MA 553: Homework 1

MA 553: Homework 1

Yingwei Wang ∗

Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Composition series

1.1 First part of Jordan Holder Theorem

Theorem 1.1 (Jordan Holder Theorem: Part One). Every finite group has a compositionseries.

Proof. Let G1 be a maximal normal subgroup of G; then G/G1 is simple. Let G2 be amaximal normal subgroup of Gl, and so on. Since G is finite, this process must end withGn = 1. Thus G > G1 > · · · > Gn = 1 is a composition series.

1.2 Composition series

Proposition 1.1. If G is a finite group and H ⊳G. Then there is a composition series ofG, one of whose term is H.

Proof. On one hand, sinceH is also a finite group, then by Theorem 1.1,H has a compositionseries H > H1 > · · · > Hn = 1.

On the other hand, consider the quotient group G/H . If G/H is not simple, then ithas a nontrivial normal subgroup, which gives a group K1 between G and H extending thenormal series H ⊳K1 ⊳G. Similarly we can consider G/K1. It indicates that we can finda normal series H = K0 ⊳K1 ⊳ · · · ⊳Km = G such that for each j ∈ [0, m], the quotientgroup Hj+1/Hj is simple.

Now we find a composition series as

1 = Hn < · · · < H1 < H < K1 · · · < Km−1 < G.

∗E-mail address : [email protected]; Tel : 765 237 7149

1

Yingwei Wang Abstract Algebra

2 Normal subgroup

Proposition 2.1. Let A,B,C be subgroups of a group G with A⊳ B and C ⊳G. Then

CA⊳ CB.

Proof. ∀ c1a ∈ CA, c2b ∈ CB, where c1, c2 ∈ C and a ∈ A, b ∈ B, then

(c2b)(c1a)(c2b)−1 = c2bc1ab

−1c−12

= c2(bc1b−1)(bab−1)c−1

2

= c2c1ac−12 , where c1 = bc1b

−1 ∈ C, a = bab−1 ∈ A,

= c2c1(ac−12 a−1)a

= c2c1c2a, where c2 = ac−12 a−1 ∈ C,

= ca ∈ CA, where c = c2c1c2 ∈ C.

It implies that CA⊳ CB.

3 Application of Isomorphism Theorem

Proposition 3.1. Let G be a finite group with G 6= 1. Suppose G has two compositionseries

1 = N0 < N1 < · · · < Nr = G,

1 = M0 < M1 < · · · < Ms = G.

(I) Show for i, j > 0 that Ni−1(Mj−1 ∩Ni)⊳Ni−1(Mj ∩Ni).(II) Use the Second Isomorphism Theorem to establish isomorphisms

Ni−1(Mj ∩Ni)

Ni−1(Mj−1 ∩Ni)∼=

Mj ∩Ni

(Mj−1 ∩Ni)(Mj ∩Ni−1)∼=

(Mj ∩Ni)Mj−1

(Mj ∩Ni−1)Mj−1. (3.1)

Proof. (I) For each i, j, we know that Mj < G and Ni < G. Thus it is obvious that

Mj−1 ∩Ni < Ni, Mj ∩Ni < Ni.

Besides, since Mj−1 ⊳Mj , then

Mj−1 ∩Ni ⊳Mj ∩Ni.

2


Furthermore, we also know that Ni−1 ⊳ Ni. By the Proposition 2.1, we can concludethat

Ni−1(Mj−1 ∩Ni)⊳Ni−1(Mj ∩Ni).

(II)Define the map

φ : Mj ∩Ni →(Mj ∩Ni)Mj−1

(Mj ∩Ni−1)Mj−1

,

byφ(y) = y(Mj ∩Ni−1)Mj−1.

It is easily seen to be a group homomorphism since Mj−1⊳Mj and Nj−1⊳Nj . Moreover,

since Mj ∩Ni−1 6 Mj ∩Ni, it is easy to see that the image of φ is(Mj∩Ni)Mj−1

(Mj∩Ni−1)Mj−1

.

Now we need to prove that ker φ = (Mj−1 ∩Ni)(Mj ∩Ni−1).On the one hand, if y ∈ Mj ∩ Ni−1, then φy is the identity coset (Mj ∩ Ni−1)Mj−1;

similarly, if y ∈ Mj−1 ∩ Ni, then φy = y(Mj ∩ Ni−1)Mj−1 ⊆ Mj−1(Mj ∩ Ni−1)Mj−1 ⊆(Mj ∩Ni−1)Mj−1 since Mj−1 ⊳Mj. Therefore, (Mj−1 ∩Ni)(Mj ∩Ni−1) ⊆ ker φ.

On the other hand, if y ∈ ker φ then y ∈ (Mj ∩ Ni−1)Mj−1, so we can write y = axwith a ∈ Mj ∩ Ni−1 and x ∈ Mj−1. Now y ∈ Ni, x ∈ Ni, so x ∈ Ni and thereforey ∈ (Mj−1 ∩Ni)(Mj ∩Ni−1), which indicates that ker φ ⊆ (Mj−1 ∩Ni)(Mj ∩Ni−1).

Then by the first isomorphism theorem, we can know that

Mj ∩Ni

(Mj−1 ∩Ni)(Mj ∩Ni−1)∼=

(Mj ∩Ni)Mj−1

(Mj ∩Ni−1)Mj−1.

In the same way we can prove another part of Eq.(3.1).

4 Second part of Jordan-Holder Theorem

Theorem 4.1 (Jordan Holder Theorem: Part Two). Let G be a finite group with G 6= 1.Then the composition factors in a composition series are unique, namely, if

1 = N0 < N1 < · · · < Nr = G,

1 = M0 < M1 < · · · < Ms = G.

are two composition series for G, then r = s and there is some permutation π of {1, 2, · · · , r}such that

Mπ(i)/Mπi−1∼= Ni/Ni−1. (4.1)

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Proof. We refine the first series by inserting between each of the r pairs Ni−1 < Ni thefollowing:

Ni−1 = Ni−1(Ni ∩M0)⊳Ni−1(Ni ∩M1)⊳ · · ·⊳Ni−1(Ni ∩Ms)⊳Ni.

Note that the subquotient Ni/Ni−1 is replaced by the s subquotients

Ni−1(Ni ∩Mj)

Ni−1(Ni ∩Mj−1). (4.2)

Similarly, the second series can be refined to replace each of the s subquotient Mj/Mj−1

by the n subquotients(Ni ∩Mj)Mj−1

(Ni−1 ∩Mj)Mj−1. (4.3)

By Eq.(3.1), we know that (4.2) is isomorphic to (4.3). So if we delete any repetitionin the refinements, we know that the sequences of subquotient are isomorphic after somepermutation.

5 Unique factorization

Theorem 5.1 (Fundamental Theorem of Arithmetic). Let n > 1 be a positive integer. Thenthere exist unique primes p1 > p2 < · · · < pk and unique positive integers r1, r2 · · · , rk suchthat n = pr11 · · · prkk .

Proof. Let G = Z/nZ, then G is a cyclic group of order n. Let d be the largest properdivisor of n and let G1 be the normal subgroup of G of size d. Then G/G1 is simple andcyclic, hence of prime order. Repeating this construction, we obtain a composition series ofG,

G = G0 ⊲G1 ⊲ · · ·⊲Gm = 1,

where for each i, Gi/Gi+1 is of prime order pi. Thus

n = |G| = |G/G1||G1/G2| · · · |Gm−1/Gm||Gm| = p1p2 · · ·pm−1(1).

By Jordan-Holder theorem (Thm 4.1), we can get the uniqueness of the prime decom-position of n.

4

MA 553: Homework 2

Yingwei Wang ∗


1 About S4 and D8

Question: Exhibit explicitly a Sylow 2-subgroup of the symmetric group S4, and show thatit is (up to isomorphism) the dihedral group D8.

Solution: Let H be a Sylow 2-subgroup of the symmetric group S4, then we can chooseH as follows

H = {(1), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)}.

Let D8 be the dihedral group of order 8, then

D8 = {e, r, r2, r3, s, sr, sr2, sr3}.

Define the isomorphism π : D8 → H by

π(e) = (1), π(r) = (1234), π(r2) = (13)(24), π(r3) = (1432),

π(s) = (24), π(sr) = (14)(23), π(sr2) = (13), π(sr3) = (12)(43).

Then it is obvious that D8∼= H .

2 Non-abelian group of order 8

Let G be a non-abelian group of order 8.


1


2.1 (a)

Question: Show that the center of G has order 2. Denote its generator by c.Solution: Let C be the center of G, then C ⊳ G and C is abelian. Since |G| = 23, by

Theorem 8 (on DF, p.125), C is nontrivial. Besides, G is non-abelian. Then we know thatthe only possibility for |C| is

|C| = 2, 22.

If |C| = 22, then G/C = 2. Hence G/C is cyclic. Let G/C =< aC >, then ∀x, y ∈ G,∃m,n ∈ N such that x = amc1, y = anc2. Then

xy = amc1anc2 = am+nc1c2 = anc2a

mc1 = yx.

It implies that G is abelian, which is a contradiction.So |C| = 2. Denote C = {1, c}.

2.2 (b)

Question: If G contains an element z 6= c of order 2, then G is isomorphic to a subgroup ofS4. Thus by Section 2.1, we know that G = D8.

Solution: Let H = {1, z}, S = {all of the left cosets of H}. Then |H| = 2, |S| = 4.Consider the action of G on S:

G× S → S,

g × xH → gxH ∈ S.

Then we find a group homomorphism σ : G → Permutation on S defined by

[σ(g)](xH) = gxH.

It indicates that G is a subgroup of S4. Since |G| = 8, we know that G ∼= D8.

2.3 (c)

Question: If c is the only order 2 element in G, then G is generated by two elements j andk of order 4 such that j2 = k2 and jk = kj−1. Hence G is isomorphic to the quaterniongroup, i.e. the multiplicative group of complex matrices generated by

(

0 1−1 0

)

&

(

0 i−i 0

)

,

2


where i2 = −1.Solution: For ∀g ∈ G, g 6= c, the possible order for g is 2, 4, 8. But in this case c is the

only order 2 element, and G is non-abelian, so we can conclude that the possible order forg is 4.

Let j be one element of order 4, then j2 is order 2, so j2 = c. It is obvious that j is notthe only element of order 4 since |G| = 8. Let k 6= j be another element of order 4, thenk2 = c.

Let H =< j >= {j, j2 = c, j3 = j−1, j4 = 1}. It is obvious that k 6= H . So we havekH = {kj, kj2, kj3, k}. Then

(kj)−1 = kj3 = kj−1 ⇒ kjkj−1 = 1 ⇒ kj = jk−1.

Now we can conclude that G = {k, j|k4 = j4 = 1, kj = jk−1}.Let

k =

(

0 1−1 0

)

& j =

(

0 i−i 0

)

,

then Q8∼= G.

2.4 (d)

Question: Is the (order 8) group of matrices of the form

1 a b0 1 c0 0 1

, (a, b, c ∈ Z/2Z)

“diheedral ”(i.e. as in (b)) or “quaternionic ”(i.e. as in (c))?Solution: Let

A =

1 1 00 1 10 0 1

, B =

1 0 00 1 10 0 1

,

and

I =

1 0 00 1 00 0 1

,

then it is easy to check that

A4 = B2 = I, AB = BA−1.

It implies that here G ∼= D8.

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3 Non-abelian group of order p3

Let G be a non-abelian group of order p3 (p an odd prime) and let C be its center.

3.1 (a)

Question: Show that the center of G/C is isomorphic to Zp × Zp where Zp is a group oforder p.

Solution: Similarly as Section 2.1, we know that the only possibility for |C| is

|C| = p, p2.

If |C| = p2, then G/C = p. Hence G/C is cyclic. Let G/C =< aC >, then ∀x, y ∈ G,∃m,n ∈ N such that x = amc1, y = anc2. Then

xy = amc1anc2 = am+nc1c2 = anc2a

mc1 = yx.

It implies that G is abelian, which is a contradiction.So |C| = p and |G/C| = p2. By Corollary 9 (on DF. p. 125), G/C is isomorphic to

either Zp2 or Zp × Zp. However, if G/C ∼= Zp2 , then G/C is cyclic. Similarly as above, itis easy to know that G is abelian, which is impossible. In conclusion, G/C is isomorphic toZp × Zp.

3.2 (b)

Question: Prove that the map f : G → G defined by f(x) = xp is a group homomorphism.Proof: First, we claim that ∀x, y ∈ G, there is a z ∈ C such that yx = xyz. Since

yx = xy(xy)−1yx = xy[y−1x−1yx],

we just need to show that z = y−1x−1yx ∈ C.Let π : G/C → Zp × Zp. Suppose

π(xC) = (m1, n1),

π(yC) = (m2, n2),

then

π(y−1x−1yxC)

= π(y−1Cx−1CyCxC)

= π(y−1C) + π(x−1C) + π(yC) + π(xC)

= (−m2,−n2) + (−m1,−n1) + (m2, n2) + (m1, n1)

= 0,

4


which indicates that y−1x−1yx ∈ C.Second, claim that the map f : G → G defined by f(x) = xp is a group homomorphism.∀x, y ∈ G, there exists a z ∈ C, such that yx = xyz. Then we have

f(xy) = (xy)p

= (xy)(xy) · · · (xy),

= x(yx)(yx) · · · (yx)y,

= x(xyz)(xyz) · · · (xyz)y,

= x2(yx) · · · (yx)y2zp−1

= · · ·

= xpypzp(p−1)/2

= xpyp

= f(x)f(y).

3.3 (c)

Question: Prove that f(G) ⊂ C, and deduce that G has at least p2 − 1 elements of order p.Proof: First, we claim that ∀x ∈ G, xp ∈ C.Since

yxp = xpyy−1x−pyxp = xpy[y−1x−pyxp],

we just need to show that y−1x−pyxp = 1.We know that ∀x, y ∈ G, there exists a z ∈ C, such that yx = xyz. So

y−1x−pyxp = y−1x−p(yx)xp−1,

= y−1x−p(xyz)xp−1,

= y−1x−p+1yxp−1z,

= y−1x−p+2yxp−2z2,

= · · · ,

= y−1x−p+pyxp−pzp,

= y−1yzp,

= zp = 1.

Second, claim that G has at least p2 − 1 elements of order p.Since f : G → G is a group homomorphism, we know that

G/ ker f ∼= f(G).

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Further, f(G) ⊂ C and |C| = p, so we know that |f(G)| ≤ p. It follows that | ker f | ≥ p2.Besides, ∀x ∈ ker f, x 6= 1, we know that the order of x is p. So G has at least p2−1 elementsof order p.

3.4 (c)

Question: Prove that G has subgroups H and K of orders p2 and p respectively, withH ∩K = e.

Proof: Choose K = C and let H ′ = G/C. It is easy to know that H ′ ∼= H , where H isa subgroup of G and H ∩ C = e.

4 Subgroup of index 2

4.1 (a)

Lemma 4.1. 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 product of m

n-cycles. Deduce that π is an odd permutation if and only if |x| is even and|G||x|

is odd.

Proof. If x is an element of order n, then

[π(x)](1) = x, [π(x)](x) = x2, · · · , [π(x)](xn−1) = xn = 1.

In other words, π(x) should include this n-cycle.

(1, x, x2, · · · , xn−1).

Similarly, for g ∈ G, g /∈< x >, we know that

[π(x)](g) = xg, [π(x)](xg) = x2g, · · · , [π(x)](xn−1g) = xng = g,

which implies that π(x) should include this n-cycle

(g, xg, x2g, · · · , xn−1g).

Since |G| = mn, we can conclude that π(x) is a product of m n-cycles.The Proposition 25 on Page 110, DF, says that “The permutation σ is odd if and only if

the number of cycles of even length in its cycle decomposition is odd. ”By this propositionand the conclusion above, we can directly know that π is an odd permutation if and only if|x| is even and |G|

|x|is odd.

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4.2 (b)

Lemma 4.2. Let G and π be the same as in the preceding Lemma. Prove that if π(G)contains an odd permutation then G has a subgroup of index 2.

Proof. First, claim that for any subgroup of Sn, either all of its elements are even permuta-tions or the number of even permutations is the same as the number of odd permutations.

If σ is an even permutation, then all of {σ, σ2, · · · , } are even permutations. It impliesthat it is possible that a subgroup of Sn only contains even permutations.

However, if σ is an odd permutation, then σ2 is even, and further, σ3 is odd. It meansthat if a subgroup of Sn contains some odd permutations, then it has to contain the samenumber of even permutations.

Second, since π : G → SG, we know that π(G) is a subgroup of Sn where n = |G|.So if π(G) contains an odd permutation, then it has the same number of odd and evenpermutation. Since A′ = An ∩ π(G) is a subgroup of π(G) of index 2, we can take π−1(A′)as the subgroup of G of index 2.

4.3 (c)

Lemma 4.3. Prove that if |G| = 2k where k is odd then G has a subgroup of index 2.

Proof. Recall the Cauchy’s Theorem:

Theorem 4.1 (Cauchy). Let G be a finite group and p be a prime. If p divides |G|, thenG has an element of order p.

Here, 2 is a prime and 2 divides |G| = 2k, so G has an element of order 2, called x. ByLemma 4.1, we know that π(x) is an odd permutation. And by Lemma 4.2, G has subgroupof index 2.

7

MA 553: Homework 3

Yingwei Wang ∗


Note: in this paper, let p > 0 be a prime number.

1 Sylow p-subgroups

Question: For any finite group H , PH denotes the set of all Sylow p-subgroups of H . LetN be a normal subgroup of the finite group G. Prove that

PH = {P ∩N |P ∈ PG}, (1.1)

PG/N = {PN/N |P ∈ PG}. (1.2)

Proof. First, let us prove Eq.(1.1).Choose P ∈ PG, and P1 = P ∩ N . It is obvious that P1 is a p-subgroup of N . Now,

consider the index of P1 in N .

[N : P1] = [N : (P ∩N)] = [PN : P ], (1.3)

[G : P ] = [G : PN ][PN : P ]. (1.4)

Since P ∈ PG, we know that [G : P ] has no factor p. By (1.4), [PN : P ] has not factorp; and by (1.3), [N : P1] has not factor p. It follows that P1 is the Sylow p-subgroups of N .

Second, let us prove Eq.(1.2).Suppose |N | = pmt, |G| = pnts, (p, st) = 1, then |G/N | = pn−ms. Choose P ∈ PG, then

|P | = pn, |P ∩N | = pr, r ≤ m.On one hand, by isomorphism theorem,

PN/N ∼= P/(P ∩N),

⇒ |PN/N | = |P/(P ∩N)| = pn−r, where n− r ≥ m− n. (1.5)


1


On the other hand, it is easy to know that

|G/N | = pn−mt, (1.6)

|PN/N | | |G/N |. (1.7)

By Eq.(1.5)-(1.7), we know that

|PN/N | = pn−m.

It implies that PN/N ∈ PG/N .

2 Normal p subgroup of G

Question: Let G be a finite group and H a normal p subgroup of G. Prove the followingassertions

2.1 (a)

H is contained in each Sylow p-subgroup of G.

Proof. Let P be any Sylow p-subgroup of G. By Sylow theorem, ∃g ∈ G such that

H ⊂ gPg−1,

⇒ g−1Hg ⊂ P.

Since H ⊳G, we know thatg−1Hg = H ⊂ P.

2.2 (b)

If K is any normal p subgroup of G, then HK is a normal p subgroup of G.

Proof. Suppose |G| = pnt, (p, t) = 1 and |H| = pr, |K| = ps, |HK| = px, where r, s, x < n.By Sylow theorem, HK < G.

Furthermore, ∀g ∈ G,

g−1HKg = g−1Hgg−1Kg = HK,

⇒ HK ⊳G.

2


2.3 (c)

The subgroup Op(G) generated by all normal p subgroup of G is equal to the intersectionof all the Sylow p-subgroup of G.

Proof. Let

Op(G) =< ∪Hi >, Hi be the normal p subgroup of G,

S = ∩Pj , Pj be the Sylow p subgroup of G.

On one hand, by Section 2.1, we know that Hi ∈ ∩Pj, for each i. So

Op(G) ⊂ S.

On the other hand, ∀g ∈ S, < g > ⊳G, which means ∃Hi such that g ∈ Hi. Then

S ⊂ Op(G).

Now we can conclude that S = Op(G).

2.4 (d)

Op(G) is the unique largest normal p-subgroup of G.

Proof. Section 2.3 tells us that every normal p-subgroup of G is contained in Op(G), whichmeans Op(G) is the largest normal p-subgroup of G.

Besides, if H is another “largest ”normal p-subgroup of G, then |H| ≥ |Op(G)|. ButH ⊂ Op(G), so H = Op(G). It means that Op(G) is the unique largest normal p-subgroupof G.

2.5 (e)

Op(G) = {1}, where G = G/Op(G).

Proof. Suppose Op(G) = J 6= {1}, then there exists a J ′ such that J ′⊳G and J = J ′/Op(G).

Besides,

|J | = pk1 ,

|J ′| = |Op(G)||J | = pk2.

It follows that J ′ is also a normal p-subgroup of G and |J ′| > |Op(G)|, which contradictswith Section 2.4.

3


3 Groups of order 20

3.1 (a)

Show that the center of the group of transformations

x 7→ ax+ b, (a, b ∈ Z5, a 6= 0) (3.1)

is trivial.

Proof. Denote G as all of the transformations in the form (3.1). Let C(G) be the center ofG. Then ∀g ∈ G, g(x) = ax+ b; ∀c ∈ C(G), c(x) = acx+ bc. We have

gc = cg,

⇒ gc(x) = cg(x), ∀x,

⇒ g(acx+ bc) = c(ax+ b), ∀x,

⇒ aacx+ abc + b = aacx+ acb+ bc, ∀x,

⇒ abc + b = acb+ bc,

⇒ (a− 1)bc = b(ac − 1), ∀a, b ∈ Z5, a 6= 0,

⇒ ac = 1, bc = 0,

⇒ c = identity.

Besides, let r ∈ G such that r(x) = x + 1 and s ∈ G such that s(x) = 2x. Then it iseasy to check that r5 = 1, s4 = 1, and srs−1 = r2. So in this case, G ∼= F20.

3.2 (b)

Let ζ = e2πi/5, a fifth root of unity. Is the group generated by the complex matrices(

0 ii 0

)

and

(

ζ 00 ζ−1

)

isomorphic to the group in (a)?

Proof. No.Let

A =

(

0 ii 0

)

, and B =

(

ζ 00 ζ−1

)

.

It is easy to check that

A4 = I, B5 = I, A−1BA−1 = B−1.

It follows that this group ∼= Z5 ⋊ Z4.

4


3.3 (c)

Show that the Sylow 5-subgroup of a group of order 20 is normal.

Proof. Let G be a group and |G| = 20 = 4 ∗ 5. Suppose the number of Sylow 5-subgroup ofG is k, then by Lagrange theorem and Sylow theorem, we know that k | 4∗5 and k = 5s+1.Since (5s+ 1, 5) = 1, it follows that 5s+ 1 | 4 ⇒ s = 0 ⇒ k = 1.

Now we know that there are only one Sylow 5-subgroup in G. By Corollary 20 on Page142,DF, we can conclude that this Sylow 5-subgroup is normal.

3.4 (d)

Show that there are exactly five distinct groups of order 20.

Proof. By the theorem 3 on page 158 DF, there are 2 abelian group of order 20: Z2×Z2×Z5

and Z20. But for the non-abelian groups, we have to use the “semidirect products”to findthem.

Let |G| = 20 = 22 ∗ 5, H be the Sylow 2-subgroup of G, and K be the Sylow 5-subgroupof G. Besides, we know that K ⊳G.

Case I: H ∼= Z4 and K ∼= Z5. G should be in this form Z5 ⋊ Z4.Consider

ϕ : Z4 → Aut (K) = Z∗

5 = {1, 2, 3, 4}.

If ϕ1 is trivial, then Z5 ⋊ Z4 = Z5 × Z4 = Z20.Suppose ϕ2(1) = 2. Let x = (0, 1), y = (1, 0), then x4 = y5 = 1. Besides,

xy = (0, 1)(1, 0)

= (0 + ϕ2(1)(1), 1 + 0)

= (2, 1),

y2x = (1, 0)(1, 0)(0, 1)

= (1 + ϕ2(0)(1), 0 + 0)(0, 1),

= (2, 0)(0, 1),

= (2 + ϕ2(0)(0), 0 + 1),

= (2, 1),

which means that xy = y2x⇒ xyx−1 = y2. So in this case, G ∼= F20.Suppose ϕ3(1) = 3. Then it is easy to check that in this case, G ∼= F20.

5


Suppose ϕ4(1) = 4. Let x = (0, 1), y = (1, 0), then x4 = y5 = 1. Besides,

xy = (0, 1)(1, 0)

= (0 + ϕ4(1)(1), 1 + 0)

= (4, 1),

yxy = (1, 0)(4, 1)

= (1 + ϕ4(0)(4), 0 + 1),

= (1 + 4, 1),

= (0, 1),

which means yxy = x⇒ x−1yx = y−1. So in this case G ∼= Z5 ⋊ Z4.Case II: H ∼= Z2 × Z2 and K ∼= Z5. G should be in this form Z5 ⋊ (Z2 × Z2).Consider

ϕ : Z2 × Z2 → Aut (K) = Z∗

5 = {1, 2, 3, 4}.

If ϕ1 is trivial, then Z5 ⋊ (Z2 × Z2) = Z5 × (Z2 × Z2) = Z10 × Z2.Let a = (1, 0), b = (0, 1), then H = Z2 × Z2 =< a, b >. Suppose ϕ2(a) = 2, ϕ2(b) = 3.

Choose r = (1, 1, 0), s = (0, 1, 0), then it is easy to know that r10 = 1 and s2 = 1. Besides,

sr = (0, 0, 1)(1, 1, 0)

= (0 + ϕ2(1, 0)(1), 1 + 1, 0 + 0)

= (2, 0, 0),

rsr = (1, 1, 0)(2, 0, 0)

= (1 + ϕ2(1, 0)(2), 1 + 0, 0 + 0),

= (1 + 4, 1, 0),

= (0, 1, 0),

which means rsr = s⇒ rs = sr−1. So in this case G ∼= D12.It is easy to know that for any other ϕ, no new kinds of group appear.In summary, the distinct groups of order 20 are as follows. Let Z5 =< y >,Z4 =< x >

,Z2 × Z2 =< a > × < b >. Then

1. Z20;

2. Z10 × Z2;

3. Z5 ⋊ϕ1Z4 where ϕ1(x)(y) = y−1;

4. F20∼= Z5 ⋊ϕ2

Z4 where ϕ2(x)(y) = y2;

6


5. D20∼= Z5 ⋊ψ (Z2 × Z2) where ψ(a)(y) = y−1 and ψ(b)(y) = y.


Show that the four different groups of order 30 have 1, 3, 5, 15 elements of order 2, respec-tively.

Proof. By the structure theorem, there is only one abelian group of order 30. Now G has 1or 10 Sylow 3-subgroups and 1 or 6 Sylow 5-subgroups. If it has 10 Sylow 3-subgroups and6 Sylow 5-subgroups, it has 45 elements, so it must have a unique Sylow 3- or 5-subgroup.

If there is a unique Sylow 3-subgroup, let c be an element of order 3 so |C(c)| = 15 or30. If |C(c)| = 15, then C(c)is generated by an element of order 15. Otherwise we have anelement of order 5 commuting with an element of order 3 to get our element of order 15.

If there is a unique Sylow 5-subgroup, then |C(c)| = 10, 15, or 30. This similarly givesus an element of order 15. Now we get three possible groups:

< a, b | a15 = b2 = 1, ba = a4b >,

< a, b | a15 = b2 = 1, ba = a11b >,

< a, b | a15 = b2 = 1, ba = a−1b > .


There are 5 distinct groups of order 12, namely Z12, Z2 × Z2 × Z3, A4, D12 and Z3 ⋊ Z4.Similarly as 2(d), Let Z3 =< y >,Z4 =< x >,Z2 × Z2 =< a > × < b >. Then

1. Z12;

2. Z6 × Z2;

3. A4∼= (Z2 × Z2)⋊φ Z3 where φ(y)(a) = b and φ(y)(b) = a;

4. Z3 ⋊ϕ Z4 where ϕ(x)(y) = y−1;

5. D12∼= Z3 ⋊ψ (Z2 × Z2) where ψ(a)(y) = y−1 and ψ(b)(y) = y.

For each of the following groups, determine which of the groups in that example it isisomorphic to.

7


5.1 (a)

The multiplicative group of matrices of the form(

a b0 c

)

(a, b, c ∈ Z3, ac 6= 0).

Proof. Let

r =

(

2 20 2

)

, and s =

(

1 00 2

)

.

Then it is easy to check that

r6 = s2 = 1, rs = sr−1.

It implies that this group ∼= D12.

5.2 (b)

The multiplicative group generated by the complex matrices(

0 ii 0

)

, and

(

ω 00 ω2

)

(i2 = 1, ω3 = 1, ω 6= 1).

Proof. Let

A =

(

0 ii 0

)

, and B =

(

ω 00 ω2

)

.

Then it is easy to check that

A4 = I, B3 = I, A−1BA = B−1.

It follows that this group ∼= Z3 ⋊ Z4.

5.3 (c)

The transformation of the form x 7→ ax+ b, (a 6= 0), of the field F4 into itself.

Proof. This group ∼= A4. See the classification of groups of order 12.

5.4 (d)

The dihedral group D12

Proof. D12∼= Z3 ⋊ψ (Z2 × Z2). See the classification of groups of order 12.

8


5.5 (e)

A non-abelian semidirect product of a group of order 4 by a group of order 3.

Proof. This group ∼= Z3 ⋊ϕ Z4. See the classification of groups of order 12.

6 Groups of order p3

There are two non-abelian groups of order p3. The one is

G1 =< x, y | xp = yp2

= 1, xyx−1 = y1+p > .

The other one is

G2 =< x, a, b | xp = ap = bp = 1, xa = ax, xb = bx, x = aba−1b−1 > .

Let p be an odd prime. Consider the group, of order p3, consisting of all matrices

1 a b0 1 c0 0 1

with a, b, c ∈ Z/p. To which of the two non-abelian groups intentioned above is this groupisomorphic?

Proof. Let

A =

1 1 00 1 00 0 1

, B =

1 0 00 1 10 0 1

, X =

1 0 10 1 00 0 1

.

Then it is easy to know that

Xp = Ap = Bp = I, XA = AX,XB = BX,X = ABA−1B−1.

It follows that this group ∼= G2.

9

MA 553: Homework 4

Yingwei Wang ∗


1 Normal Sylow p-subgroup

Lemma 1.1. Let H be a normal subgroup of a finite group G, and let N ⊳ H be a Sylow

subgroup of H. Prove that N ⊳G.

Proof. Since H ⊳G, we know that gHg−1 = H , ∀g ∈ G. Besides, N ⊂ H , we can get

gNg−1 ⊂ gHg−1 = H, ∀g ∈ G.

By Corollary 20 on Page 142 DF, that N ⊳H is a Sylow p-subgroup of H means that Nis the unique Sylow subgroup of H . Besides, by Sylow theorem, all of the Sylow p-subgroupsof H are conjugate in H . Then

gNg−1 = N, ∀g ∈ G.

It follows that N ⊳G.

2 Nilpotent groups

Lemma 2.1. Let G be a finite group. Let N be a normal subgroup such that G/N is

nilpotent. Suppose that for every Sylow subgroup P of G, PN is nilpotent. Prove that G is

nilpotent.


1


Proof. First, claim that PN ⊳G.Since P is a Sylow subgroup of G and N ⊳G, we know that PN/N is a Sylow subgroup

of G/N . Besides, G/N is nilpotent, so we know that PN/N⊳G/N . It follows that PN⊳G.Second, claim that P is a Sylow subgroup of PN and P ⊳ PN .Since PN < G and P is a Sylow subgroup of G, we know that P is also a Sylow subgroup

of PN . Besides, PN is is nilpotent, so we know that P ⊳ PN .Third, by Lemma 1.1, the above two claims imply that P ⊳G. Then we know that every

Sylow subgroup P of G is normal in G, which means G is nilpotent.

3 Abelian groups

Lemma 3.1. Let n be an integer > 1. Prove the following classification: every group of

order n is abelian if and only if n = pα1

1pα2

2· · · pαr

r , where p1, p2, · · · , pr are distinct primes,

αi = 1 or 2 for all i ∈ {1, · · · , r} and pi does not divide pαj

j − 1 for all i and j.

Proof. First, suppose |G| = n = pα1

1pα2

2· · · pαr

r where p1, p2, · · · , pr are distinct primes,αi = 1 or 2 for all i ∈ {1, · · · , r} and pi does not divide p

αj

j − 1 for all i and j. We want toshow that |G| is abelian.

Let us do induction by r.If r = 1, then |G| = pα1

1, where α1 = 1 or 2. If α1 = 1, then G is cyclic; if α1 = 2, then

G ∼= Zp21or G ∼= Zp1 × Zp1. Anyway, here G is abelian.

Suppose that every group of width at most k which satisfies the above conditions for |G|is abelian. Let G be a group of width r+1 such that |G| = n satisfies the above conditions.Now every proper subgroup of G has width r at most, so that by the induction hypothesis,every proper subgroup of G is abelian. We know that every finite group whose every propersubgroup is abelian is solvable. So G is solvable.

Since G is solvable, we can find H < G such that H ⊳ G and G/H has prime order,which means G/H ∼= Zp. Let P be a Sylow p-subgroup of G. H is the direct product ofits Sylow subgroups, and the Sylow subgroups of H are normal in H . In particular, H isgenerated by elements of prime and prime squared order. Let x ∈ H have prime or primesquared order and let y ∈ P . If < xy >= G, then G is cyclic, hence abelian. If < xy > isa proper subgroup of G, then yx = xy, so that < x >< CG(y). Anyway, we can conclude< x >< CG(y), which implies H < CG(P ) ⇒ HP = G < CG(P ) ⇒ P < Z(G). Similarly,HP = G < C(H), so that H < C(G). Thus HP = G < C(G), and G is abelian.

Second, suppose every group of order n = pα1

1pα2

2· · · pαr

r is abelian. Without loss ofgenerality, if α1 ≥ 3, let Q be a nonabelian group of order pα1

1. (We may take Q to be the

direct product of a nonabelian group of order p3 and a cyclic group.) Then Q × Zpα2

2···p

αrr

is nonabelian of order n. Thus ai ∈ {1, 2} for all i. Now suppose (again without loss of

2


generality) that p1 divides pα2

2. If α2 = 1, then we may construct a nonabelian group Q of

order p1p2. Then Q×Zpα3

3···p

αrr

is a nonabelian group of order n. If a2 = 2, we may construct

a nonabelian group of order pq2. By taking the direct product with a cyclic group, thereexists a nonabelian group of order n. It implies that pi ∤ p

αj

j − 1 for all i and j.Now we can conclude that every group of order n is abelian, then n should satisfy the

conditions enumerated in the problem statement.

4 Classification of finite order

Question: Describe explicitly some groups of order 1, 163, 225 such that every group of thatorder is isomorphic to precisely one of those you have described.

Solution: Since 1, 163, 225 = 52 × 7× 172 × 23, the elementary factors are

{5, 5, 7, 17, 17, 23},

{5, 5, 7, 172, 23},

{52, 7, 17, 17, 23},

{52, 7, 172, 23}.

The invariant factors are

{5× 17, 5× 7× 17× 23},

{5, 5× 7× 172 × 23},

{17, 52 × 7× 17× 17× 23},

{52 × 7× 172 × 23}.

The invariant factor decompositions of the abelian groups of order 1, 163, 225 are asfollows

Z85 × Z13685,

Z5 × Z232645,

Z17 × Z68425,

Z1163225.

Besides, by Lemma 3.1, we know that there is no nonabelian group of order 1, 163, 225.

3


5 Simple groups

5.1 (a)

Lemma 5.1. Show that a simple group which has a subgroup of index n > 2 is isomorphic

to a subgroup of the alternating group An.

Proof. Let G be a simple group and H < G such that [G : H ] = n > 2. Consider the actionof G on the left coset of H ,

ϕ : G → Sn.

Since G is simple, kerϕ = 1 Thus ϕ is injective. It follows that G is isomorphic to asubgroup of Sn. Identify G with its image in Sn we write G < Sn. If G is not contained inAn, then GAn = Sn and

|G||An|

|G ∩An|= Sn,

which implies that G∩An is of index 2 in G and therefore is a normal subgroup of G, whichcontradicts that G is simple. Hence, G < An.

5.2 (b)

Question: What is the smallest index [An : G] occurring for a subgroup G & An?Solution:

For n = 1, 2, An is trivial.For n = 3, A3 only has trivial subgroup.For n = 4, A4 has a proper normal subgroup K ⊳ A4, where

K = {(1), (12)(34), (13)(24), (14)(23)}.

It follows that [A4 : K] = 3.For n ≥ 5, we know that An is simple. Suppose An has a subgroup of index k, then by

Lemma 5.1, An is isomorphic to a subgroup of Ak. It is obvious that |An| ≤ |Ak| ⇒ n ≤ k.So For n at least 5, the index of a subgroup of An is at least n.

5.3 (c)

Question: Show that there is not simple group of order 112.Solution: Let G be a simple group of order 112. We know that 112 = 24 × 7. Suppose

there are n2 Sylow 2 subgroup of G. Sylow’s theorems imply that n2 = 7. (n2 6= 1 since G

4


is simple.) Let H be one of the Sylow 2 subgroups of G. By Lemma 5.1, G is isomorphicto a subgroup of A7.

However, since |G| = 24 × 7 and |A7| = 23 × 3 × 5 × 7, we know that |G| ∤ |A7|, whichmeans G can not be the subgroup of A7.

Therefore there is no simple group of order 112.

5.4 (d)

Question: Show that there is not simple group of order 120.Solution: Let G be a simple group of order 120. We know that 120 = 23×3×5. Suppose

there are n5 Sylow 5 subgroup of G. Sylow’s theorems imply that n5 = 6. Let H be oneof the Sylow 5 subgroups of G, and its normalizer is NG(H). Then [G : N(H)] = 6. ByLemma 5.1, G is isomorphic to a subgroup of A6.

On one hand, since [G : NG(H)] = 6, then |NG(H)| = 20. On the other hand, NG(H) isa subgroup of A6.

Claim that there is no subgroup of A6 of order 20.By the table of groups of small order on page 168 DF, we know that there are 5 distinct

groups of order 20, namely Z20, Z10 × Z2, D20, Z5 ⋊ Z4 and F20. It is easy to check thatnone of them is the subgroup of A6.

This contradiction proves that there is no subgroup of A6 of order 20 and therefore thereis no simple group of order 120.

5.5 (e)

Question: Is every group of order 120 solvable?Solution: No.There are solvable as well as non-solvable groups of this order. All the non-solvable

groups have alternating group A5 and cyclic group Z2 as the composition factors in theircomposition series.

For example, S5 is a non-solvable of order |S5| = 120. We know that

{e} < A5 < S5

is a composition series for S5 (note that S5/A5∼= Z2, which is simple). The Jordan-Holder

Theorem tells us that any composition series for S5 has quotient groups A5 and Z2, and sothere is no composition series with abelian quotients.

5



Question: Find the distinct abelian groups of order 2800 having different number of elementsof order 28.

Solution: Since 2800 = 24 × 52× 7, there are 10 distinct abelian groups. We will discussthem respectively.

1. Z16 × Z25 × Z7, the number of elements of order 28 is 2× 6 = 12;

2. Z16 × Z5 × Z5 × Z7, the number of elements of order 28 is 2× 6 = 12;

3. Z2 × Z8 × Z25 × Z7, the number of elements of order 28 is 2× 2× 6 = 24;

4. Z2 × Z8 × Z5 × Z5 × Z7, the number of elements of order 28 is 2× 2× 6 = 24;

5. Z4 × Z4 × Z25 × Z7, the number of elements of order 28 is 12× 6 = 72;

6. Z4 × Z4 × Z5 × Z5 × Z7, the number of elements of order 28 is 12× 6 = 72;

7. Z2 × Z2 × Z4 × Z25 × Z7, the number of elements of order 28 is 2× 2× 2× 6 = 48;

8. Z2×Z2×Z4 ×Z5×Z5×Z7, the number of elements of order 28 is 2× 2× 2× 6 = 48;

9. Z2 × Z2 × Z2 × Z2 × Z25 × Z7, the number of elements of order 28 is 0;

10. Z2 × Z2 × Z2 × Z2 × Z5 × Z5 × Z7, the number of elements of order 28 is 0.

In conclusion, there are exactly 2 distinct abelian groups of order 2800 having exactly12 elements of order 28; there are exactly 2 distinct abelian groups of order 2800 havingexactly 24 elements of order 24; there are exactly 2 distinct abelian groups of order 2800having exactly 48 elements of order 28; there are exactly 2 distinct abelian groups of order2800 having exactly 72 elements of order 28; there are exactly 2 distinct abelian groups oforder 2800 having exactly 0 elements of order 28.

6

MA 553: Homework 5

Yingwei Wang ∗


Note: in this paper, let M be a fixed commutative monoid with cancelation; [a, b] meansthe least common multiple (LCM) of a and b; (a, b) means the greatest common divisor(GCD) of a and b.

1 LCM and GCD

Lemma 1.1. Prove that if [a, b] exists, then for all c,

1. [ac, bc] = [a, b]c.

2. (a, b) exists, and (ac, bc) = (a, b)c.

Proof. Let d = [a, b], then

a | d, ab | d ⇒ ac | dc, bc | dc ⇒ dc is a common multiple of ac and bc.

Let x be any common multiple of ac and bc, then

ac | x, bc | x,

⇒ x = ace = bcf,

⇒x

c| a,

x

c| b,

⇒ d = [a, b] |x

c,

⇒ dc | x.

It follows that dc = [a, b]c is the least common multiple of ac and bc.By the proposition in class, we know that if [a, b] exists, then

(a, b) =ab

[a, b].


1


Let g = (a, b). Suppose y be any common divisor of ac and bc, then

y | ac, y | bc. (1.1)

Let us focus on y. On one hand, if y ∤ c, then by (1.1), we know that

y | a, y | b,

⇒ y | g,

⇒ y | gc.

On the other hand, if y | c, then it is obvious that y | gc.Anyway, we can get y | gc. It follows that gc = (a, b)c is the greatest common divisor of

ac and bc.

2 LCM and GCD

Lemma 2.1. Prove that [a, b] exists ⇔ (ac, bc) exists for all c.

Proof. ⇒): By Lemma 1.1 (Part two), we know that if [a, b] exists then (a, b) exists and(ac, bc) = (a, b)c exists for all c.

⇐): Since for ∀c, (ac, bc) | ac and (ac, bc) | bc, we know that (ac, bc) | abc. Now let usdenote

d =abc

(ac, bc). (2.1)

Since (ac, bc) | bc, we can rewrite d as

d = abc

(ac, bc),

which means a | d. Similarly, we can get b | d. It follows that d is a common multiple of aand b.

Choose c = 1, then we know that (a, b) exists. Besides, similarly as Lemma 1.1 (Partone), it is easy to know that (ac, bc) = (a, b)c for all c. Now we can rewrite d as

d =ab

(a, b). (2.2)

Leta = a1(a, b), b = b1(a, b),

then, (a1, b1) = 1.

2


Besides, suppose x is any common multiple of a and b, then

x = ae = bf, (2.3)

⇒ x = a1(a, b)e = b1(a, b)f, (2.4)

⇒ a1e = b1f. (2.5)

Now we claim that a1 | f . Since

a1

(a1, f)| a1,

a1

(a1, f)|

b1f

(a1, f),

⇒a1

(a1, f)|

(

a1,b1f

(a1, f)

)

= 1

⇒ a1 = (a1, f),

⇒ a1 | f.

Now we know that

f = a1f1,

⇒ x = bf = b1(a, b)a1f1. (2.6)

Besides,

d =ab

(a, b)= a1(a, b)b1. (2.7)

By Eqs.(2.6) and (2.7), we know that

d | x.

It implies that d is the least common multiple of a and b.

3 Prime

Lemma 3.1. If a is prime and a does not divide b, then [an, b] = anb for all n.

Proof. First, claim that if n = 1, then [a, b] = ab.It it obvious that ab is a common multiple of a and b. Suppose x is any common multiple

of a and b, then

x = ae = bf,

⇒ a | x = bf,

⇒ a | b, or a | f, (since a is prime)

⇒ a | f, (since a ∤ b)

⇒ abf | xf, (since bf = x)

⇒ ab | x.

3


It follows that ab is the least common multiple of a and b.Second, if [an, b] = anb, claim that [an+1, b] = a[an, b] = an+1b.Since

[an+1, b] | [an+1, ab] = a[an, b],

we know that a[an, b] is a common multiple of an+1 and b.Besides, suppose x is any common multiple of an+1 and b, then

[an, b] | [an+1, b] | y,

⇒ anb | y.

Since an+1 | y and a ∤ b, we know further that

an+1b | y.

It implies that an+1b is least common multiple of a and b.By induction, we can get the conclusion.

4 Unit

Lemma 4.1. Suppose a is a unit or a product of prime. Prove that

1. (a, b) and [a, b] exist for all b.

2. If (b, c) exists then (ab, ac)=a(b,c).

Proof. 1. If a is a unit, then it is obvious that (a, b) = a and [a, b] = b for all b.

If a is a prime, then it is easy to know that

[a, b] =

{

b if a | b,ab if a ∤ b.

If a = p1p2 · · · pn, where each pi is a prime, i = 1, · · · , n, then claim that

[a, b] = LCM {[p1, b], · · · , [pn, b]} ,

where

[pi, b] =

{

b if pi | b,pib if pi ∤ b.

By induction, we know that [a, b] exists.

By the proposition in class, we know that if [a, b] exists, then

(a, b) =ab

[a, b].

4


2. It is obvious that

(b, c) | b, (b, c) | c,

⇒ a(b, c) | ab, a(b, c) | ac,

⇒ a(b, c) is a common divisor of ab and ac.

If g is any common divisor of ab and ac, then

ge = ab, gf = ac. (4.1)

If a is a unit, then

a−1ge = b, a−1gf = c,

⇒ a−1g | (b, c),

⇒ g | a(b, c),

which implies that a(b, c) is the GCD of ab and ac.

If a is a prime, then by Eq.(4.1), we know that either g | a or g | b&g | c. It followsthat g | a(b, c), which means a(b, c) is the GCD of ab and ac.

If a is a product of prime, then by induction, we can get the conclusion.

5 GCD

Lemma 5.1. Assume that (x, y) exists for all x, y ∈ M .

1. Prove that if (a, b) = 1, then (ai, bj) = 1 for all (i, j).

2. Without assuming (a, b) = 1, deduce from Part 1 that

(a, b)n = (an, bn)

for all n.

3. Prove that if (a, b) = 1 and ab = cn then a ∼ (a, c)n and b ∼ (b, c)n.

Proof. 1. In class, we know this lemma:

Lemma 5.2. If (ac, bc) exists, and (a, b) = (a, c) = 1, then (a, bc) = 1.

5


Here, by Lemma 5.2, we know that

(a, b) = 1,

⇒ (a, b2) = 1,

⇒ (a, bj) = 1,

⇒ (a2, bj) = 1,

⇒ (ai, bj) = 1,

for ∀i, j.

2. We know that(

a

(a, b),

b

(a, b)

)

= 1,

⇒

(

an

(a, b)n,

bn

(a, b)n

)

= 1, (by Part 1)

⇒ (an, bn) = (a, b)n,

for all n.

3. On one hand, by Part 2, we know that

(a, c)n = (an, cn)

⇒ a | (a, c)n. (since a | an, a | cn) (5.1)

On the other hand, (a, c)n = (an, cn) = (an, b) = (an, ab) = a(an−1, b). By Part 1,

(a, b) = 1,

⇒ (an−1, b) = 1,

⇒ a(an−1, b) | a,

⇒ (a, c)n | a. (5.2)

Now, by Eqs.(5.1) and (5.2), we can conclude that

a ∼ (a, c)n.

Similarly, we can conclude thatb ∼ (b, c)n.

6

MA 553: Homework 6

Yingwei Wang ∗


1 Definition of ring

In each case, decide whether the given structure forms a ring. If it is not a ring, determinewhich of the ring axioms hold and which fail:

1.1 (a)

U is an arbitrary set, and R is the set of subsets of U . Addition and multiplication ofelements of R are defined by the rules

A+B = A ∪B,

A · B = A ∩B.

Solution: This is a ring.

1. (R,+) is an abelian group:

A+B = A ∪ B = B ∪ A = B + A.

2. (R, ·) is associative:

(A · B) · C = A ∩B ∩ C = A ∩ (B ∩ C) = A · (B · C).

3. the distributive laws hold:

A · (B + C) = A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C) = (A · B) + (A · C),

(A +B) · C = (A ∪B) ∩ C = (A ∩ C) ∪ (B ∩ C) = (A · C) + (B · C).


1


1.2 (b)

U is an arbitrary set, and R is the set of subsets of U . Addition and multiplication ofelements of R are defined by the rules

A+B = (A ∪ B)− (A ∩ B),

A · B = A ∩ B.

Solution: This is a ring.


A+B = (A ∪B)− (A ∩ B) = (B ∪ A)− (B ∩A) = B + A.

2. (R, ·) is associative:

(A · B) · C = A ∩B ∩ C = A ∩ (B ∩ C) = A · (B · C).

3. the distributive laws hold:

A · (B + C) = A ∩ [(B ∪ C)− (B ∩ C))] = (A ∩B) ∪ (A ∩ C)− A ∩B ∩ C,

A · B + A · C = (A ∩B) + (A ∩ C) = (A ∩B) ∪ (A ∩ C)−A ∩B ∩ C,

⇒ A · (B + C) = A · B + A · C.

Similarly, we can get

(A+B) · C = (A · C) + (B · C).

1.3 (c)

R is the set of continuous functions R → R. Addition and multiplication are defined bythe rules

[f + g](x) = f(x) + g(x),

[f ◦ g](x) = f(g(x)).

Solution: This is not a ring.


[f + g](x) = f(x) + g(x) = g(x) + f(x) = [g + f ](x).

2


2. (R, ·) is associative:[(f ◦ g) ◦ h](x) = f(g(h(x))) = [f ◦ (g ◦ h)](x).

3. but the distributive laws do not hold:

f ◦ [g + h](x) = f(g(x) + h(x)),

[f ◦ g + f ◦ h](x) = f(g(x)) + f(h(x)),

⇒ f ◦ [g + h] 6= f ◦ g + f ◦ h.

2 Cubic root ring

Let ζ = e2πi/3.

2.1 (a)

Show thatZ[ζ ] := {a+ bζ | a, b ∈ Z}

is a subring of the field C of complex numbers.Solution: It is easy to know that

ζ = e2πi/3 = −1

2+

√3

2i,

ζ2 = e4πi/3 = −1

2−

√3

2i = −1− ζ.

Let z1 = a1 + b1ζ, z2 = a2 + b2ζ ∈ Z[ζ ], then

z1 + z2 = (a1 + a2) + (b1 + b2)ζ ∈ Z[ζ ],

and

z1z2 = (a1 + b1ζ)(a2 + b2ζ),

= a1a2 + b1b2ζ2 + (a1b2 + a2b1)ζ,

= a1a2 + b1b2

(

−1

2−

√3

2i

)

+ (a1b2 + a2b1)ζ,

= a1a2 − b1b2 − b1b2

(

−1

2+

√3

2i

)

+ (a1b2 + a2b1)ζ,

= (a1a2 − b1b2) + (a1b2 + a2b1 − b1b2)ζ ∈ Z[ζ ].

It follows that Z[ζ ] is a subring of C.

3


2.2 (b)

Let the norm of s ∈ Z[ζ ] be defined to be

N(s) := ss,

where s is the complex conjugate of s. Show that for s, t ∈ Z[ζ ],

N(st) = N(s)N(t).

Solution: It is easy to know that

ζζ = |ζ |2 = 1,

ζ + ζ = −1.

Let s = a1 + b1ζ, t = a2 + b2ζ ∈ Z[ζ ], then

N(s) = ss,

= (a1 + b1ζ)(a1 + b1ζ),

= a21+ b2

1|ζ |2 + a1b1(ζ + ζ),

= a21+ b2

1− a1b1.

Similarly, we can getN(t) = a2

2+ b2

2− a2b2.

By Part (a), we know that

st = (a1a2 − b1b2) + (a1b2 + a2b1 − b1b2)ζ.

It follows that

N(st) = (a1a2 − b1b2)2 + (a1b2 + a2b1 − b1b2)

2 − (a1a2 − b1b2)(a1b2 + a2b1 − b1b2),

= (a1a2)2 + (a1b2)

2 + (a2b1)2 + (b1b2)

2 − a1b1a2

2− a1b1b

2

2− a2

1a2b2 − a2b2b

2

1+ a1a2b1b2,

= (a21+ b2

1− a1b1)(a

2

2+ b2

2− a2b2),

= N(s)N(t).

2.3 (c)

Show thats ∈ Z[ζ ] ⇒ s ∈ Z[ζ ].

4


Solution: Let s = a + bζ ∈ Z[ζ ], then

s = a+ bζ,

= a+ b

(

−1

2−

√3

2i

)

,

= (a− b)− b

(

−1

2+

√3

2i

)

,

= (a− b)− bζ ∈ Z[ζ ].

2.4 (d)

Show that s is a unit in Z[ζ ] if and only if N(s) = 1.Solution: On one hand, if ss−1 = 1, then by part (b),

N(1) = N(ss−1) = N(s)N(s−1),

⇒ 1 = N(s)N(s−1),

⇒ N(s−1) =1

N(s).

But both N(s) and N(s−1) are in N, which implies N(s) = N(s−1) = 1.On the other hand, if N(s) = 1, let s = a + bζ ∈ Z[ζ ], then

N(s) = a2 + b2 − ab = 1. (2.1)

So

s−1 =1

a + bζ,

=1

(a− b/2) +√3b/2i

,

=(a− b/2)−

√3b/2i

(a− b/2)2 + 3b2/4,

=a− b

a2 + b2 − ab− b

a2 + b2 − abζ.

By Eq.(2.1), we can gets−1 = (a− b)− bζ ∈ Z[ζ ].

5


2.5 (e)

Show that the group of units in Z[ζ ] consists of all 6-th roots of unity in C.Solution: First, claim that all 6-th roots of unity are in Z[ζ ].

ξ1 = epi/3 =1

2+

√3

2i = 1 + ζ ∈ Z[ζ ],

ξ2 = e2pi/3 = ζ ∈ Z[ζ ],

ξ3 = −1 ∈ Z[ζ ],

ξ4 = e4pi/3 = ζ2 = −1 − ζ ∈ Z[ζ ],

ξ5 = e5pi/3 =1

2−

√3

2i = −ζ ∈ Z[ζ ],

ξ6 = 1 ∈ Z[ζ ].

Second, by part (d), just need to show that if ξ is one of the 6-th roots of unity in C,then N(ξ) = 1. We know that

N(ξj) = ξj ξj = |ξj|2 = 1, j = 1, 2, 3, 4, 5, 6.

3 Nilpotent

An element x of a ring R is called nilpotent if some power of x is zero. Prove that if x isnilpotent, then 1 + x is a unit in R.

Solution: If x is nilpotent, then ∃p ∈ N, such that

xp = (−x)p = 0.

For 1 + x, suppose (1 + x)y = 1, then

y =1

1 + x,

=

∞∑

k=0

(−x)k,

=

p−1∑

k=0

(−x)k ∈ R.

It implies that 1 + x is a unit in R.

6


4 Unit ideal

Prove or disprove: If an ideal I contains a unit, then it is the unit ideal.Solution: Yes.Since ring R contains a unit u, then we know that u−1 is in R.Since u ∈ I and u−1 ∈ R, then uu−1 = 1 ∈ I.But then for any r ∈ R, 1 ∈ I, and thus 1r = r ∈ I.Thus I is all of R.Note: That is why the ideal of R which is all of R is called the “unit ideal ”– it is the

only ideal that could possibly contain a unit.

5 Polynomial ring

Prove that if two rings R,R′ are isomorphic, then so are the polynomial rings R[x] andR′[x].

Solution: Suppose we have an isomorphic

φ : R → R′,

such thatr → φ(r), ∀r ∈ R.

Now we can define a mapφ : R[x] → R′[x],

such that for any p =∑n

k=0akx

k,

φ(p) =

n∑

k=0

φ(ak)xk.

An easy verification proves that φ is a ring isomorphism.

6 Automorphism

Let R be a ring, and let f(y) ∈ R[y] be a polynomial in one variable with coefficients inR. Prove that the map R[x, y] → R[x, y] defined by

φ : x → x+ f(y), y → y, (6.1)

7


is an automorphism of R[x, y].Solution: It is obvious that the map φ : R[x, y] → R[x, y] defined by (6.1) is homo-

morphism. We just need to define the inverse:

φ−1 : x → x− f(y), y → y, (6.2)

which implies that φ−1 is also a homomorphism.It follows that φ is an automorphism.

7 Automorphism

Determine all automorphisms of the polynomial ring Z[x].Solution: Suppose

φ : Z[x] → Z[x]

be an automorphism of Z[x].First, it is easy to know that φ(1) = 1, which means φ(c) = c for all constant c. Hence,

φ is completely determined by φ(x), which can be regarded as an element in Z[x].Second, suppose deg(φ(x)) = d, then for non-constant polynomials f(x) ∈ Z[x],

deg(φ(f(x))) > d,

since f(x) is a linear combination of the power of x. However, since φ is an automorphism,it must be surjective, so there exists non-constant f(x) ∈ Z[x] such that φ((f(x)) = x.Hence, it must be the case that d = 1, which is to say that

φ(x) = αx.

Third, suppose α = pq where p, q ∈ Z, and

φ(g(x)) = qx,

then

φ(x) = αx = pqx, p = φ(p), qx = φ(g(x)),

⇒ φ(x) = φ(p)φ(g(x)) = φ(pg(x)).

However, the only values of p and g for which this could be satisfied are

p = ±, g(x) = ±x.

Finally, there are only two automorphisms of Z[x], φ1 and φ2, where

φ1(x) = x, φ2 = −x.

8


8 Quotient ring

Find a simpler description for each of the following rings.

8.1 (a)

Z[x]/(x2 − 3, 2x+ 4).

Solution: My answer is

Z[x]/(x2 − 3, 2x+ 4) ∼= {a+ bi | a = 0, 1 & b ∈ Z}.

For any p(x) = anxn + · · · + a1x + a0 ∈ Z[x], if deg p(x) = n ≥ 2, we can choose

q1(x) = anxn−2 such that

p(x) = q1(x)(x2 − 3) + r1(x), (8.1)

where deg r1(x) ≤ n− 1.Keep doing this process, we can get

p(x) = (q1 + · · · qk)(x2 − 3) + rk(x), (8.2)

where deg rk(x) ≤ 1. Supposerk(x) = a1x+ a0.

If a1 = 0, 1, then it is fine; if a1 ≥ 2, then we can get

a1 = 2s+ t,

where t = 0, 1.So we have

rk(x) = s(2x+ 4) + tx+ (a0 − 4s), (8.3)

where t = 0, 1.By Eqs.(8.2)-(8.3), we can get

p(x) = (q1 + · · · qk)(x2 − 3) + s(2x+ 4) + tx+ (a0 − 4s). (8.4)

It is implies that

Z[x]/(x2 − 3, 2x+ 4) ∼= {a+ bi | a = 0, 1 & b ∈ Z}.

9


8.2 (b)

Z[i]/(2 + i).

Solution: My answer isZ[i]/(2 + i) ∼= Z5.

For any a+ bi ∈ Z[i], we know that

a + bi = b(2 + i) + (a− 2b). (8.5)

Besides, we know that(2− i)(2 + i) = 5.

If (a− 2b) < 5, then it is fine; if (a− 2b) ≥ 5, then we can do further:

a− 2b = 5q + r = (2 + i)(2− i)q + r, (8.6)

where r = 0, 1, 2, 3, 4.By Eqs.(8.5)-(8.6), we can get

a+ bi = (b+ 2q − qi)(2 + i) + r, (8.7)

where r = 0, 1, 2, 3, 4.Now it is clear that

Z[i]/(2 + i) ∼= Z5.

10

MA 553: Homework 7

Yingwei Wang ∗


1 Fractions of ring

Let p be a prime ideal in an integral domain R, and let M consist of all elements in R lyingoutside p (so that M is a multiplicative submonoid of R). In this case it is customary todenote the ring of fractions RM by Rp .

1.1 (a)

Show that if q is any prime ideal in Rp, then q ∩R is a prime ideal in R, contained in p;and that one obtains in this way a one-one correspondence between all prime ideals in Rp

and those prime ideals in R which are contained in p.

Solution: We know that the homomorphism

θ : R → Rp

is given by

θ(r) =r

1.

It implies that for any r ∈ R, we can regard r as an element in Rp.Suppose q is any prime ideal in Rp, then ∀x, y ∈ q ∩ R, either x ∈ q or y ∈ q. But

x, y ∈ R, so either x ∈ q ∩ R or y ∈ q ∩R. It implies that q ∩R is a prime ideal in R.Besides, since the kernel of θ is p, so q ∩R is contained in p.Now we have a one-one correspondence between all prime ideals in Rp and those prime

ideals in R which are contained in p:

φ : q ↔ q ∩ R.


1


1.2 (b)

Definition 1.1. A commutative ring R is called a local ring if it has a unique maximal

ideal.

Show that Rp is a local ring.

Solution: We know that every maximal ideal is prime. But by part (a), we also knowthat any prime ideal q in Rp, q ∩ R is a prime ideal in R and contained in p.

Then φ−1(p ∩R) is the unique maximal ideal in Rp, which means Rp is a local ring.

2 Category

Let R be a commutative ring and let M and N be commutative monoids. With coordi-natewise multiplication, M ×N is then also a commutative monoid.

2.1 (a)

Consider the category T of triples (S, µ, ν) such that S is a commutative R-algebra andµ : M → S and ν : N → S are monoid homomorphisms, maps between such triples beingdefined in the obvious way.

Find monoid homomorphisms µ1 : M → (R[M ])[N ] and ν1 : N → (R[M ])[N ], µ2 :M → R[M×N ] and ν2 : N → R[M×N ], such that both ((R[M ])[N ], µ1, ν1) and (R[M×

N ], µ2, ν2) are initial objects in T ; and deduce that there is an R-algebra isomorphism

α : (R[M ])[N ] → R[M ×N ],

such that

α

(

∑

n∈N

(∑

m∈M

rmnm)n

)

=∑

(m,n)∈M×N

rmn(m,n).

Solution: First, we haveϕ1 : M → R[M ],

in whichϕ1(m) = polynomial with value 1 at m and 0 eleswhere

Second, we also haveφ1 : R[M ] → (R[M ])(N),

2


in whichφ1(r) = fr,

wherefr(e) = r, fr(n) = 0, n 6= e.

Now just define µ1 : M → (R[M ])[N ] in a natural way µ1 = φ1 ◦ ϕ1, which means

µ1(m) = m. (2.1)

Note that the m in the right hand side of the Eq.(2.1) should be regarded as an elementin (R[M ])[N ].

Similar definition can be done to ν1 : N → (R[M ])[N ] in which

ν1(n) = n. (2.2)

Note that the n in the right hand side of the Eq.(2.2) should be regarded as an elementin (R[M ])[N ].

Besides, we have these homomorphisms in natural way

ϕ2 : M → M ×N,

andφ2 : M ×N → R[M ×N ].

So we can define µ2 : M → R[M ×N ] as µ2 = φ2 ◦ ϕ2, which means

µ2(m) = (m, 1N). (2.3)

Similarly, we can define ν2 : N → R[M ×N ], which means

ν2(n) = (1M , n). (2.4)

Suppose the universal property of monoid algebras holds here, for any S in this cate-gory T, we can define f : (R[M ])[N ] → S as

f

(

∑

n∈N

(∑

m∈M

rmnm)n

)

=∑

m∈M,n∈N

rmnµ1(m)ν2(n). (2.5)

Besides, we can define g : R[M ×N ] → S as

g

∑

(m,n)∈M×N

rmn(m,n)

=∑

m∈M,n∈N

rmn(µ2(m), ν2(n)). (2.6)

3


It follows that both ((R[M ])[N ], µ1, ν1) and (R[M × N ], µ2, ν2) are initial objects inT. That is to say there exists a unique R-algebra homomorphism

α1 : (R[M ])[N ] → R[M ×N ],

and a unique R-algebra homomorphism

α2 : R[M ×N ] → (R[M ])[N ].

It follows that α1 = α−12 , which means there is an R-algebra isomorphism

α : (R[M ])[N ] → R[M ×N ].

2.2 (b)

Explain carefully how the isomorphism α specializes to give isomorphisms of polynomialrings, such as

(R[W,X ])[Y, Z] → R[W,X, Y, Z].

Solution: By part(a), we just need to choose M = WX and N = Y Z. and define αin this way

α

(

∑

y∈Y,z∈Z

(

∑

w∈W,x∈X

rwxwx

)

yz

)

=∑

w∈W,x∈X,y∈Y,z∈Z

rwxyz(wx, yz),

which meansα : (R[W,X ])[Y, Z] → R[WX × Y Z]

is an isomorphism.Besides, we know that there is a natural isomorphism

π : WX × Y Z → WXY Z,

in whichπ(wx, yz) = wxyz.

Then we have an isomorphism

π : R[WX × Y Z] → R[W,X, Y, Z]

4


defined by

π

(

∑


rwxyz(wx, yz)

)

=∑


rwxyzπ(wx, yz).

Now we know that

β : (R[W,X ])[Y, Z] → R[WX × Y Z] → R[W,X, Y, Z],

where β = π ◦ α is an isomorphism.

2.3 (c)

Let φ : M → N be a monoid homomorphism, and let θ : R[M ] → R[N ] be the cor-responding R-algebra homomorphism (given by the universal property of R[M ], so thatθ(∑

rmm) =∑

rmφ(m)). Show that the kernel of θ is generated by the set of elementsof the form 1.m− 1.m with φ(m) = φ(m).

Solution: Since 0 is obvious in the kernel of θ, we can assume rm 6= 0 such that

θ(

∑

rmm)

=∑

rmφ(m) = 0.

Let f =∑

rmm ∈ R[M ].For each n ∈ N , if there are two element m,m′ such that φ(m) = φ(m′), then

∑

rmθ(m) =∑

(rm + rm′)θ(m).

It follows that in order to get θ(f) = 0 ∈ R[N ], we should let

rm + rm′ = 0,

which meansf = rm(m−m′).

Besides, if there are three elements m1, m2, m3 ∈ M such that φ(m1) = φ(m2) =φ(m3) = n, then similarly, in order to get θ(f) = 0 ∈ R[N ], we have

rm1 + rm2 + rm3 = 0 ⇒ rm3 = −rm1 − rm2.

Sof = rm1m1 + rm2m2 + rm3m3 = rm1(m1 −m3) + rm2(m2 −m3).

By induction we know that the kernel of θ is generated by the set of elements of theform 1.m− 1.m with φ(m) = φ(m).

5


3 Commutative monoid with cancellation

Show that a monoid M is a commutative monoid with cancellation if and only if thereexists an injective monoid homomorphism from M into an abelian group.

Solution: On one hand, suppose there exist an injective monoid homomorphism

φ : M → R,

where R is an abelian group.Then φ(M) is a submonoid of R. Now

φ : M → φ(M) ⊂ R,

is an isomorphism. Since R is abelian, we know that M is a commutative monoid withcancellation.

On the other hand, suppose M is a commutative monoid with cancellation, we justneed to do the “localization ”with respect to M .

Find a ring R such that M is a multiplication submonoid of R. Define a equivalencerelation on pairs (r,m) ∈ R×M ,

(r1, m1) ≡ (r2, m2) if r1m2 = r2m1.

Denoter/m := equivalence class of (r,m).

Then define

r1m1

+r2m2

=r1m2 + r2m1

m1m2

,

r1m1

·r2m2

=r1r2m1m2

.

Now defineλ : R → RM

by

λ(r) =r

1,

such that λ(m) is a unit for every element m ∈ M .Now λ is a ring injective homomorphism.

6

MA 553: Homework 8

Yingwei Wang ∗


1 Quadratic norm-Eculidean domain

1.1 (a)

Let α and β be rational numbers with |α| ≤ 1/2, and let m > 0 be an integer such that

α2 −mβ2 = −1 − δ, (1.1)

where 0 ≤ δ < 1. Set

ǫ =

{

1, if α ≥ 0−1, if α < 0.

(1.2)

Show that if m is not of the form 5n2(n ∈ Z) then

|(α+ ǫ)2 −mβ2| < 1. (1.3)

Solution: It is easy to know that

|(α+ ǫ)2 −mβ2| = |α2 + 2αǫ+ ǫ2 −mβ2| = |2αǫ− δ|. (1.4)

By the definition of ǫ, we know that if 0 ≤ α < 1/2, then

2αǫ = 2α ∈ (0, 1),

⇒ (2αǫ− δ) ∈ (−1, 1),

⇒ |2αǫ− δ| < 1.


1


Similarly, if −1/2 < α < 0, then

2αǫ = −2α ∈ (0, 1),

⇒ (2αǫ− δ) ∈ (−1, 1),

⇒ |2αǫ− δ| < 1.

Besides, if α = ±12, thanks to the fact that m is not of the form 5n2(n ∈ Z), then

α2 − mβ2 = −1 − δ, where δ ∈ (0, 1) (note that here δ 6= 0). Now we can also get theinequality (1.3).

1.2 (b)

Deduce that Z[ω] is norm-Euclidean when ω =√6 or ω =

√7.

Solution: Let ω2 − q = 0, then q = 6 or 7. For any α, β ∈ Q, e, f ∈ Z, then

N((α + βω)− (e+ fω)) = (α− e)2 − q(β − f)2. (1.5)

First, choose e1 such that |α− e1| ≤ 1/2; second, choose f such that

(α− e1)2 − q(β − f)2 = −1 − δ,

where 0 ≤ δ < 1; third, choose ǫ such that

ǫ =

{

1, if α− e1 ≥ 0−1, if α− e1 < 0.

Let e = e1 − ǫ ∈ N. Then by part (a), we can get

|(α− e)2 − q(β − f)2| < 1,

where q = 6 or 7.It follows that Z[ω] is norm-Euclidean when ω =

√6 or ω =

√7.

1.3 (c)

Deduce that Z[ω] is norm-Euclidean when ω2 − ω + q = 0 with q = −4,−5,−7.

2


2 Primes in polynomial ring

Let R be a UFD, with fraction field K. Suppose you already have computer algorithms forfactoring into primes in R and in the polynomial ring K[X ]. Describe briefly how you wouldinstruct a computer to factor into primes in R[X ].

Solution: Choose f(x) ∈ R[X ], and factor f(x) into primes in K[X ], we can get

f(x) = c(x− a1/b1)(x− a2/b2) · · · (x− an/bn).

In order to have f(x) ∈ R[X ], we should have

c = dn∏

j=1

bj .

Now it is clear that

f(x) = d(b1x− a1)(b2x− a2) · · · (bnx− an).

Factoring d into primes in R, we can get

d = d1d2 · · · dn.Now, we can factor f(x) into primes in R[X ],

f(x) = d1d2 · · · dn(b1x− a1)(b2x− a2) · · · (bnx− an).

Note that aj and bj should not have any common factor for any j.

3 Prime and irreducible polynomial

Let k be a field, x, y, and z indeterminates.

3.1 (a)

Let f(x) and g(x) be relatively prime polynomials in k[x]. Show that in the polynomial ringk(y)[x], f(x)− yg(x) is irreducible.

Proof. We know that k(y)[x] ∼= k(x)[y]. So we can see f(x)− yg(x) as an element in k(x)[y].If it is not irreducible, then

f(x)− yg(x) = (a(x)y + b(x))c(x),

⇒ f(x) = a(x)c(x), g(x) = b(x)c(x),

which contradicts with the fact that f(x) and g(x) are relatively prime.

3


3.2 (b)

Prove that in k(y, z)[x], the polynomial

f(x, y, z) = x4 − yzx3 + (y2z2 − y)x2 + (y2z − y)x+ y2z,

is irreducible.

Proof. We can rewrite f(x, y, z) as

f(x) = x2y2z2 + (xy2 + y2 − yx3)z − yx2 − yx+ x4,

= (xy)2z2 + y[(x+ 1)y − x3]z − x[(x+ 1)y − x3],

= a2z2 + a1z + a0

Choose P be the ideal generated by (x+1)y−x3. By part (a), P is a prime ideal. Besides,a1, a0 ∈ P while a2 /∈ P and a0 /∈ P 2. By Eisenstein theorem, we know that f(x, y, z) isirreducible.

4 Polynomial ring

Let R be an integral domain with fraction field K, let R[X ] be a polynomial ring, and let aand b be nonzero elements in R.

4.1 (a)

If R is a UFD and P ⊂ R[X ] is a prime ideal with P ∩R = (0), then P is a principal ideal.

Proof. There is a natural map π : R[X ] → K[X ] defined by

π(f(x)) = f(x).

If P is a prime ideal in R[X ], then P is a principal ideal in K[X ], which means

P =

(

f(x)

d

)

,

where f(x) ∈ R[X ].

4


It implies that ∀hinP ,

h(x) =f(x)

dg(x),

= f(x)g(x)

d.

Since f, h ∈ R[X ], we know that g(x)d

∈ R[X ]. It follows that P is generated by f(x). Hence,P is a principle ideal.

4.2 (b)

aR ∩ bR = abR iff the ring R[X ]/(aX − b) is an integral domain.

Proof. On one hand, suppose aR ∩ bR = abR. Then we know that aR ∩ bR = [a, b]R soab = [a, b], which means (a, b) = 1. It follows that aX − b is irreducible.

Let P = (aX − b). Then for any fg ∈ P , f(X)g(X) = (aX − b)h(X). So we have eitheraX − b | f(X) or aX − b | g(X). It implies that P is a prime ideal. Hence, R[X ]/(aX − b)is an integral domain.

On the other hand, if R[X ]/(aX−b) is an integral domain, then (aX−b) is a prime ideal.So aX − b is irreducible. It follows that (a, b) = 1, which implies that aR ∩ bR = abR.

4.3 (c)

If c = aq = bp is a nonzero common multiple of a and b then c is an l.c.m. of a and b iffpX − q is a prime element in R[X ].

Proof. On one hand, if c = [a, b], then (p, q) = 1. By part(b), we know that R[X ]/(pX − q)is an integral domain. It follows that pX − q is a prime element in R[X ].

On the other hand, if (pX − q) is a prime ideal. So pX − q is irreducible. It follows that(p, q) = 1, which implies that c = aq = bp = [a, b].

4.4 (d)

An l.c.m. [a, b] exists iff the kernel of the R-homomorphism φ : R[X ] → R[ ba] ⊂ K taking X

to bais a principal ideal.

5


Proof. If f(X) ∈ ker(φ), then

φ(f(X)) = f(b/a) = 0,

which means b/a is a root of f(x) in K. So

f(X) = (aX − b)g(X). (4.1)

Now we can prove the conclusion.

On one hand, if [a, b] exists, then (a, b) exists, and(

a(a,b)

, b(a,b)

)

= 1. By Eq.(4.1), we can

get

f(X) = (aX − b)g(X),

=

(

a

(a, b)X − b

(a, b)

)

(a, b)g(x),

= (pX − q)g(x),

where p = a(a,b)

, q = b(a,b)

. By part (c), we know that the kernel of φ is a principle ideal.

On the other hand, if the kernel of φ is a principle ideal, which is generated by (pX − q)(Since aX − b ∈ ker(φ), so the generator should be in this form.) By part (b) and (c), wehave (p, q) = 1. So

aX − b = c(px− q),

⇒ a = cp, b = cq.

Now cpq = [a, b].

5 Quotient ring

5.1 (a)

Prove that if x 6= 0 and y are elements in a UFD such that x2 divides y2, then x divides y.

Proof. In a UFD, we have

x = pα1

1 pα2

2 · · · pαn

n ,

y = pβ1

1 pβ2

2 · · ·pβn

n ,

where {pj} are primes and αj , βj ≥ 0.

6


Now x2 | y2 means

2αj ≤ 2βj ,

⇒ αj ≤ βj ,

⇒ x | y.

5.2 (b)

Let k be a field. In the quotient ring R = k[X, Y, Z]/(Y 2 − X2Z) let x = X and y = Y bethe natural images of X and Y . Show that x2 divides y2 in R, but x does not divide y.

Proof. By the definition of the quotient ring, we know that

x = X = f(X, Y, Z)(Y 2 −X2Z) +X, y = Y = g(X, Y, Z)(Y 2 −X2Z) + Y,

⇒{

x2 = (Y 2 −X2Z)(f 2Y 2 − fX2Z + 2f) +X2,y2 = (Y 2 −X2Z)(g2Y 2 − gX2Z + 2g) + Y 2 = (Y 2 −X2Z)(g2Y 2 − gX2Z + 2g + 1) +X2Z,

⇒ x2 | y2.

But it is obvious that x can not divides y since Y can not be written in the form (Y 2 −X2Z)q + rX .

5.3 (b)

Is R an integral domain? (Why?)

Proof. No. Since if R is an integral domain, then the cancelation law holds. But by part (b),we know that it does not holds.

6 Fermat equation

Find all solutions in positive integers of the equation

y3 = x2 + 4. (6.1)

7


6.1 (a)

a+ bi is divisible by 1 + i ⇔ a− b is even.

Proof. On one hand, suppose a+ bi is divisible by 1 + i, then

a+ bi = (1 + i)(c+ di) = (c− d) + (c+ d)i,

⇒ a− b = 2d.

which implies that a− b is even.On the other hand, if a − b is even, then it is easy to know that a + b is also even. We

can suppose that

a− b = 2k,

a + b = 2l,

which implies that

a = l − k

b = l + k.

Now we can find l + ki such that

(1 + i)(l + k) = (l − k) + (l + k)i = a+ bi,

which means a+ bi is divisible by 1 + i.

6.2 (b)

If y3 = x2 + 4, (x, y ∈ Z), then

(x+ 2i, x− 2i) =

{

1 if x is odd,(1 + i)3 if x is even,

, (6.2)

Proof. If x = 2k + 1, then

(x+ 2i, x− 2i) = (2k + 1 + 2i, 2k − 1− 2i) = 1.

If x = 2k, then

(x+ 2i, x− 2i) = (2k + 2i, 2k − 2i) = −2 + i = (2i)(1 + i) = (1 + i)3.

8


6.3 (c)

If y3 = x2 + 4, then x+ 2i = in(a+ bi)3, for some n, a, b.

Proof. By part (b), we know that

(x+ 2i, x− 2i) = z3,

where z = 1 or z = 1 + i.It follows that

(

x+ 2i

z3,x− 2i

z3

)

= 1,

Now we can get

y3 = (x+ 2i)(x− 2i),

⇒( y

a2

)3

=

(

x+ 2i

z3

)(

x− 2i

z3

)

,

⇒ x+ 2i

z3= (a+ bi)3,

⇒ x+ 2i = (z(a + bi))3 = in(a+ bi)3.

Let n = 0, then

x+ 2i = (a + bi)3 = (a3 − 3ab2) + (3a2b− b3)i,

⇒ 3a2b− b3 = 2,

⇒ b(3a2 − b2) = 2,

⇒ b = 1, a = ±1; or b = −2, a = ±1,

⇒ x = 2, y = 2; or x = 11, y = 5.

So the equation (6.1) has two solutions over positive integers:

{

x = 2,y = 2,

or{

x = 11,y = 5,

9

MA 553: Homework 9

Yingwei Wang ∗


1 Quadratic norm-Eculidean domain

Let ω ∈ C satisfy ω2 − pω + q = 0 where p and q are integers such that p2 − 4q is not thesquare of an integer. The norm of a+ b ∈ Z[ω] is

N(a+ bω) : = (a + bω)(a+ bω),

= (a + bω)(a+ b(p− ω)).

It was shown that if (a, b) = 1, then the natural map is an isomorphism

Z/(N(a + bω))Z → Z[ω]/(N(a+ bω))Z[ω].

Prove that for any a, b ∈ Z, |N(a + bω)| is the cardinality of Z[ω]/(N(a + bω))Z[ω]

Proof. First, I want to show that

φ : Z⊕ Z → Z[ω]

is a group isomorphism. Just define φ as

φ(a, b) = a+ bω,

where (a, b) ∈ Z⊕ Z, a+ bω ∈ Z[ω].Then it is easy to know that φ is a group isomorphism.Second, I want to show that

ϕ : Z[ω] → (a+ bω)Z[ω]


1


is also a group isomorphism. Here, for any x+ yω ∈ Z[ω], we define

ϕ(x+ yω) = (a + bω)(x+ yω),

= ax+ byω2 + (ay + bx)ω,

= ax+ by(pω − q) + (ay + bx)ω,

= (ax− bqy) + (bpy + ay + bx)ω.

Note that a and b does not equal to 0 at the same time.It is obvious that ϕ is a homomorphism. Besides, since (a+ bω)Z[ω] ⊂ Z[ω], we just need

to show that ϕ is injective, which means kerϕ = 0. For any x+ yω ∈ kerϕ,

ϕ(x+ yω) = (a + bω)(x+ yω) = 0,

⇒ x+ yω = 0,

since a+ bω 6= 0.Now we have this group isomorphism:

Z ⊕ Z ∼= Z[ω] ∼= (a+ bω)Z[ω]. (1.1)

Third, we know that

Z[ω]

(a+ bω)Z[ω]=

Z[ω]

(a/(a, b) + bω/(a, b))Z[ω]· (a/(a, b) + bω/(a, b))Z[ω]

(a+ bω)Z[ω](1.2)

Since(

a

(a,b), b

(a,b)

)

= 1, we know that

cardinality ofZ[ω]

(a/(a, b) + bω/(a, b))Z[ω]=

|N(a + bω)|(a, b)2

. (1.3)

Besides, by the isomorphism (1.1), we know that

(a/(a, b) + bω/(a, b))Z[ω] ∼=(

a

(a, b)

)

Z ⊕(

b

(a, b)

)

Z,

(a+ bω)Z[ω] ∼= aZ ⊕ bZ.

It follows that

cardinality of(a/(a, b) + bω/(a, b))Z[ω]

(a+ bω)Z[ω]= cardinality of

(

a(a,b)

)

Z ⊕(

b(a,b)

)

Z

aZ ⊕ bZ= (a, b)2.

(1.4)

2


By Eqs.(1.2)-(1.4), we can get

cardinality ofZ[ω]

(a+ bω)Z[ω],

= cardinality ofZ[ω]

(a/(a, b) + bω/(a, b))Z[ω]× cardinality of

(a/(a, b) + bω/(a, b))Z[ω]

(a+ bω)Z[ω],

=|N(a+ bω)|

(a, b)2× (a, b)2 = |N(a + bω)|.

2 Square in Z/π

Assume that Z[ω] (as in section 1) is a UFD. Let π be a Z-prime. Suppose there are integersa and b, not both divisible by π, such that π divides a2 + pab + qb2. Show that there areintegers c and d such that π = ±(c2 + pcd + qd2). Deduce from this that if e = 1 or e = 2,then π is of the form x2 + ey2 ⇔ −e is a square in Z/π.

Proof. We know that

π | a2 + pab+ qb2,

⇒ π | (a+ bω)(a+ bω), but π ∤ a+ bω, π ∤ a+ bω,

⇒ π is not a prime,

⇒ π is not irreducible,

⇒ π is reducible,

⇒ π = ±(c + dω)(c+ dω),

⇒ π = ±(c2 + pcd+ qd2).

Furthermore, if π = 2, then for e = 1, we have

2 = x2 + y2,

⇔ x = 1, y = 1,

⇔ −1 is a square in Z/2.

Similarly, for e = 2, we have

2 = x2 + 2y2,

⇔ x = 0, y = 1,

⇔ −2 is a square in Z/2.

3


For the case that π ≥ 3, we claim that π = x2 + ey2 | a2 + b2, but not both a and bdivisible by π, where e = 1. For e = 1, we know that x2 + y2 | 4x2 + 4y2. So we can choosea = 2x, b = 2y, then it is fine.

Similarly, for π ≥ 3 and e = 2, we can choose a = 2x, b = 2y, then π = x2 + ey2 | a2 + b2,but not both a and b divisible by π.

By previous conclusion, we know that

x2 + ey2 ⇔ −e is a square in Z/π.

3 Primes

Let ω 6= −1 be a complex number satisfying ω3 = −1. We showed in class that Z[ω] is aEuclidean domain.

3.1 (a)

Let p > 3 be an odd prime in Z. Show that:

p ≡ 1( mod 6) ⇔ −1 has three cubic roots in Z/π ⇔ −3 is a square in Z/π. (3.1)

Proof.

p ≡ 1( mod 6),

⇔ |(Z/π)∗| divided by 6,

⇔ (Z/π)∗ has an element of order 6,

⇔ −1 has three cubic roots in Z/π. (3.2)

Besides,

p ≡ 1, ( mod 6),

⇔ p = 6n+ 1,

⇔ (Z/π)∗ has an element of order 6n,

⇔ −3 is a square in Z/π, since (−3)3n ≡ 1, ( mod p) (3.3)

By Eq.(3.2)-(3.3), we can get (3.1).

4


3.2 (b)

Prove that every prime p > 0 in Z of the form p = 6n + 1 can be represented in the formp = a2 + ab+ b2(a > b > 0) in one and only one way.

Proof. Consider the UDF Z[ω] where ω satisfies

ω2 + ω + 1 = 0.

By part (a), we know that −3 is a square in Z/p. Hence for some a, b ∈ Z,

p = a2 + ab+ b2.

Furthermore, if we set a > b > 0, then the a and b should be unique.

3.3 (c)

Prove that every prime p > 0 in Z of the form p = 6n + 1 can be represented in the formp = a2 + 3b2(a, b > 0) in one and only one way.

Proof. Consider the UDF Z[ω] where ω satisfies

ω2 + 3 = 0.

Since −12 = −3 × 22 and −3 is a square in Z/p, we know that for some a, b ∈ Z,

p = a2 + 3b2.

Furthermore, if we set a > b > 0, then the a and b should be unique.

3.4 (d)

Prove that every odd prime p in Z factors into primes in Z[√−3]. What about p = 2?

Proof. Let p be a prime and p = 2n+ 1. Consider n by different cases.

1. If n = 3k, then p = 6k+1. By part (c), we know that p factors into primes in Z[√−3].

2. If n = 3k + 1, then p = 6k + 3 = 3(2k + 1). Now p is not a prime, which is impossible.

3. If n = 3k + 2, then p = 6k + 5. Now p can also factor into primes in Z[√−3].

5

MA 553: Homework 10

Yingwei Wang ∗


1 Field extension

Question: Let K ⊂ L be a field extension of finite degree, and let f ∈ K[X ] be an irreduciblepolynomial whose degree is relatively prime to [L : K]. Show that f is irreducible in L[X ].

Proof. Suppose f is reducible in L[X ]. But we know that f is irreducible in K[X ]. So ∃α ∈ Lsuch that f(x) is the minimal polynomial of α in K. Consider F = K(α), then

K ⊂ F ⊂ L. (1.1)

Let [L : K] = m, deg(f) = n, then [F : K] = n, by (1.1),

m = [L : F ]n,

which contradicts with (m,n) = 1.Thus f should be irreducible in L[X ].

2 Factor into irreducibles

2.1 (a)

Question: Factor X16 −X into irreducible polynomials over F4 and over F8.Solution: In the textbook, we know that over F2, the X16 − X can be factored into

irreducible polynomials as

X16−X = X(X− 1)(X2+X +1)(X4+X +1)(X4+X3+1)(X4+X3+X2+X +1). (2.1)

In order to get the factorization over F4, we just need to do factoring based on (2.1).


1


Suppose F4 = {0, 1, α, α2}. It is easy to know that over F4,

X2 +X + 1 = (X − α)(X − α2), (2.2)

X4 +X + 1 = (X2 −X + α)(X2 +X + α2), (2.3)

X4 +X3 + 1 = (X2 + αX + α)(X2 − α2X + α2), (2.4)

X4 +X3 +X2 +X + 1 = (X2 − αX + 1)(X2 + α2X + 1). (2.5)

Now, we know that over F4,

X16 −X

= X(X − 1)(X − α)(X − α2)(X2 −X + α)(X2 +X + α2)

(X2 + αX + α)(X2 − α2X + α2)(X2 − αX + 1)(X2 + α2X + 1).

For F8, we know that

F8 = F2[X ]/(X3 +X2 + 1),

= {0, 1, x, x+ 1, x2, x2 + 1, x2 + x, x2 + x+ 1}.

2.2 (b)

Question: Use Maple to factor X80 − 1 mod 3. How many irreducible polynomials of degree4 are there in Z3[X ]? What about degree 3?

Solution: The answer is

X80 − 1 = X(X + 2)

(

1 +

71∑

k=5

Xk

)

.

There are 18 irreducible polynomials of degree 4 are there in Z3[X ]. And there are 8irreducible polynomials of degree 4 are there in Z3[X ].

2.3 (c)

Question: Display the set of all irreducible polynomials of degree 2 over a field of cardinality9.

Solution:

2


3 Irreducible polynomials over finite field

3.1 (a)

Question: Show that the polynomials f(X) = X3 + X2 + 1 and g(X) = X3 + X + 1 areirreducible over the field F4 with four elements.

Proof. We know that F2 = {0, 1}, and it is easy to check that both f(X) and g(X) areirreducible in F2[X ], whose degrees are 3. Besides, [F4 : F2] = 2. According to Problem 1, weknow that f(X) and g(X) are irreducible in F4[X ].

3.2 (b)

Question: Describe explicitly an isomorphism between the field F4[X ]/(f(X)) and F4[Y ]/(g(Y )),both of which have cardinality 64.

Solution: The isomorphism is given by

Y = X + 1.

We just need to check that α(x) = x + 1 is a root of f(X) = X3 + X2 + 1 in the fieldF4[X ]/(g(X)).

f(x+ 1) = (x+ 1)3 + (x+ 1)2 + 1,

= x3 + 4x2 + 5x+ 3,

= x3 + x+ 1,

= 0.

4 Minimal polynomial

Question: Determine the minimal polynomial for α =√3 +

√5 over each of the following:

4.1 (a): Q.

Solution:

α =√3 +

√5,

⇒ α2 = 8 + 2√15,

⇒ α2 − 8 = 2√15,

⇒ α2 − 16α2 + 64 = 60,

⇒ α2 − 16α2 + 4 = 0,

which means the minimal polynomial of α over Q is

X4 − 16X2 + 4.

3


4.2 (b): Q(√5).

Solution: Suppose α2 + aα + b = 0, where a = a1 + a2√5, b = b1 + b2

√5. It is easy to find

that a1 = 0, a2 = −2, b1 = 2, b2 = 0. Then the minimal polynomial of α over Q√5 is

X2 − 2√5X + 2.

4.3 (c): Q(√10).

Solution: This case is similary to part (a). The minimal polynomial of α over Q√10 is

X4 − 16X2 + 4.

4.4 (d): Q(√15).

Solution: Since α2 = 8 + 2√15, the minimal polynomial of α over Q

√15 is

X2 − 8− 2√15.

5 Algebraic extension

Let F be a field, and let α be an element which generates a field extension of F of degree 5.Prove that α2 generates the same extension.

Proof. First, claim that α2 /∈ F and further [F (α2) : F ] > 1.If α2 ∈ F , then f(X) = X2 − α2 would be the minimal polynomial of α over F , which

implies [F (α) : F ] = 2. It contradicts with the fact that F (α) is a field extension of F ofdegree 5.

Seccond, since αinF (α) and F (α) is a field, then α2 ∈ F (α). So F (α2) ⊂ F (α). Now wehave

F ⊂ F (α2) ⊂ F (α)

⇒ [F (α) : F ] = [F (α) : F (α2)][F (α2) : F ],

⇒ 5 = [F (α) : F (α2)][F (α2) : F ].

Since [F (α2) : F ] > 1, we can get

[F (α) : F (α2)] = 1, [F (α2) : F ] = 5.

It implies that α2 generates the same extension as α.

4



Decide whether or not i =√−1 is in the field

6.1 (a): Q(√−2).

No.

6.2 (b): Q( 4√−2).

No.

6.3 (c): Q( 4√−4).

Yes.Since

( 4√−4)2 =

√−4 = 2

√−1,

⇒√−1 =

1

2( 4√−4)2 ∈ Q( 4

√−4).

6.4 (b): Q(α) where α3 + α = −1.

No.Since we know that [Q(α) : Q] = 3 and [Q(

√−1) : Q] = 2. If

√−1 ∈ Q(α), then we have

Q ⊂ Q(√−1) ⊂ Q(α),

⇒ [Q(α) : Q] = [Q(α) : Q(√−1)][Q(

√−1) : Q],

⇒ 3 = [Q(α) : Q(√−1)] 2,

which is impossible.


Question: Let α, β be complex numbers of degree 3 over Q, and let K = Q(α, β). Determinethe possibilities for [K : Q].

Solution: It is easy to know that [K : Q] ≤ 9. Besides,

Q ⊂ Q(α) ⊂ Q(α, β),

⇒ [K : Q] = [Q(α, β) : Q(α)][Q(α) : Q],

⇒ [K : Q] = [Q(α, β) : Q(α)] 3.

It follows that the possibilities for [K : Q] is

3, 6, 9.

5


8 Straightedge and compass construction

Determine whether or not the regular 9-gon is constructible by ruler and compass.Solution: It is imposible to consctruct the regular 9-gon.To construct a regular 9-gon is the same as constructing the complex number ζ = e2π/9.

Then the minimal polynomial for ζ over Q is

f(X) = X6 +X3 + 1.

It follows that [Q(ζ) : Q] = ϕ(9) = 6. But we can not find any integer k such that 2k = 6. Soit is imposible to consctruct the regular 9-gon.

9 Finite field

9.1 (a)

Let F be a finite field of characteristic p, and let ϕ : F → F be the map defined by ϕ(x) = xp.Show that ϕ is an automorphism of F .

Proof. First, it is easy to know that ∀x, y ∈ F ,

ϕ(x+ y) = (x+ y)p = xp + yp = ϕ(x) + ϕ(y),

ϕ(xy) = xpyp = ϕ(x)ϕ(y).

It implies that ϕ is a homomorphism.Second, since ϕ(x) = xp = 0 ⇒ x = 0, we know that kerϕ = 0.Now we can conclude that ϕ is an automorphism of F .

9.2 (b)

Show that every automorphism of F is a power of ϕ.

Proof. First, considering only the multiplicative structure, claim that any automorphism mustbe of the form

xxa,

for some fixed a < |F |.Second, write a = peb with (p, b) = 1. Claim that every element of F is a root of the

polynomial(Xpe + 1)b −Xa − 1

(of degree ¡ |F |). Further, we can conclude that b = 1.

6

MA 553: Homework 11

Yingwei Wang ∗


1 Splitting field

Determine the splitting field and its degree over Q.

1.1 x4 − 2.

We know that

f(x) = x4 − 2 = (x− 4√2)(x+

4√2)(x− i

4√2)(x+ i

4√2).

Let K = Q( 4√2, i) = Q( 4

√2)(i).

On one hand, K contains all the roots of f(x) = x4− 2. Hence it contains the splittingfield of f(x).

On the other hand, since i 4√2 and 4

√2 are contained in the splitting field, then i = i

4√2

4√2

should also be contained in the splitting field, which means the splitting field contains Ksince K is the smallest field containing Q, i and 4

√2.

Now we can conclude that K = Q( 4√2, i) is the splitting field of f(x) = x4 − 2.

Besides, the minimal polynomial of i over Q( 4√2) is p1(x) = x2 + 1; while the minimal

polynomial of 4√2 over Q is p2(x) = x4 − 2. Now we have

[K : Q] = [Q(4√2, i) : Q(

4√2)][Q(

4√2) : Q] = 2× 4 = 8.


1


1.2 x4 + 2.

Suppose α is a root of g(x) = x4 + 1, then

α = cos(π

4

)

+ i sin(π

4

)

=

√2

2(1 + i).

It is easy to check that α2 = i and (α3)2 = −i. We know that

f(x) = x4 + 2 = (x2 − i√2)(x2 + i

√2) = (x− 4

√2α)(x+

4√2α)(x− 4

√2α3)(x+

4√2α3).

Let K = Q( 4√2α, i) = Q( 4

√2α)(i) and S = the splitting field of f(x) = x4 + 2.

On one hand, K contains all of the roots of f(x), which means K ⊇ S.On the other hand,

4√2α,

4√2α3 ∈ S,

⇒4√2α3

4√2α

= α2 = i ∈ S,

⇒ K = Q(4√2α, i) ⊆ S,

since K is the smallest field containing Q, 4√2α and i.

Now we can conclude that K = Q( 4√2α, i) is the splitting field of f(x) = x4 + 2.

Besides, the minimal polynomial of 4√2α over Q(i) is p1(x) = x4+2; while the minimal

polynomial of i over Q is p2(x) = x2 + 1. Now we have

[K : Q] = [Q(4√2α, i) : Q(i)][Q(i) : Q] = 4× 2 = 8.

1.3 x4 + x2 + 1.

We know thatf(x) = x4 + x2 + 1 = (x2 + x+ 1)(x2 − x+ 1).

Suppose ω = −12+

√32i, then ω2 = −1

2−

√32i. It is easy to know that ω and ω2 are roots

of g1(x) = x2 + x+ 1.

Besides, −w = 12−

√32i and −ω2 = 1

2+

√32i. It is easy to know that −ω and −ω2 are

roots of g2(x) = x2 − x+ 1.Now we know that

f(x) = (x− ω)(x− ω2)(x+ w)(x+ w2).

2


It is easy to check that the splitting field of f(x) is

K = Q(ω).

It follows that[K : Q] = 2.

1.4 x6 − 4.

We know thatf(x) = x6 − 4 = (x3 − 2)(x3 + 2).

Let ω = −12+

√32i, then ω2 = −1

2−

√32i. It is easy to know that ω3 = 1. Then

f(x) = (x− 3√2)(x− ω

3√2)(x− ω2 3

√2)(x+

3√2)(x+ ω

3√2)(x+ ω2 3

√2). (1.1)

Let K = Q( 3√2, ω). On one hand, since K contains all roots of f(x), it must contain

the splitting field of f(x).

On the other hand, the splitting field contains 3√2 and ω 3

√2; hence ω = ω

3√2

3√2is contained

in the splitting field. It follows that the splitting field contains K since K is the smallestfield containing Q, 3

√2 and ω.

Now we know that K = Q( 3√2, ω) is the splitting field of f(x) = x6 − 4.

Furthermore, the minimal polynomial of ω over Q( 3√2) is p1(x) = x2 + x+1; while the

minimal polynomial of 3√2 over Q is p2(x) = x3 − 2. Now we have

[K : Q] = [Q(3√2)(ω) : Q(

3√2)][Q(

3√2) : Q] = 2× 3 = 6.

2 Splitting field

Lemma 2.1. Let K be a finite extension of F . Prove that K is a splitting field over F if

and only if every irreducible polynomial in F [x] that has a root in K splits completely in

K[x].

Question: Let K1 and K2 be finite extensions of F contained in the field K, and assumeboth are splitting field over F .

3


2.1 (a)

Prove that their composite K1K2 is a splitting field over F .

Proof. Since K1 and K2 are finite extensions of F , we can suppose

K1 = F (α1, α2, · · · , αn),

K2 = F (β1, β2, · · · , βm).

Then it is easy to know that

K1K2 = F (α1, α2, · · · , αn, β1, β2, · · · , βm).

By Lemma 2.1, we just need to show that every irreducible polynomial in F [x] thathas a root in K1K2 splits completely in K1K2[x].

Suppose there exists a polynomial f(x) ∈ F [x] which has a root αiβj ∈ K1K2, but itcan not splits completely in K1K2[x]. Then

f(x) =∏

(x− αiβj)f(x)

Then we can find an irreducible polynomial f(x) ∈ F [x] such that f has a root in K1

but can not split in K1[x], which contradicts with the fact that K1 is a splitting field overF .

Therefore, K1K2 is a splitting field over F .

2.2 (b)

Prove that K1 ∩K2 is a splitting field over F .

Proof. Let f(x) be irreducible in F [x] which has a root γ ∈ K1 ∩ K2. By Lemma 2.1,f(x) splits completely over both K1 and K2. Therefore, f(x) splits splits completely overK1 ∩K2, which means K1 ∩K2 is a splitting field over F .

3 Wilson Theorem

Prove that an integer p > 1 is prime if and only if (p− 1)! ≡ −1( mod p).

4


Proof. Let F be a finite field with char(F) = p and |F| = pn. So |F∗| = pn − 1 whereF∗ = F \ {0}. Let

f(x) = xpn−1 − 1. (3.1)

On one hand, by Lagrange’s theorem, if α ∈ F∗, then the order of α divides pn − 1. Soαpn−1 = 1 or αpn−1 − 1 = 0. It implies that ∀α ∈ F∗, f(α) = 0.

On the other hand, since the degree of f(x) is pn − 1, then f(x) has at most pn − 1roots.

No we can conclude that f(x) splits completely over the field F, which means

f(x) =∏

α∈F∗

(x− α). (3.2)

Let x = 0 in (3.2), we can get

∏

α∈F∗

α = (−1)pn

=

{

1, p = 2−1, p is odd .

(3.3)

Now we can prove the conclusion that an integer p > 1 is prime if and only if (p−1)! ≡−1( mod p).

If p = 2, the it is obviously true.If p is odd and n = 1, then F∗ ∼= Z∗

p = {1, 2, 3, · · · , p− 1; mod p}. By (3.3), we knowthat

(p− 1)! ≡ −1( mod p). (3.4)

Conversely, if (3.4) is true, then

kp− 1 = (p− 1)!.

If p > 3, then (p− 1)! is even, so p must be odd.If p is not a prime, say

p = rs.

Thenrs | (p− 1)!,

since r < p− 1, s < p− 1.However, by (3.4), we know that

rs | (p− 1)! + 1,

which meansrs | 1.

It is impossible. So p must be a prime.

5


4 Even polynomials

Let f(X) ∈ Q[X ] be an irreducible polynomial of degree n > 2 such that

f(X) = f(−X). (4.1)

Prove that the Galois group of f is not the symmetric group Sn.

Proof. By the (4.1), we know that the degree of f(X) is an even number, say n = 2k.Besides, since the f(X) is irreducible over Q, then the roots of f(X) should appear inpairs, say

S = {w1,−w1, w2,−w2, · · · , wk,−wk},where |S| = n = 2k.

Let G be the Galois group of f , then G ⊂ Sn. If σ ∈ G such that σ1(w1) = w2 (or wesay if σ ∈ Sn contains a two cycle (12)) then σ should satisfy σ(−w1) = −w2 since σ isan automorphism of the splitting field of f . It implies that the permutation (12) ∈ Sn but(12) /∈ G, which means G $ Sn.

5 Galois group

Let K be the splitting field over Q of f(x) = x4 − 2x2 − 1.

5.1 (a)

Determine the Galois group of K/Q.

Proof. The resolvent cubic for f(x) is

g(x) = x3 + 4x2 + 8x,

= x(x2 + 4x+ 8),

= x(x+ 2 + 2i)(x+ 2− 2i).

The discriminant of g(x) isD = −322.

Just need to consider whether f(x) is reducible over Q(32i) = Q(i).

6


We know that

f(x) = x4 − 2x2 − 1,

= (x2 + (√2− 1))(x2 − (1 +

√2)),

=(

x+ (√2− 1)i

)(

x− (√2− 1)i

)

(

x−√

1 +√2

)(

(x+

√

1 +√2

)

,

where i =√−1.

It is clear that f(x) is irreducible over Q(i). It follows that the Galois group of K/Q is

D8 = {1, (12)(34), (13)(24), (14)(23), (12), (34), (1324), (1423)}.

5.2 (b)

Show that the only three subfields of K having degree 2 over Q are Q(√−1),Q(

√2) and

Q(√−2).

Proof. Let

r1 = (√2− 1)i =

√−2 −

√−1,

r2 = −(√2− 1)i = −

√−2 +

√−1,

r3 =

√

1 +√2,

r4 = −√

1 +√2.

Let σ = (1324), τ = (12). It is easy to know that D8 has only three subgroup of order4,

S = {1, σ, σ2, σ3},T = {1, σ2, τ, σ2τ},U = {1, σ2, στ, σ2τ}.

The corresponding subfields are

KS = Q(√−1),

KT = Q(√2),

KU = Q(√−2).

7


6 Resolvent cubic

Let k be a field of characteristic 6= 2, and let f ∈ k[X ] be an irreducible polynomial ofdegree 4. If r1, r2, r3 and r4 are the roots of f (in some splitting field), then the polynomialg whose roots are

p1 = r1r2 + r3r4,

p2 = r1r3 + r2r4,

p3 = r1r4 + r2r3.

is called the resolvent cubic of f .

6.1 (a)

Show that the the discriminant of f is the same as that of g.

Proof. It is easy to get that

p1 − p2 = (r1 − r4)(r2 − r3), (6.1)

p1 − p3 = (r1 − r3)(r2 − r4), (6.2)

p2 − p3 = (r1 − r2)(r3 − r4). (6.3)

By the definition of the discriminant, we can know that discriminant of f is the sameas that of g.

6.2 (b)

Let G ⊂ S4 be the galois group of f , and let V ⊳ S4 be the unique normal subgroup oforder 4. Prove that the fixed field T of V ∩G is a splitting field of g.

Proof. It is easy to know that

V = {(1), (12)(34), (13)(24), (14)(23)}.

Since G < S4 and V ⊳ S4, by the second isomorphism theorem,

V ∩G⊳G.

Let L = k[r1, r2, r3, r4], then

T = LG∩V = {x ∈ L | gx = x, ∀g ∈ G ∩ V }.

8


Let S = k[p1, p2, p3] be the splitting field of g over k and denote the galois group of S/kas GS.

On one hand, by the definition of V , we know that any permutation in V fixes p1, p2, p3,so GS ⊇ G ∩ V . On the other hand, if σ ∈ G but σ /∈ G ∩ V , then σ moves at least one ofthe p1, p2, p3, which means σ /∈ GS. Therefore, GS = G ∩ V .

By fundamental theorem of Galois theory, we can conclude that T = S.

6.3 (c)

Let t = [T : k] (see (b)). Prove that G = S4, A4 or V according as t = 6, 3 or 1. What arethe possibilities for G when t = 2?

Proof. First of all, by the fundamental theorem of Galois theory, we know that

t = [T : k] = [G : V ∩G].

Now let us consider t = 6, 3, 1 or 2 respectively.

1. t = 6 or 3.

In these two cases, since |G| = t|V ∩ G|, 3 is a divisor of |G|. Besides, 4 is also adivisor of G. Hence, |G| is a multiply of 12. Notice that |S4| = 24. So G must be S4

or A4. Furthermore,

[S4 : S4 ∩ V ] = [S4 : V ] = 24/4 = 6,

[A4 : A4 ∩ V ] = [A4 : V ] = 12/4 = 3.

It follows that if t = 6 then G = S4 while if t = 3 then G = A4.

2. t = 1.

In this case, G = G ∩ V , so G < V . Besides, |G| is a multiple of 4 and |V | = 4, wehave G = V .

3. If t = 2, then G = D8 or Z4.

In this case, |G| = 2|G ∩ V |. Since |V | = 4, we have |G ∩ V | = 1, 2, 4.

(a) If |G ∩ V | = 1, then |G| = 2, which contradicts with the face that |G| is amultiple of 4.

(b) If |G ∩ V | = 2, then |G| = 4, which implies that G = Z4.

9


(c) If |G ∩ V | = 4, then |G| = 8. But a subgroup of S4 of order 8 is a Sylow2-subgroup, and all of such subgroup are conjugate and therefore isomorphic.One of theses subgroups is D8.

6.4 (d)

Can the roots of f(X) = X4 +X − 5 ∈ Q[X ] be constructed with ruler and compass?

Solution: The resolvent cubic of f(X) is

g(X) = X3 + 20X + 1,

which is irreducible over Q. Besides, the discriminant is D = 32027, which is a prime. Sothe galois group of f(x) is S4.

Let L be the splitting field of f over Q, then [F : Q] = 24 is not a power of 2. It followsthat roots of f(X) can not be constructed with ruler and compass.

6.5 (e)

Lemma 6.1. Suppose that f(x) = x4 + ax2 + b ∈ Q[x] is irreducible and G is its galois

group. Then

1. If√b ∈ Q, then G ∼= V , the Klein 4-group;

2. If√a2 − 4b

√b ∈ Q, then G ∼= C, the cyclic group Z4;

3. otherwise, G ∼= D8.

Proof. By Problem 4 of this homework, we know that G 6= S4. Furthermore, G 6= A4 sincewe have shown that (12) /∈ G. The possible G’s are V, C,D8. Suppose the roots of f(x)are α, β,−α,−β which satisfy the following relations:

αβ =√b,

α2 − β2 =√a2 − 4b,

α3β − βα3 =√a2 − 4b

√b.

Now we have the discussion here.

10


1. If√b ∈ Q, then αβ ∈ Q. Let σ ∈ G be such that σ(α) = β, then σ(β) =

√b

σ(α)= α.

Similarly, if σ(α) = −β, then σ(−β) = α; Finally, if σ(α) = −α, then σ(β) = −β.Thus every element of G has order 2. It implies that G ∼= V , the Klein 4-group.

2. If√a2 − 4b

√b ∈ Q, then α3β − βα3 ∈ Q. Let σ ∈ G be such that σ(α) = β. If

σ(β) = α, thenσ(α3β − βα3) = β3α− αβ3,

which is impossible. Therefore σ(β) = −α. It implies that σ is of order 4, whichmeans G ∼= C, the cyclic group Z4.

3. We know that the splitting field must containQ(√b), Q(

√a2 − b) andQ(

√a2 − 4b

√b).

The irreducibility of the polynomial implies that√a2 − 4b /∈ Q . Therefore if

√b,√

a2 − 4b√b /∈ Q, the splitting field contains at least three subfields of degree 2.

Hence the either G ∼= K4 or G ∼= D8. However, if G ∼= K4, then αβ is fixed by anyelement of G. Since

√b /∈ Q, the only possibility is G ∼= D8.

Using the Lemma 6.1, we can determine the galois group for the minimal polynomialover Q of these numbers.

6.5.1√

3 + 2√2

x =

√

3 + 2√2,

⇒ x2 = 3 + 2√2,

⇒ x2 − 3 = 2√2,

⇒ (x2 − 3)2 = 8,

⇒ x4 − 6x2 + 1 = 0.

So the minimal polynomial is x4 − 6x2 + 1.Besides, since 1 ∈ Q, the galois group is V , the Klein 4-group.

11


6.5.2√

7 + 2√10

x =

√

7 + 2√10,

⇒ x2 = 7 + 2√10,

⇒ x2 − 7 = 2√10,

⇒ (x2 − 7)2 = 40,

⇒ x4 − 14x2 + 9 = 0.

So the minimal polynomial is x4 − 14x2 + 9.Besides, since

√9 ∈ Q, the galois group is V , the Klein 4-group.

6.5.3√

5 + 2√5

x =

√

5 + 2√5,

⇒ x2 = 5 + 2√5,

⇒ x2 − 5 = 2√5,

⇒ (x2 − 5)2 = 20,

⇒ x4 − 10x2 + 5 = 0.

So the minimal polynomial is x4 − 10x2 + 5.Besides, since

√102 − 4× 5

√5 = 20 ∈ Q, the galois group is C, the cyclic group Z4.

6.5.4√

5 + 2√21

x =

√

5 + 2√21,

⇒ x2 = 5 + 2√21,

⇒ x2 − 5 = 2√21,

⇒ (x2 − 5)2 = 84,

⇒ x4 − 10x2 − 49 = 0.

So the minimal polynomial is x4 − 10x2 − 49.Besides, since

√−49 /∈ Q and

√102 + 4× 49

√−49 /∈ Q, the galois group is D8.

12

MA 553: Homework 12

Yingwei Wang ∗


1 Problem 40(c), Page 622

Question: Express the symmetric function as a polynomial in the elementary symmetricfunctions.

f(x1, x2, x3) = x2

1x2

2+ x2

1x2

3+ x2

2x2

3.

Solution: Since n = 3, we know that

s1 = x1 + x2 + x3,

s2 = x1x2 + x1x3 + x2x3,

s3 = x1x2x3.

By the procesure in Exercise 38, we know that the first step is to compute

f(x1, x2, x3)− s01s22

= x2

1x2

2+ x2

1x2

3+ x2

2x2

3− (x1x2 + x1x3 + x2x3)

2,

= −2x2

1x2x3 − 2x1x

2

2x3 − 2x1x2x

2

3.

The second step is

f(x1, x2, x3)− s01s22+ 2s1s

0

2s3,

= −2x2

1x2x3 − 2x1x

2

2x3 − 2x1x2x

2

3+ 2(x1 + x2 + x3)x1x2x3,

= 0.

Now we know thatf(x1, x2, x3) = s2

2− 2s1s3.


1


2 Problem 48, Page 623

Determine the splitting field K for the polynomial

f(x) = x6 − 2x3 − 2,

over Q.

2.1 (a)

Prove that f(x) is irreducible over Q with roots the three cube roots of 1±√3.

Proof. By Eisenstein’s criterion, we know that f(x) is irreducible over Q.Besides, it is easy to get

f(x) = (x3 − 1−√3)(x3 − 1 +

√3). (2.1)

Let α =3

√√3 + 1, β =

3

√√3− 1 and ω = −1

2+

√

3

2i, then

f(x) = (x− α)(x− ωα)(x− ω2α)(x+ β)(x+ ωβ)(x+ ω2β). (2.2)

3 (b)

Prove that K contains the field Q(√−3) of 3rd roots of unity and containsQ(

√3), and hence

contains the biquadratic field F = Q(i,√3). Furthermore, conclude that K is an extension of

the field L = Q( 3√2, i,

√3).

Proof. First of all,

(

α

β

)3

=

√3 + 1√3− 1

= 2 +√3 ∈ K,

⇒ Q(√3) ∈ K.

Second,

α, ωα ∈ K,

⇒ ωα

α= ω = −1

2+

√−3

2∈ K,

⇒ Q(√−3) = Q(

√3i) ∈ K.

Now we know that K ⊃ F = Q(i,√3).

2


Furthermore,

αβ =3

√

(√3 + 1)(

√3− 1) =

3√2 ∈ K,

⇒ L = Q(3√2, i,

√3) ∈ K,

3.1 (c)

Prove that [L : Q] = 12 and that K is obtained from L by adjoining the cube root of anelement in L, so that [K : Q] = 12 or 36.

Proof. It is easy to know that

[L : Q] = [Q(3√2, i,

√3) : Q(i,

√3)][Q(i,

√3) : Q(i)][Q(i) : Q],

= 3× 2× 2 = 12.

Besides, since K is obtained from L by adjoining the cube root of 1 +√3 ∈ L, then

[K : Q] = 12 or 36.

3.2 (d)

Prove that if [K : Q] = 12 then K = Q( 3√2, i,

√3) and that Gal(K/Q) is isomorphic to the

direct product of the cyclic group of order 2 and S3. Furthermore, if [K : Q] = 12 then thereis a unique real cubic subfield in K, namely 3

√2.

Proof. We know that K ⊃ L ⊃ Q and [L : Q] = 12. So if [K : Q] = 12 then K = L =Q( 3

√2, i,

√3).

It is easy to know that L is the splitting field of g(x) over Q, where g(x) defined by

g(x) = (x3 − 2)(x2 − 3).

The galois group of x3 − 2 is S3 while the galois group of x2 − 3 is Z2. It follows that

Gal (K/Q) = Z2 × S3.

Furthermore, [(Z2 × S3) : (Z2 ×Z2)] = 3 and the fixed field of Z2 ×Z2 is Q( 3√2). So 3

√2 is

the unique real cubic subfield in K if [K : Q] = 12.

3


3.3 (e)

Show that3

√

2 +√3,

3

√

2−√3 ∈ R are both elements of K. Show that γ =

3

√

2 +√3 +

3

√

2−√3 is a real root of the irreducible cubic equation x3 − 3x − 4, whose discriminant is

−2244. Conclude that the galois closure of Q(γ) contains Q(i) so in particular Q(γ) 6= 3√2.

Proof. First of all,

α

β=

3

√

2 +√3 ∈ K,

β

α=

3

√

2−√3 ∈ K,

⇒ γ =3

√

2 +√3 +

3

√

2−√3 ∈ K.

Besides, the minimal polynomial of γ over Q is

h(x) = x3 − 3x− 4.

It is easy to check that h(x) has one real root and two complex roots. So the Galois closureof Q(γ) contains Q(i). In particular, Q(γ) 6= 3

√2.

3.4 (f)

Determine all the elements of G = Gal (K/Q) explicitly and in particular show that G isisomorphic to S3 × S3.

Proof. Form part (e), we know that |G| = [K : Q] = 36.

4


4 Finite fields

Let L ⊃ K be finite fields, c := |K|, and let f(X) ∈ K[X ] be irreducible, of degree e dividing[L : K]. Show that there is an a ∈ L such that in L[X ],

f(X) = (X − a)(X − ac)(X − ac2

) · · · (X − ace−1

).

How many such a are there?

Proof. First, suppose c = p is a prime. Then K = Fp, L = Fpm and L/K is a Galois extension.Besides, the Galois group Gal(L/K) is cyclic and a generator is the pth power map:ϕp : t → tp.

Since e | [L : K], the splitting field of f(x) over K is contained in L. Let a ∈ L is a root off(x), then other roots can be obtained from a by applying Gal(K(a)/K) to this root. Sincethe Galois group is generated by the pth power map, the roots of f(x) are a, ap, ap

2

, · · · . Oncewe reach ap

e

, we have cycled back to the start ape

= a since K(a) ∼= K/f has order pe.The polynomial f is separable since its roots lie in a Galois extension K(α)/K. Besides,

since its degree is d, its different roots must be

a, ap, ap2

, · · · , ape−1.

Second, suppose c = pn and [L : K] = d Then K = Fc and L = Fcd. Similarly, L/K is alsoa Galois extension and the Galois group GalL/K is cyclic with generator the cth power mapϕc : t → tq. Similarly, we can show that if f(x) is irreducible with degree e | d, then there isa ∈ L such that a is a root of f and the full set of roots is

a, ac, ac2

, · · · , ace−1.

Since K(a) ⊂ L and [K(a) : K] = e, [L : K] = d, the number of such a should bed− e + 1.

5

MA 553: Homework 1

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