain an element of order 2.

Hence, all three roots are real.

14.2.14 Show that Q(√

2 +√

2) is a cyclic quartic field, i.e. is a Galois extension ofdegree 4 with cyclic Galois group.

Let α =√

2 +√

2, so α2 = 2 +√

2 which is α2 − 2 =√

2. Squaring both sides we get(α2 − 2)2 = 2, or α4 − 4α2 + 4 = 2. So we have α4 − 4α2 + 2 = 0. Therefore, the polynomialf(x) = x4 − 4x2 + 2 = 0 has α as a root. f(x) is irreducible by Eisenstein at p = 2.

Since f(x) is an irreducible polynomial over Q hence separable. Find all roots of f(x). Letf(x) = x4 − 4x2 + 2 = 0, we have x2 = 2 ±

√2. Taking the square root both sides we get

x = ±√

2±√

2 and these are the roots of f(x).

49

Now, show that Q(√

2 +√

2) is the splitting field of f(x). Let K be the splitting field of f(x).

K must contains all roots hence apparently contain Q(√

2 +√

2). On the other hand, need to

prove that all roots are in K. Let β =√

2−√

2, since β =√

2−√

2 =√

2−√

2

√2+√

2√2+√

2=

√2√

2 +√

2. But α2 = 2 +

√2 gives

√2 = α2 − 2. Hence, β =

√2√

2 +√

2=α2 − 2

α= α − 2

α,

so β is in Q(√

2 +√

2). Hence all roots ±α and ±β are in K. Thus, K = Q(√

2 +√

2) is thesplitting field of f(x).

Since K is the splitting field of the separable polynomial f(x), so K/Q is Galois extension. LetG = Gal(K/Q). Any automorphism in G is completely determined by its action on α. Define

σ : α 7→ β(= α − 2

α). So we have α

σ−→ α − 2

α

σ−→ (α − 2

α) − 2

α− 2α

=α2 − 2

α− 2

α2−2α

=

√2

α− 2α√

2=

2− 2α2

√2α

=2− 2(2 +

√2)

√2(√

2 +√

2)= −

√2 + 2√

2 +√

2= −α σ−→ −α − 2

−α= −α +

2

α=

−α2 + 2

α= −√

2

α=

√2√

2 +√

2= −

√2−√

2σ−→√

2 +√

2 = α

In summary, we have

ασ−→ α− 2

α

σ−→ −α σ−→ βσ−→ α

i.e. σ4 = 1. Therefore, the Galois group G is isomorphic to cyclic group of order 4. i.e.

Q(√

2 +√

2) is a cyclic quartic field.

14.3.6 Suppose K = Q(θ) = Q(√D1,√D2) with D1, D2 ∈ Z, is a biquadratic extension

and that θ = a + b√D1 + c

√D2 + d

√D1D2 where a, b, c, d ∈ Z are integers. Prove

that the minimal polynomial mθ(x) for θ over Q is irreducible of degree 4 over Qbut is reducible modulo every prime p. In particular show that the polynomialx4 − 10x2 + 1 is irreducible in Z[x] but is reducible modulo every prime.

(1) Prove that there are no biquadratic extensions over finite fields.

Suppose K is a biquadratic extension over Fp for prime p, by definition of biquadratic exten-sion, [K : Fp] has degree 4 and the Gal(K/Fp) is isomorphic to Klein 4 group V4 (exercise14.2.15). But note that for any finite field K of dimension 4 over Fp, then the field K has p4

elements and is precisely the splitting field for xp4−x by the discussion on 13.5 (page 549-550).

And the finite field, denoted by Fp4 , is unique up to isomorphism. Hence K ∼= Fp4 . But theGalois group of finite field extension Fp4/Fp is a cyclic group of order 4. Since Klein 4 groupis not isomorphic to cyclic group of order 4, a contradiction. Therefore, we proved that thereare no biquadratic extensions over finite fields. �

Prove that the minimal polynomial for θ over Q is irreducible of degree 4 over Q.

Since the biquadratic field extension has degree 4 (exercise 13.2.8), so [Q(θ) : Q] = degmθ(x) = 4. And by definition, the minimal polynomial is irreducible. Hence mθ(x) is ir-reducible of degree 4.

50

However, mθ(x) is reducible modulo every p. Suppose not, suppose it is irreducible mod p.But since mθ(x) is the minimal polynomial, so we have [Fp(θ) : Fp] = deg mθ(x) = 4 which isnot possible since we just proved that there are no biquadratic extensions over finite fields.

(2) Show that x4 − 10x2 + 1 is irreducible in Z but not irreducible modulo every prime.

In exercise 13.2.7 in the first homework, we have proved that the polynomial x4 − 10x2 + 1 isirreducible in Q[x] and has ±

√2 ±√

3 as roots. By Corollary 6 (the corollary of the GaussLemma) in Section 9.3, then x4 − 10x2 + 1 is irreducible over Z[x].

However, x4 − 10x2 + 1 is reducible modulo every prime p. Suppose it is irreducible, thenx4 − 10x2 + 1 is the minimal polynomial for θ =

√2 +√

3 (it is monic and has θ as a root).Hence [Fp(θ) : Fp] = deg mθ(x) = 4 which is a biquadratic extension over Fp. Again this isnot possible since we just proved that there are no biquadratic extensions over finite fields.

14.3.7 Prove that one of 2,3 or 6 is a square in Fp for every prime p. Concludethat the polynomial

x6 − 11x4 + 36x2 − 36 = (x2 − 2)(x2 − 3)(x2 − 6)

has a root modulo p for every prime p but has no root in Z.

If 2 or 3 is a square in Fp, then we are done since the polynomial has a root mod p. If neither2 nor 3 is a square in Fp, want to prove that 2 · 3 = 6 is a square in Fp.

By Proposition 18 in section 9.5, the multiplicative group F×p of nonzero elements of Fp is

cyclic. Let F×p = 〈σ〉. Every element in F×p is σk for some integer k. So 2 = σm and 3 = σn,

where m,n are integers. Note that σk is a square if and only if k is even.

But we assume that 2, 3 are not squares, so m,n are both odds. Thus, we have 2 · 3 = σmσn =σm+n which is a square in Fp since k+m must be even. This implies that 6 is a square in Fp.i.e. the polynomial has a root mod p for every prime p.

However, this polynomial has no root in Z[x] since all roots ±√

2,±√

3 and ±√

6 are not in Z.

14.4.1 Determine the Galois closure of the field Q(√

1 +√

2).

Let α =√

1 +√

2, then α2 = 1 +√

2. So α2 − 1 =√

2. Squaring both sides we have(α2 − 1)2 = α4 − 2α2 + 1 = 2 hence α4 − 2α2 − 1 = 0. So α satisfies the polynomial

f(x) = x4 − 2x2 − 1 over Q. To find all roots of f(x), solving for x we get x = ±√

1±√

2.

Now, α is in R, however the other root√

1−√

2 is in C. So the field Q(√

1 +√

2) is clearlynot a splitting field of f(x) hence not a Galois extension of Q.

f(x) is irreducible over Q by Eisenstein at p = 2, hence is separable (Corollary 34 in Sec

13.5). So Q(√

1 +√

2) is a finite (since α satisfies an irreducible polynomial of degree 4) andseparable extension (by Corollary 39). To find the Galois closure, need to find the splitting

field of f(x). Let β =√

1−√

2. As mentioned earlier β is in C, more precisely, β =i

αthrough

the following computation.51

1

α=

1√1 +√

2

√1−√

2√1−√

2=

√1−√

2

i= −i

√1−√

2 = −iβ

Hence, all roots of f(x) are ±α and ±β and the splitting field of f(x) is therefore Q(α, i). SoK = Q(α, i) is Galois over Q since it is the splitting field of the separable polynomial f(x) inQ. Now, since in a fixed algebraic closure of K any other Galois extension of Q containing

Q(√

1 +√

2) contains K.

14.4.3 Let F be a field contained in the ring of n× n matrices over Q. Prove that[F : Q] ≤ n.

Proof

By exercise 13.2.19, the ring of n× n matrices over Q contains fields of degree n over Q. LetF denote one of such fields. Since F sits inside an n × n matrix ring over Q, F/Q is a finiteextension. By Theorem 25, F/Q is a simple extension, so F = Q(θ) for some θ ∈ F .

Now, since [Q(θ) : Q] = deg mθ(x). And the minimal polynomial must divide the characteris-tic polynomial which has degree n for an n× n matrix over Q. Hence, the degree of minimalpolynomial can not be greater than n. i.e. [Q(θ) : Q] =degmθ(x) ≤ n. Therefore, we provedthat [F : Q] ≤ n.

14.5.4 Let σa ∈ Gal (Q(ξn)/Q) denote the automorphism of the cyclotomic field ofnth roots of unity which maps ξn to ξan where a is relatively prime to n and ξn is aprimitive nth root of unity. Show that σa(ξ) = ξa for every nth root of unity.

σ : ξn 7→ ξan is a homomorphism with (a, n) = 1. Let ξ be any nth root of unity, then ξ = ξmnfor some positive integer m. Hence we have

σa(ξ) = σa(ξmn ) = σa(ξn)m = (ξan)m = ξamn = (ξmn )a = ξa

Therefore, σa(ξ) = ξa for every nth root of unity.

14.5.10 Prove that Q( 3√

2) is not a subfield of any cyclotomic field over Q.

Suppose that Q( 3√

2) is a subfield of a cyclotomic field over Q. i.e. there is a field extensions

Q ⊆ Q(3√

2) ⊆ Q(ξn)

for some positive integer n. Since the Galois group of the cyclotomic field extension Q(ξn)/Qis isomorphic to the multiplicative group (Z/nZ)× which is a cyclic group of order ϕ(n) henceis abelian. So every subgroup of the Galois group is normal subgroup.

Let H be the corresponding subgroup of the field Q( 3√

2), so H is normal in G. By GaloisTheory, the Q( 3

√2) is Galois over Q. But this is a contradiction since Q( 3

√2) is not Galois over

Q because the 3√

2 satisfies the irreducible polynomial f(x) = x3 − 2 over Q, however, Q( 3√

2)is not a splitting field of f(x) since the root ξ3 is not contained in Q( 3

√2).

Therefore, Q( 3√

2) is not a subfield of any cyclotomic field over Q.

52

14.5.11 Prove that the primitive nth roots of unity form a basis over Q for thecyclotomic field of nth roots of unity if and only if n is squarefree.

By the discussion on the bottom of page 598, suppose n = pa11 pa22 · · · p

akk is the decomposition

of n into distinct prime powers. Since ξpa22 ···p

akk

n is a primitive pa11 th root of unity. The fieldK1 = Q(ξpa11

) is a subfield of Q(ξn). Similarly, each of the fields Ki = Q(ξpaii), i = 1, 2, · · · , k

is a subfield of Q(ξn). Hence K1K2 · · ·Kk ⊆ Q(ξn). On the other hand, the composite of thethe fields contains the product ξpa11

ξpa22· · · ξpakk , which is the primitive nth root of unity, hence

contains Q(ξn). i.e. K1K2 · · ·Kk ⊇ Q(ξn). Thus, the composite field is Q(ξn). i.e.

Q(ξn) = K1K2 · · ·Kk = Q(ξpa11) · · ·Q(ξpakk

)

Since the extension degrees [Ki : Q] = ϕ(paii ), i = 1, 2, · · · , k and ϕ(n) = ϕ(pa11 )ϕ(pa22 ) · · ·ϕ(pakk ),the degree of the composite of the fields Ki is precisely the product of the degrees of the Ki.i.e.

[Q(ξn) : Q] = [Q(ξpa11) · · ·Q(ξpakk

) : Q] = [Q(ξpa11) : Q] · · · [Q(ξpakk

) : Q]

Now, prove that the primitive nth roots of unity is linearly independent over Q if and only ifn is squarefree.

(⇐)

If n is square free, then n = p. In this case, {1, ξp, ξ2p , · · · , ξ

p−2p } is linearly independent since

its minimal polynomial over Q, ξp has degree p−1. Multiplying by ξp we get {ξp, ξ2p , · · · , ξ

p−1p },

this is still linearly independent hence they also forms a basis of Q(ξp)/Q.

(⇒)Suppose n is not square free, i.e. n = pa for a ≥ 2. Then the primitive pa-th roots of

unity is ξkpa where (k, p) = 1. Let ω = ξpa−1

pa , then ωp = 1 which implies that ωp − 1 =

(ω − 1)(1 + · · ·+ ωp−1) = 0. But ω 6= 1, hence we have

0 = 1 + ξpa−1

pa + (ξpa−1

pa )2 + · · ·+ (ξpa−1

pa )p−1 = 1 + ξpa−1

pa + ξ2pa−1

pa + · · ·+ ξ(p−1)pa−1

pa

Multiplying ξpa gives

ξpa + ξpa−1+1pa + ξ2pa−1+1

pa + · · ·+ ξ(p−1)pa−1+1pa

i.e.

ξpa = −(ξpa−1+1pa + ξ2pa−1+1

pa + · · ·+ ξ(p−1)pa−1+1pa )

This proves that the primitive pa-th roots of unity for a ≥ 2 is linearly dependent. There-fore, the primitive nth roots of unity {1, ξn, ξ2

n, · · · , ξn−1n } is a basis if and only if n is squarefree.

14.5.12 Let σp denote the Frobenius automorphism x 7→ xp of the finite field Fq ofq = pn elements. Viewing Fq as a vector space V of dimension n over Fp we canconsider σp as a linear transformation of V to V . Determine the characteristicpolynomial of σp and prove that the linear transformation σp is diagonalizable overFp if and only if n divides p− 1, and is diagonalizable over the algebraic closure ofFp if and only if (n, p) = 1.

For each x ∈ Fp, we have xpn − x = 0. So the transformation σp satisfies the polynomial

Xn − 1 = 0. Since this polynomial is of degree n, it must be the characteristic polynomial.The linear transformation σp is diagonalizable over Fp if and only if Xn − 1 splits into linear

53

factors (not necessarily distinct) in Fp, and this is true if and only if Fp contains all nth rootsof unity. And this is true if and only if F×p contains a cyclic group of order n, and is true if

and only if n | |F×p | = p− 1.

The linearly transformation σp is diagonalizable over Fp =⋃n≥1 Fpn if and only if Xn − 1

is separable (guarantee all roots are distinct?). And Xn − 1 is separable if and only if it isrelatively prime to nXn−1, which is true if and only if nXn−1 is nonzero in Fp, i.e. p - n, or(p, n) = 1.

14.6.17 Find the Galois group of x4 − 7 over Q explicitly as a permutation groupon the roots.

Since x4− 7 = (x+ i 4√

7)(x− i 4√

7)(x+ 4√

7)(x− 4√

7). So ± 4√

7, ±i 4√

7 are the four roots of thepolynomial x4 − 7 over Q. Prove that the splitting field is Q(i, 4

√7).

Let K be the splitting field of f(x) over Q. K contains all roots hence contain the ratio of tworoots 4

√7, i 4√

7 which is i. Hence K ⊇ Q(i, 4√

7). On the other hand, all roots above are in thisfield Q(i, 4

√7). So K ⊆ Q(i, 4

√7), hence K = Q(i, 4

√7).

Consider the extension of fields

Q ⊆ Q(4√

7) ⊆ Q(i,4√

7)

The field extension Q( 4√

7)/Q has degree at most 4 since 4√

7 satisfies the polynomial x4 − 7.But x4 − 7 is irreducible by Eisenstein at p = 7 hence is the minimal polynomial of 4

√7 over

Q. Therefore, the extension Q( 4√

7)/Q has degree precisely 4. Furthermore, the extensionQ(i, 4

√7)/Q( 4

√7) clearly has degree 2 since i is not contained in the field Q( 4

√7). Hence by

multiplication rule of fields extension we have

[Q(i,4√

7) : Q] = [Q(i,4√

7) : Q(4√

7)] · [Q(4√

7) : Q] = 2 · 4 = 8

Hence, the Galois group G = Gal(Q(i, 4√

7)/Q) has degree 8.

Any automorphism in G is completely determined by its action on the generators 4√

7 and i.First note that the 4th root of unity is i. So we define

σ:

{4√

7→ i 4√

7i→ i

and τ :

{4√

7→ 4√

7i→ −i

Since σ : 4√

7 7→ i 4√

7 7→ i2 4√

7 7→ i3 4√

7 7→ 4√

7 and σ fixes i hence σ4 = 1. On the other hand, τfixes 4

√7 and τ : i 7→ −i 7→ i so τ2 = 1.

Also, στ :

{4√

7→ 4√

7→ i 4√

7i→ −i→ −i

And τσ3:

{4√

7→ i 4√

7→ i 4√

7→ i2 4√

7→ i3 4√

7→ i 4√

7i→ i→ i→ i→ −i

So στ = τσ3. Hence the Galois group G is isomorphic to a Dihedral group of order 8D8 = {σ, τ | σ4 = τ2 = 1, στ = τσ3}.

To describe the elements of the Galois group explicitly as a permutation group on the roots,denote the roots of x4 − 7 by α1 = 4

√7, α2 = i 4

√7, α3 = − 4

√7 and α4 = −i 4

√7. Or simply

54

denote 1 = 4√

7, 2 = i 4√

7, 3 = − 4√

7 and 4 = −i 4√

7. So as a permutation group, σ = (1234),τ = (24). Further computation gives

σ2 = (1234)(1234) = (13)(24)σ3 = (1234)(13)(24) = (1432)τσ = (24)(1234) = (14)(23)τσ2 = (24)(13)(24) = (13)τσ3 = (24)(1432) = (12)(34)

In summary, we have G = {1, (13), (24), (13)(24), (12)(34), (14)(23), (1234), (1432)} which isisomorphic to D8.

15.1.2 Show that each of the following rings are not Noetherian by exhibiting anexplicit infinite increasing chain of ideals:(a) the ring of continuous real valued functions on [0, 1].(b) the ring of all functions from any infinite set X to Z/2Z.

(a) Let R be the ring of continuous real valued functions on [0, 1]. And let Jm be the collectionof closed sub-intervals on [0, 1] such that Ji = [1

2 −12i ,

12 + 1

2i ], then J1 ⊃ J2 ⊃ · · · Jn ⊃ · · · .Now, let Ii be subsets of R as follows:

I1 = {f ∈ R | f = 0 on J1}I2 = {f ∈ R | f = 0 on J2}

...

In = {f ∈ R | f = 0 on Jn}...

Prove that all Ii, i = 1, 2, · · · are ideals of R.

- Ii is nonempty since zero map is a continuous function.- Ii is closed under multiplication: for f, g ∈ Ii, both f = 0 and g = 0 on Ji. So we havefg = 0 on Ji, hence fg ∈ Ii.- Ii is closed under subtraction: for f, g ∈ Ii, clearly f − g = 0. So f − g ∈ Ii.Hence I1 is a subring.

Furthermore, for any function h ∈ R and f ∈ Ii, we have hf = hf = 0 on Ji. So hf and fhare in Ii. Therefore, we proved that Ii is an ideal for all i.

Now, the collection of {Ji} is an infinite descending chain of sub-intervals in [0, 1]. This givesus a collection of ideals vanishing on {Ji} which is an infinite ascending chain. i.e.

I1 ⊂ I2 ⊂ I3 ⊂ · · ·is an infinite ascending chain of ideals.

Hence R is not Noetherian.

(b) Let R be the ring of all functions f : X → Z/2Z. And let Ji be the collection of points inX such that Ji = X − {xα1 , xα2 , · · · , xαi} where xαj ∈ X and αj ∈ I an index set. Then we

55

obtain an descending chain of subsets J1 ⊃ J2 ⊃ · · · Jn ⊃ · · · in X.

Now, let Ii be subsets of R as follows:

I1 = {f ∈ R | f = 0 on J1}I2 = {f ∈ R | f = 0 on J2}

...

In = {f ∈ R | f = 0 on Jn}...

Same argument as in (a) will prove that all Ii, i = 1, 2, · · · are ideals of R.

- Ii is nonempty since zero map is a continuous function.- Ii is closed under multiplication: for f, g ∈ Ii, both f = 0 and g = 0 on Ji. So we havefg = 0 on Ji, hence fg ∈ Ii.- Ii is closed under subtraction: for f, g ∈ Ii, clearly f − g = 0. So f − g ∈ Ii.Hence I1 is a subring.

Furthermore, for any function h ∈ R and f ∈ Ii, we have hf = hf = 0 on Ji. So hf and fhare in Ii. Therefore, we proved that Ii is an ideal for all i.

Now, the collection of {Ji} is an infinite descending chain in X. This gives us a collection ofideals vanishing on {Ji} which is an infinite ascending chain. i.e.

I1 ⊂ I2 ⊂ I3 ⊂ · · ·is an infinite ascending chain of ideals.

Hence R is not Noetherian.

15.1.5 Suppose M is a Noetherian R-module and ϕ : M →M is an R-module endo-morphism of M . Prove that ker(ϕn)∩ image(ϕn) = 0 for n sufficiently large. showthat if ϕ is surjective, then ϕ is an isomorphism.

(1) Prove that ker(ϕn)∩ image(ϕn) = 0 for n sufficiently large.

⊇ is clear.⊆ For an element a ∈ kerϕn∩imageϕn, there exists an element b ∈ M such that ϕn(b) = a.Also, ϕn(a) = 0. This implies that ϕn(ϕn(b)) = ϕ2n(b) = 0. So we get b ∈ kerϕ2n = kerϕn

since M is a Noetherian R-module so that the ascending chain of ideals kerϕ stabilize forsufficiently large n.

Therefore, ϕn(b) = 0 = a hence a = 0. This proves the containment ⊆. Hence, ker(ϕn)∩image(ϕn) = 0 for n sufficiently large.

(2) Prove that if ϕ is surjective, then ϕ is an isomorphism.

If ϕ is surjective, ϕn is surjective so imageϕn = M . But we just proved that ker(ϕn)∩image(ϕn) = 0 for n sufficiently large. This means that ker(ϕn) ∩M = 0 for n sufficiently

56

large. Hence kerϕn = 0. Note that ker(ϕ) ⊆ ker(ϕ2) ⊆ · · · , hence kerϕ = 0. Therefore, ϕ isinjective and ϕ is an isomorphism.

15.1.7 Prove that submodules, quotient modules, and finite direct sums of Noe-therian R-modules are again Noetherian R-modules.

Let M be a Noetherian R-module and N be a submodule of M . If N ′ is a submodule of N ,since N ′ is also a submodule of M , hence is finitely generated. So N is Noetherian. Everysubmodule of M/N has the form L/N where L is a submodule of M with N ⊂ L ⊂M . SinceM is Noetherian, L is finitely generated, and the reduction of those generators mod N willgenerate L/N as an R-module. Finally, 0→M1 →M1 ⊕M2 →M2 → 0 is an exact sequence,if M1 and M2 are Noetherian, by 15.1.6 M1⊕M2 is Noetherian. Then by induction we are done.

15.1.8 If R is a Noetherian ring, prove that M is a Noetherian R-module if andonly if M is a finitely generated R-module.

Suppose {x1, · · · , xn} is generators for M . Define a ring homomorphism ϕ : Rn → M by(a1, · · · , an) 7→ a1x1 + · · ·+ anxn. Then ϕ is surjective and we have an exact sequence

0→ kerϕ→ Rn →M → 0.

Since R is Noetherian, Rn is Noetherian by 15.1.7 (direct sum of Noetherian modules is Noe-therian). Then by 15.1.6 M is Noetherian.

The other direction is clear since from the definition of a Noetherian module, an R-modulethat is Noetherian has to be finitely generated.

15.1.11 Suppose R is a commutative ring in which all the prime ideals are finitelygenerated. Proves that R is Noetherian.(a) Prove that if the collection of ideals of R that are not finitely generated isnonempty, then it contains a maximal element I, and that R/I is a Noetherianring.(b) Prove that there are finitely generated ideals J1 and J2 containing I withJ1J2 ⊆ I and that J1J2 is finitely generated.(c) Prove that I/J1J2 is a finitely generated R/I−submodule of J1/J1J2.(d) Show that (c) implies the contradiction that I would be finitely generated overR and deduce that R is Noetherian.

(a) The proof is basically similar to the proof of Proposition 11 in 7.4 which shows that in aring with identity every ideal is contained in a maximal idea. Let S 6= φ be the collection ofideals of R that are not finitely generated. Then S is partially ordered by inclusion. If C isa chain in S, defined I to be the union of all ideals in C. i.e. I =

⋃A∈C A. The same proof

as Proposition 7.4 shows that I is an ideal and actually I is a proper ideal. Hence each chainhas an upper bound in S. By Zorn’s Lemma, S has a maximal element I. So I is not finitelygenerated.

This implies that all ideals in the quotient R/I are finitely generated. Hence by Theorem 2,R/I is Noetherian.

57

(b) By assumption all prime ideals of R are finitely generated, so the maximal element I in Sfound in (a) is not finitely generated hence is not prime. So there are two elements a, b of R\Isuch that ab is contained in I. Let J1 = I + (a) and J2 = I + (b), then J1, J2 contains I. Sinceboth J1 and J2 are not contained in I, so they are not in S hence are finitely generated. Sotheir product J1J2 is also finitely generated.

(c) Since (b) proved that J1J2 ⊆ I ⊆ J1 ⊆ R, then by Lattice Isomorphism Theorem we have

1 ⊆ I/J1J2 ⊆ J1/J1J2 ⊆ R/J1J2

so I/J1J2 is a R−submodule of J1/J1J2 and can be made into a R/I-submodule by definingthe action of R/I on I/J1J2 by (r+ I)m = rm for each m ∈ I/J1J2 and r ∈ R. Hence I/J1J2

is a R/I-submodule of J1/J1J2.

Now, since in (a) we proved that R/I is Noetherian, then R/I is a Noetherian module overitself. Hence its quotient (R/I)/(J1J2/I) ∼= R/J1J2 is Noetherian by exercise 15.1.7. Thenfrom above, I/J1J2 is a submodule of R/J1J2, hence again Noetherian. Thus, I/J1J2 is aNoetherian R/I-submodule of J1/J1J2. By exercise 15.1.8, I/J1J2 is finitely generated.

(d) In (c) we prove that I/J1J2 is a finitely generated R/I-submodule of J1/J1J2. Since J1J2

is finitely generated by (b). But this implies that I is a finitely generated R-submodule byexercise 10.3.7. i.e. I is a finitely generated ideal of R. This is a contradiction since I is anideal in the set S consisting of all ideals that are not finitely generated. Hence we concludethat R is Noetherian.

15.1.26 Let V = Z(xz − y2, yz − x3, z2 − x2y) ⊆ A3.(a) Prove that the map ϕ : A1 → V defined by ϕ(t) = (t3, t4, t5) is surjective mor-phism.(b) Describe the corresponding k-algebra homomorphism ϕ : k[V ] → k[A1] explic-itly.(c) Prove that ϕ is not an isomorphism.

(a) ϕ is a morphism since there are polynomials f1 = t3, f2 = t4 and f3 = t5 ∈ k[A1] such thatϕ(t) = (f1(t), f2(t), f3(t)) = (t3, t4, t5) for all t ∈ A1.

To prove that ϕ is surjective, assume that (x, y, z) 6= (0, 0, 0) (if one of x, y, z is zero, the othertwo must be zero. hence V = A3). So let t = y

x . Then we have

ϕ(y

x) =

((y

x)3, (

y

x)4, (

y

x)5)

=

(y3

x3,y4

x4,y5

x5

)Since by definition V = Z(xz − y2, yz − x3, z2 − x2y) = {(x, y, z) ∈ A3 | fi = 0}, wheref1 = xz − y2, f2 = yz − x3 and f3 = z2 − x2y. Solving fi = 0 for i = 1, 2, 3 we obtain xz = y2,yz − x3 and z2 = x2y. These give us the followings

y3

x3=

(xz)y

x3=zy

x2=x3

x2= x

y4

x4=x2z2

xyz=xz

y=y2

y= y

58

y5

x5=x2z2y

yzx2= z

Hence we haveϕ(t) = ϕ(

y

x) = (x, y, z)

so that ϕ is surjective.

(b) The morphism ϕ : A1 → V induces an associated k-algebra homomorphism ϕ : k[V ] →k[A1] defined by ϕ(f) = f ◦ ϕ. And by definition, k[V ] = k[A3]/(I) = k[x, y, z]/(xz − y2, yz −x3, z2−x2y). So explicitly, ϕ maps f to f ◦ϕ(t) = f(t3, t4, t5). i.e. ϕ : f(x, y, z) 7→ f(t3, t4, t5).Or

ϕ:

x 7→ x3

y 7→ x4

z 7→ x5

(c) Note that the image of ϕ is the subalgebra k[x3, x4, x5] = k + x3k[x] of k[x]. In particular,ϕ is not surjective (for example, x2 is not in k+x3k[x]), hence ϕ is not an isomorphism. Thenby Theorem 6(4), ϕ is not an isomorphism.

15.2.3 Prove that the intersection of two radical ideals is again a radical ideal.

Lemma Let I and J be ideals in the ring R. Then rad (I ∩ J) =rad I∩ rad J .

Proof of Lemma: rad (I ∩ J) = {a ∈ R | ak ∈ I ∩ J for some k ≥ 1} = {a ∈ R | ak ∈I for some k ≥ 1} ∩ {a ∈ R | ak ∈ I for some k ≥ 1} = rad I ∩ rad J . �.

So now, let I, I be two radical ideals. So rad I = I and rad J = J . Prove that I∩J is a radicalideal. i.e. prove that rad(I∩J) = I∩J . Indeed, rad (I∩J)=rad I∩ rad J (by Lemma) =I∩J .

Hence rad (I ∩ J) = I ∩ J , the intersection of two radicals is a radical ideal.

15.2.4 Let I = m1m2 be the product of the ideals m1 = (x, y) and m2 = (x− 1, y − 1)in F2[x, y]. Prove that I is a radical ideal. Prove that (x3 − y2) is a radical ideal inF2[x, y].

First prove that I is a radical ideal, i.e. prove that rad I = I. It’s obviously that I ⊆ rad I.It remains to show that

Since F2[x, y]/m1∼= F2 so m1 is a maximal ideal. Similarly F2[x, y]/m2

∼= F2. So both m1 andm2 are maximal ideals hence are prime, so they are radical by Corollary 13.

Now, rad (I) = rad (m1m2) = rad (m1) ∩ rad (m2) = m1 ∩m2 since m1,m2 are radical.

But m1 and m2 are comaximal in R since for any r ∈ R, r is a polynomial of variables x, ywith coefficient either 0 or 1. But each term xiyj for i > 0, j > 0 can be written as

xiyj = (x− 1)xi−1yj + (y − 1)xi−1yj−1 + xi−1yj−1

hence xiyj ∈ (x, y) + (x− 1, y − 1).

59

If i = j = 0, i.e. the polynomial is a constant 1, then 1 = (x−1)+x+2 ∈ (x, y)+(x−1, y−1).On the other hand, x ∈ (x, y) + (x− 1, y− 1) and y ∈ (x, y) + (x− 1, y− 1). Hence m1 = (x, y)and m2 = (x−1, y−1) are comaximal i.e. m1 +m2 = R so m1m2 = m1∩m2 by exercise 7.3.34.

Return to what we just proved earlier that rad I = m1 ∩m2. Now since m1 ∩m2 = m1m2.Hence we have rad I = m1m2 = I. Namely, I is a radical ideal.

Finally, the ideal (x3− y2) is prime in k[x, y] for any field k hence prime in F2[x, y]. Therefore,(x3 − y2) is a radical ideal in F2[x, y].

15.2.9 Prove that for any field k the map Z in the Nullstellansatz is always sur-jective and the map I in the Nullstellansatz is always injective. Give examples(over a field k that is not algebraically closed) where Z is not injective and I isnot sujective.

(1) Prove that the map Z is always surjective and the map I is always injective.

By Properties of Z in 15.1, if V = Z(I) is an algebraic set, then V = Z(I(V )). i.e. Z · I is anidentity map which implies that I is injective and Z is surjective.

(2) Example that I is not surjective.

Let I = (x2 + 1) be an ideal of R[x]. I is a maximal ideal since R[x]/I = C. So I is primehence is a radical ideal. But I has no zeros in R, so the map I is not surjective.

(3) Example that Z is not injective

Let I1 = (1) and I2 = (x2 + 1) be ideals of R[x]. Then I1 6= I2. However, Z(I1) = Z(I2) = φ.Hence Z is not injective.

15.2.28 Prove that each of the following rings have infinitely many minimal primeideals, and that (0) is not the intersection of any finite number of these.(a) the infinite direct product ring Z/2Z× Z/2Z× · · · .(b) k[x1, x2, · · · ]/(x1x2, x3x4, · · · , x2i−1x2i, · · · ), where x1, x2, · · · are independent vari-ables over the field k.

(a) Consider the ideals I1 = 〈(0, 1, 1, · · · )〉, I2 = 〈(1, 0, 1, · · · )〉, Ii = 〈(1, · · · , 1, 0, 1, · · · )〉. i.e. Iiis an ideal generated by an element with all 1s on all coordinates except 0 in the ith coordinate,in the ring R = Z/2Z× Z/2Z× · · · . These ideals are prime since R/Ii = Z/2Z is an integraldomain. Also, these ideals are minimal. If there is a prime ideal J with J ⊂ Ii and J 6= Ii,then J must have two or more of zeros, for example, if J has two zeros in the coordinates. i.e.J = (1, · · · , 1, 0, 1, · · · , 1, 0, 1, · · · ) then R/J ∼= Z/2Z × Z/2Z which is not an integral domainsince there is a zero divisor when in the product (i.e. a = (1, 0) and b = (0, 1) are nonzero inZ/2Z × Z/2Z, but ab = 0). Similar for the case that more than two zeros in the coordinateof ideal J . So J can not be prime, a contradiction. Therefore, Ii for i = 1, 2, · · · are minimalprime ideals in R.

Now prove that (0) is not the intersection of any finite number of these ideals. i.e. (0) 6=⋂Ni=1 Ii

for some positive integer N . Indeed, the element (0, · · · , 0, 1, 1, 1, · · · , ) with first N position

60

all 0s is contained in the intersection. Hence (0) is not the intersection of any finite number ofthese ideals.

(b) First, consider the following ideals(x1, x2, x3, x7, · · · ) generated by xi where i is odd.(x2, x4, x6, x8, · · · ) generated by xi where i is even.(x1, x4, x5, x8, · · · ) the generator have indexes rotating between odd and even.(x2, x3, x6, x7, · · · ) same as above....

In general, let Ik = (xk12 , xk34 , · · · ) where xk12 is a choice from {x1, x2} and xk34 is chosenfrom {x3, x4} and so on. So there are infinitely many combinations of choices hence there areinfinitely many such ideals in R = k[x1, x2, · · · ].

Ik are all prime since if we let J = (x1x2, x3x4, · · · , x2i−1x2i, · · · ), by the Third IsomorphismTheorem we have

R/J

Ik/J∼=R

Ik

which is an integral domain (for example, k[x1, x2, · · · ]/(x1, x3, x5, · · · ) ∼= k[x2, x4, x6, · · · ] is anintegral domain). Hence Ik/J is a prime ideal in R/J .

Now, claim that Ik/J are minimal. Since Ik is an ideal such that each coordinate was chosenform the pair x2i−1, x2i, so suppose there is an ideal I in R such that I/J strictly containedin Ik/J , i.e. I/J ⊂ Ik/J , then I/J must have zero in at least one of the coordinates, say x2i−1

x2i = 0. But note that the quotient I/J just means that J is contained in I. And we knowthat J = (x1x2, x3x4, · · · , x2i−1x2i, · · · ). Hence this is a contradiction since J can not sit insideI. So we proved that Ik/J is the minimal ideal.

Finally, clearly (0) is not the intersection of any finitely many of these ideals Ik/J . Suppose

(0) =⋂Nk=1 Ik/J , but we know the product of finitely many ideals is contained in their product.

i.e.∏Nk=1 Ik/J ⊆

⋂Nk=1 Ik/J . Hence we have

∏Nk=1 Ik/J = 0 which is not possible since no

coordinate in Ik/J is zero for all k.

15.2.33 Let I = (x2, xy, xz, yz) in k[x, y, z]. Prove that a primary decomposition ofI is I = (x, y) ∩ (x, z) ∩ (x, y, z)2, determine the isolated and embedded primes of I,and find rad I.

(1) Prove that (x, y), (y, z) and (x, y, z)2 are primary.

Since k[x, y, z]/(x, y) ∼= k[z] which is an integral domain, so (x, y) is prime hence is primary.Similarly, k[x, y, z]/(y, z) ∼= k[x], so (y, z) is primary. The ideal (x, y, z)2 is primary since it isa power of maximal ideal (x, y, z) (since k[x, y, z]/(x, y, z) ∼= k is a field).

(2) Find isolated and embedded prime ideals.

As above k[x, y, z]/(x, y) ∼= k[z] is an integral domain, so (x, y) is prime. But prime ideals areradical, so rad(x, y) = (x, y). Similarly, rad(y, z) = (y, z). i.e. (x, y) and (y, z) are themselvesthe associated prime ideals. Since (x, y) and (y, z) do not contain any other prime ideals of I

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so they are isolated prime ideals. The other associated prime ideal is rad(x, y, z)2 = (x, y, z).But (x, y, z) is clearly not an isolated prime since it does contain other ideal (e.g. (x, y)), henceis the embedded prime ideal.

(3) Prove that I = (x2, xy, xz, yz) = (x, y) ∩ (y, z) ∩ (x, y, z)2.

The ideal I1 = (x, y) =∑

fx+gy for f, g ∈ k[x, y, z] and it consists of elements {x, y, xy, yz, xzx2, y2, xyz, · · · }. Similarly, the ideal I2 = (x, z) consists of elements {x, z, xz, xy, yz, x2, z2, xyz, · · · }and I3 = (x, y, z)2 consists of elements {x2, y2, z2, xy, yz, xz, xyz, · · · }.

To see the intersection I1∩I2∩I3, note that x is contained in I1, I2 but not I3, hence is not con-tained in the intersection. Similarly, elements y, z are not contained in the intersection. Nowlook at the possible degree-two elements, we see that xy, yz, xz are clearly in the intersection.In addition, x2 is also in the intersection, but not y2 nor z2. All degree 3 or higher monomialcan be generated by these elements. Hence we conclude that the intersection of these threeideals is {x2, xy, yz, xz}. i.e. I = (x2, xy, xz, yz) = (x, y) ∩ (y, z) ∩ (x, y, z)2

(4) Find the rad I.

The radical ideal of I is the intersection of the isolated primes of I by Corollary 22. HenceradI = (x, y) ∩ (x, z) which is (x, yz). The reason (x, y) ∩ (x, z) = (x, yz) is similar to (3).We had (x, y) = {x, y, xy, yz, xz x2, y2, xyz, · · · } and (x, z) = {x, z, xz, xy, yz, x2, z2, xyz, · · · },hence (x, y) ∩ (x, z) = (x, yz) since all degree 3 or higher elements can be generated by thesetwo generators.

15.2.39 Fix an element a in the ring R. For any ideal I in the ring R let Ia = {r ∈R | ar ∈ I}.(a) Prove that Ia is an ideal and Ia = R if and only if a ∈ I.(b) Prove that (I ∩ J)a = Ia ∩ Ja for ideals I and J .(c) Suppose that Q is a P -primary ideal and that a 6∈ Q. Prove that Qa is a P -primary ideal and that Qa = Q if a 6∈ P .

(a) First, prove that Ia is an ideal.- Ia is nonempty: since aI ⊆ I, so I ⊆ Ia, hence Ia is nonempty.- Ia is closed under multiplication: for r, s ∈ Ia, then r, s ∈ R and ar, as ∈ I. So a(rs) =(ar)s ∈ I, hence rs ∈ Ia. Thus, Ia is closed under multiplication.- Ia is closed under subtraction: for r, s ∈ Ia, then a(r − s) = ar − as ∈ I hence r − s ∈ Ia.Hence Ia is a subring.

Now for any element c ∈ R, and r ∈ Ia, we have (ac)r = a(cr) ∈ I since cr ∈ I. So cIa ⊆ Iafor any c ∈ R. Similarly we can prove that Iac ⊆ Ia. Hence Ia is an ideal.

Prove that Ia = R if and only if a ∈ I.This is clear since if Ia = R, then aR ⊆ I which implies that a = a · 1 ∈ I. On the other hand,a ∈ I gives aR ⊆ I by definition of ideal. Hence Ia = R.

(b) Since ar ∈ I ∩ J if and only if ar ∈ I and ar ∈ J . Therefore, we have(I ∩ J)a = {r ∈ R | ar ∈ I ∩ J} = {r ∈ R | ar ∈ I and ar ∈ J}

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= {r ∈ R | ar ∈ I} ∩ {r ∈ R | ar ∈ J} = Ia ∩ Ja

(c) If Q is P -primary, then radQ = P . First, prove that Qa is primary. Since Qa = {r ∈ R |ar ∈ Q}. If xy ∈ Qa and x 6∈ Qa, want to prove that yn ∈ Qa for some positive integer n.Since xy ∈ Qa, then a(xy) = (ax)y ∈ Q. But Q is P -primary, and x 6∈ Qa which implies thatax 6∈ Q so yn ∈ Q for some positive integer n. So we have ayn ∈ Q. Therefore Qa is primary.

Now prove that Qa is P -primary. i.e. radQa = P .

Since radQa = {r ∈ R | rk ∈ Qa for some k ≥ 1} = {r ∈ R | ark ∈ Q for some k ≥ 1}. Butsince a 6∈ P and Q is contained in radQ = P , hence a 6∈ Q. This implies that (rk)l ∈ Q forsome positive integer l (because ark ∈ Q). Thus, radQa ⊆ rad Q. On the other hand, if r ∈radQ, then rk ∈ Q for some positive integer k, so ark ∈ Q, hence rk ∈ Qa. Hence, radQ ⊆radQa. Therefore, we have radQa = radQ = P .

Finally, prove that Qa = Q if a 6∈ P . One containment is clear, Q ⊆ Qa. For the othercontainment, if r ∈ Qa, then ar ∈ Q. By assumption a 6∈ Q, we claim that r ∈ Q. Supposeon the contrary r 6∈ Q, but ar ∈ Q and Q is primary, so ak ∈ Q ⊆ P (since Q is primary,P =radQ is the smallest prime ideal containing Q). Now, ak = ak−1a ∈ P , but a 6∈ P , soak−1 must be in P . Then breaking down to next stage ak−2a ∈ P obtaining that ak−1 ∈ P .Eventually we will end up getting a ∈ P which is a contradiction since a is not in P . Hencewe prove that r ∈ Q. Therefore, Qa = Q if a 6∈ P .

15.2.40 With notation as in the previous exercise, suppose I = Q1∩ · · · ∩Qm is min-imal primary decomposition of the ideal I and let Pi be the prime ideal associatedto Qi.(a) Prove that Ia = (Q1)a∩· · ·∩ (Qm)a and that rad(Ia) =rad((Q1)a)∩· · · ∩ rad((Qm)a).(b) Prove that rad(Ia) is the intersection of the prime ideals Pi for which a 6∈ Qi.(c) Prove that if rad(Ia) is a prime ideal then rad(Ia) = Pj for some j.(d) For each i = 1, · · · ,m, prove that rad(Ia) = Pi for some a ∈ R.(e) Show from (c) and (d) that the associated primes for a minimal primary de-composition are precisely the collection of prime ideals among the ideals rad(Ia)for a ∈ R, and conclude that they are uniquely determined by I independent ofthe minimal primary decomposition.

(a) Since I = Q1 ∩ · · · ∩ Qm, clearly Ia = (Q1 ∩ · · · ∩ Qm)a. Part (b) in previous exerciseproved that (Q1 ∩ Q2)a = (Q1)a ∩ (Q2)a by induction on m we have (Q1 ∩ · · · ∩ Qm)a =(Q1)a ∩ · · · ∩ (Qm)a. Hence Ia = (Q1)a ∩ · · · ∩ (Qm)a.

Now, since Ia = (Q1)a ∩ · · · ∩ (Qm)a, we have

rad(Ia) = rad((Q1)a ∩ · · · ∩ (Qm)a) = rad((Q1)a) ∩ · · · ∩ rad((Qm)a)

by exercise 15.2.2(c).

(b) For a 6∈ Qi, by previous exercise, (Qi)a is P -primary, so rad(Qi)a = Pi is prime. As-sume that among these Qi, i = 1, 2, · · · ,m there are k of them (k ≤ m) such that a isnot contained in these Qi. Renumber these indices to be 1, 2, · · · , k. Then by (a) we haverad(Ia) =rad((Q1)a) ∩ · · · ∩ rad((Qk)a) i.e. rad(Ia) = P1 ∩ · · · ∩ Pk, the intersection of prime

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ideals Pi for which a 6∈ Qi.

(c) If rad(Ia) is a prime ideal, it is irreducible. Therefore, rad(Ia) = Pj for some j.

(d) In a minimal primary decomposition, no primary ideal contains the intersection of theremaining primary ideals, so for each i we can pick an element a in the intersection of Qj forall j 6= i but a 6∈ Qi. i.e. a ∈

⋂j 6=iQj but a 6∈ Qi. Then by (b), for this element a we have

rad(Ia) = Pi.

(e) From all above, we proved that the associated primes Pi for a minimal primary decom-position are precisely the collection of prime ideals among the ideals rad(Ia) for a ∈ R, i.e.{rad(Ia) | a ∈ R} = {Pi}. Hence these prime ideals are uniquely determined by I independentof the minimal primary decomposition.

16.1.4 Prove that an Artinian integral domain is a field.

Let R be an Artinian integral domain, and a 6= 0 an element in R. We have a descending chainof ideals (a) ⊇ (a2) ⊇ · · · . Since R is Artinian, the descending chain stabilizes. i.e. there isn ∈ N such that (an) = (an+1). Hence, there is an element r ∈ R such that an = an+1r. Thismeans, an = an · ar which implies that 1 = ar since a 6= 0 and R is an integral domain so Rhas the cancellation property.

Therefore an Artinian integral domain is a field.

16.1.8 Let M be a maximal ideal of the ring R and suppose that Mn = 0 for somen ≥ 1. Prove that R is Noetherian if and only if R is Artinian.

Consider the filtration R ⊇M ⊇ · · · ⊇Mn−1 ⊇Mn = 0, observe that each successive quotientM i/M i+1, i = 0, · · · , n − 1 is a module over the field R/M . i.e. each of these quotients is afinite dimensional vector space over the field F . This implies that M i/M i+1 is Noetherian aswell as Artinian by exercise 16.1.7.

15.1.6 and 16.1.6 say that if 0 → M ′ → M → M ′′ → 0 is an exact sequence of R-modules.Then M is a Noetherian (Artinian) R-module if and only if M ′ and M ′′ are Noetherian (Ar-tinian) R-modules.

we have

0 −→M i+1 −→M i −→M i/M i+1 −→ 0

0 −→M −→ R −→ R/M −→ 0

Now, if R is Noetherian, its submodule M and quotient module are also Noetherian by exer-cise 15.1.7. Hence by this exercise again the submodule M2 of M and the quotient M2/M areNoetherian. Inductively, we have all M i and M i/M i+1 Noethterian for all i = 1, · · · , n − 1.But we just proved that M i/M i+1 is Noetherian if and only if they are Artinian. Then byexercise 16.1.6 the short exact sequence in Artinian version we can prove that R is Artinian.The other direction is exactly the same argument.

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Therefore we conclude that R is Noetherian if and only if R is Artinian.

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g=z g is cyclic then g is abelian.ccheng/courses/dummit_foote.pdf · 3.4.11 prove that if h is a...

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