Algebra Prelim Fall 2014

Harini Chandramouli

Problem 1. Exhibit a non-abelian group of order n3 for ever 2 ≤ n ∈ Z.

Solution. Consider the Heisenberg group:    1 a c   H(n) = 0 1 b a, b, c ∈ Z/nZ .  0 0 1 

We can see that |H(n)| = n3 since there are n many choices for a, b, and c. The 3-by-3 identity matrix, I3, is clearly in H(n) and so we have an identity element. Inverses are given by: 1 a c−1 1 −a −ab − c 0 1 b = 0 1 −b  . 0 0 1 0 0 1

Now we want to show that H(n) is closed. Let a1, a2, b1, b2, c1, c2 ∈ Z/nZ, then       1 a1 c1 1 a2 c2 1 a1 + a2 c2 + a1b2 + c1 0 1 b1 0 1 b2 = 0 1 b1 + b2  ∈ H(n) 0 0 1 0 0 1 0 0 1

Thus, H(n) is a group. Now we want to show it is not abelian. Consider

1 1 0 1 0 0 X := 0 1 0 Y := 0 1 1 . 0 0 1 0 0 1

We see that X,Y ∈ H(n).

1 1 0 1 0 0 1 1 1 XY = 0 1 0 0 1 1 = 0 1 1 0 0 1 0 0 1 0 0 1 1 0 0 1 1 0 1 1 0 YX = 0 1 1 0 1 0 = 0 1 1 , 0 0 1 0 0 1 0 0 1 so XY 6= YX and thus H(n) is a non-abelian group of order n3.

1 Problem 2. Let p, q be primes with p < q and q 6= 1 mod p. Show that every group of order pq is abelian.

Proof. Let G be a group such that |G| = pq.

By the Sylow Theorems, the number of p-Sylow subgroups, np, has the following two prop- erties:

np = 1 mod p np q.

So since q 6= 1 mod p, we know that np = 1. So there is one p-Sylow subgroup, call it Hp. By the Sylow Theorems, we know that Hp must be self-congruent and hence normal in G. We also know that |Hp| = p and so by Lagrange’s Theorem, we know that Hp is cyclic.

By the Sylow Theorems, the number of q-Sylow subgroups, nq, has the following proper- ties:

nq = 1 mod q nq p.

So since p < q, we know that this means that nq = 1. Let Hq represent our q-Sylow subgroup. By the same logic as above, Hq is normal in G, cyclic, and |Hq| = q.

Since |Hp| and |Hq| are relatively prime, we know that Hp ∩ Hq = {e}. Since Hp and Hq are normal subgroups of G, we know that HpHq is a normal subgroup of G such that

|Hp| · |Hq| |HpHq| = = pq. |Hp ∩ Hq|

Thus, HpHq = G. So any element of G can be written as a combination of elements in Hp −1 and Hq. Let a ∈ Hp and b ∈ Hq. Since Hp is normal, b ab ∈ Hp. Since Hq is normal, −1 a ba ∈ Hq. Thus,

−1 −1 −1 −1 b a ba ∈ Hp ∩ Hq =⇒ b a ba = e =⇒ ab = ba.

Thus, G is abelian.

2 Problem 3. Let S,T be linear operators on a ﬁnite-dimensional complex vector space V , and ST = TS. Show that S,T have a common eigenvector.

Proof. Since C is algebraically closed, we know that the minimum polynomial of T on V has a root, λ ∈ C, that is an eigenvalue of T with a corresponding, non-trivial eigenvector for T .

Let Vλ represent the λ-eigenspace. That is, Vλ = v ∈ V T v = λv .

So ∀ v ∈ Vλ, since S and T commute,

T (Sv) = T Sv = ST v = Sλv = λSv = λ(Sv).

So S stabilizes Vλ. That is, S maps vectors in Vλ to vectors in Vλ. So S is a linear operator on Vλ.

Let f(x) represent the minimum polynomial of S on Vλ. Since C is algebraically closed, we know that f(x) has a root, γ ∈ C, that is an eigenvalue of S on Vλ with corresponding non-trivial eigenvector, u ∈ Vλ. Notice that γ is also an eigenvalue for S on all of V since f(x) must divide the minimum polynomial of S on all of V and hence u is an eigenvector for S on all of V .

Since u ∈ Vλ, we know that T u = λu. But since u is an eigenvector for S, we know that Su = γu. Thus, S and T have a common eigenvector.

3 √ √ √ Problem 4. Find all intermediate ﬁelds between Q and Q( 2, 3, 5).

√ √ √ √ √ √ Proof. First observe that Q( 2, 3, 5) is a Galois extension of Q as Q( 2, 3, 5) is a separable, splitting ﬁeld over Q. Notice h √ √ √ i h √ √ √ √ √ i h √ √ √ i h √ i Q( 2, 3, 5) : Q = Q( 2, 3, 5) : Q( 2, 3) Q( 2, 3) : Q( 2) Q( 2) : Q . √ √ √ 2 First, it is easy to see that Q( 2) : Q = 2 as 2 is the root of x − 2 but 2 ∈/ Q. √ √ √ √ √ √ Next, we will show that ( 2, 3) : ( 2) = 2. This is true iﬀ 3 ∈/ ( 2) as 3 Q Q √ √ Q is the root of x2 − 3. So suppose for contradiction that 3 ∈ Q( 2). Then, ∃ a, b ∈ Q such that √ √ 3 = a + b 2 √ 3 = a2 + 2b2 + 2ab 2 ( a2 + 2b2 = 3 =⇒ 2ab = 0

√ √ So either a = 0 or b = 0. If a = 0, then 2b2 = 3 =⇒ b = √3 ∈/ so we have a √ 2 Q 2 contradiction. If b = 0, then a = 3 =⇒ a = 3 ∈/ Q, which is also a contradiction.√ Not√ as well that if a = b = 0, then we get that 0 = 3, which is a contradiction. Thus, 3 ∈/ Q( 2) √ √ √ and Q( 2, 3) : Q( 2) = 2. √ √ √ √ √ Next, we want to show that ( 2, 3, 5) : ( 2, 3) = 2. This will be true iﬀ √ √ √ √ Q Q √ 2 5√∈/ √Q( 2, 3)√ as √ 5 is the root of x √− 5. So suppose for contradiction that 5 ∈ Q( 2, 3) = Q( 2)( 3). Then ∃ a, b ∈ Q( 2) such that √ √ 5 = a + b 3 √ 5 = a2 + 3b2 + 2ab 3 ( a2 + 3b2 = 5 =⇒ 2ab = 0

2 Now, since√ a = b = 0 cannot√ hold, let us consider ﬁrst if a = 0. Then 3b = 5 where b ∈ Q( 2). Thus, b = α + β 2 where α, β ∈ Q. So, √ 3(α + β 2)2 = 5 √ 3α2 + 6β2 + 6αβ 2 = 5 ( 3α2 + β2 = 5 =⇒ 6αβ = 0

4 √ So if√α = 0, then we get that β = 5 ∈/ Q, a contradiction. If β = 0, then we get that α = √5 ∈/ , a contradiction. Thus a = 0 cannot hold. 3 Q If b = 0, then √ √ 2 a = 5 =⇒ a = 5 ∈ Q( 2). √ √ √ However,√ 5 ∈/ Q( 2) by an argument nearly identical√ to the√ argument√ as to why 3 ∈/ Q( 2) given above. Thus b = 0 cannot hold. Thus, 5 ∈/ Q( 2, 3) and h √ √ √ √ √ i Q( 2, 3, 5) : Q( 2, 3) = 2.

So we see that h √ √ √ i √ √ √

Q( 2, 3, 5) : Q = 8 =⇒ Gal(Q( 2, 3, 5/Q) = 8. √ √ √ Now any automorphism in Gal(Q( 2, 3, 5/Q) must map a root of any polynomial to another root of that polynomial. Consider √ √ √ √ √ √ α : 2 7→ − 2, 3 7→ 3, 5 7→ 5, Q fixed √ √ √ √ √ √ β : 2 7→ 2, 3 7→ − 3, 5 7→ 5, Q fixed √ √ √ √ √ √ γ : 2 7→ 2, 3 7→ 3, 5 7→ − 5, Q fixed Then the Galois group is as follows: √ √ √ Gal(Q( 2, 3, 5)/Q) = {id, α, β, γ, α ◦ β, α ◦ γ, β ◦ γ, α ◦ β ◦ γ} . We also notice that every element of the Galois group (other than the identity) is of order 2. By Lagrange’s Theorem, any proper subgroup will be of order 2 or 4. The groups of order 2 are cyclic and hence we can see they are:

hαi = {id, α} hβi = {id, β} hγi = {id, γ} hα ◦ βi = {id, α ◦ β} hα ◦ γi = {id, α ◦ γ} hβ ◦ γi = {id, β ◦ γ} hα ◦ β ◦ γi = {id, α ◦ β ◦ γ}

It is easy to see that√ these√ √ are all proper subgroups and all possible subgroups of order 2. Now since Gal(Q( 2, 3, 5)/Q) has no elements of order 4, the subgroups of order 4 will not be cyclic. Using a combinatorial argument, and facts about K4, one can show that there

5 are 7 subgroups of order 4. They are:

hα, βi = {id, α, β, α ◦ β} hα, γi = {id, α, γ, α ◦ γ} hβ, γi = {id, β, γ, β ◦ γ} hα, β ◦ γi = {id, α, β ◦ γ, α ◦ β ◦ γ} hβ, α ◦ γi = {id, β, α ◦ γ, α ◦ β ◦ γ} hγ, α ◦ βi = {id, γ, α ◦ β, α ◦ β ◦ γ} hα ◦ β, α ◦ γi = {id, α ◦ β, α ◦ γ, β ◦ γ} .

By the Fundamental Theorem of Galois Theory, there exists a one-to-one correspondence between the intermediate ﬁelds and the subgroups of the Galois group.

Now we can find the fixed elements. For the subgroups of order 2, it is easy to see that in the Galois pairing: √ √ hαi ↔ Q( 3, 5) √ √ hβi ↔ Q( 2, 5) √ √ hγi ↔ Q( 2, 3) √ √ hα ◦ βi ↔ Q( 5, 6) √ √ hα ◦ γi ↔ Q( 3, 10) √ √ hβ ◦ γi ↔ Q( 2, 15) √ √ √ √ √ √ ∼ ∼ hα ◦ β ◦ γi ↔ Q( 6, 10) = Q( 10, 15) = Q( 6, 15). √ √ √ Now we consider the groups of order 4. By applying the maps to 2, 3, 5 and combina- tions of them, we se that in the Galois pairing: √ hα, βi ↔ Q( 5) √ hα, γi ↔ Q( 3) √ hβ, γi ↔ Q( 2) √ hα, β ◦ γi ↔ Q( 15) √ hβ, α ◦ γi ↔ Q( 10) √ hγ, α ◦ βi ↔ Q( 6) √ hα ◦ β, α ◦ γi ↔ Q( 30). And these are all the intermediate fields.

6 c Problem 5. Show that every rational number is expressible as a ﬁnite sum of fractions pn with denominators which are prime powers (and c ∈ Z).

a α1 αn Proof. Let b ∈ Q where b = p1 ··· pn where each pi is distinct, 1 ≤ i ≤ n, i ∈ N. We will prove this using induction. α1 α2 base case: As n = 1 is trivial, consider n = 2. Since p1 and p2 are relatively prime, by the Euclidean algorithm, ∃c1, c2 ∈ Z such that

α1 α2 α1 α2 c2p1 + c1p2 = gcd(p1 , p2 ) = 1

α1 α2 =⇒ ac2p1 + ac1p2 = a ac2 ac1 a a =⇒ α2 + α1 = α1 α2 = ∈ Q p2 p1 p1 p2 b a So we see that b ∈ Q is written as a ﬁnite sum of fractions. ˜ α1 αk inductive step: Assume that if b = p1 ··· pk , then a c c = 1 + ··· + k . ˜ α1 αk b p1 pk

α1 αk αk+1 Now, let b = p1 ··· pk pk+1 . Then, a a = α1 αk αk+1 b p1 ··· pk pk+1 a 1 = α1 αk · αk+1 p1 ··· pk pk+1 c1 ck 1 = α1 + ··· + αk · αk+1 p1 pk pk+1 c c = 1 + ··· + k α1 ak+1 αk ak+1 p1 pk+1 pk pk+1 k X ci = pαi pak+1 i=1 i k+1 k X γi1 γi2 = + pαi pak+1 i=1 i k+1 k P γi2 γ11 γk1 i=1 = α1 + ··· + αk + ak+1 . p1 pk pk+1 And thus by the Principle of Mathematical Induction, every rational number is expressible c as the ﬁnite sum of fractions pn with denominators which are prime powers.

7 Problem 6. Show that x5 + y7 + z11 is irreducible in C[x, y, z].

Proof. Let f(x, y, z) := x5 + y7 + z11. Notice that C[x, y, z] ⊆ C(x, y)[z]. If f(x, y, z) is irreducible in C(x, y)[z] then it will be irreducible in C[x, y, z]. In C(x, y)[z], f(x, y, z) is a degree 11 polynomial in z with constant term x5 + y7.

∼ 5 7 Consider C(x, y) = C(x)[y]. We want to ﬁnd a polynomial that divides x + y whose square does not divide x5 + y7 so that we can use Eisenstein’s Criterion. Since C(x)[y] is a

unique factorization domain, ∃ p(y), an irreducible non-unit, such that p(y) x5 + y7.

We want to show that (p(y))2 - x5 + y7. Suppose for contradiction that (p(y))2 x5 + y7. ∂ 5 7 6 Then this would imply that p(y) ∂y (x + y ) = 7y . Then, y x5 + y7 − · 7y6 = x5 7

must also be divisible by p(y). However, x5 is a unit in C(x)[y] and thus we have a contradiction so (p(y))2 - x5 + y7.

So we have that p(y) x5 + y7,(p(y))2 - x5 + y7, and p(y) - 1, the coeﬃcient of z11 in f(x, y, z). Thus, by Eisenstein’s Criterion, f(x, y, z) is irreducible in C(x, y)[z] and hence is irreducible in C[x, y, z].

8 Problem 7. Show that the set of nilpotent elements in a commutative ring R is an ideal.

n Proof. Let I := x ∈ R x = 0 for some n ∈ N . We want to show that I is an ideal and so we need to show that I is an additive subgroup and that ∀ x ∈ I and ∀ r ∈ R, xr, rx ∈ I. Let x, y ∈ I such that xn = ym = 0 for some m, n ∈ N.

First we wil show that I is an additive subgroup. We know that

n+m X n + m (x + y)n+m = xkyn+m−k k k=1 by the Binomial Theorem. We want to show that (x + y) ∈ I. Well if k ≥ n, then xk = xn · xk−n = 0. Similarly, if k < n, then yn+m−k = yn−k · ym = 0. Thus, (x + y) ∈ I. Notice that 0 = 01 so 0 ∈ I. Additionally, (−x)n = (−1)n · xn = 0 and so (−x) ∈ I. Thus, I is an additive subgroup.

Since R is commutative, we see that

(rx)n = rxrx ··· rx = rr ··· rxx ··· x = rnxn = 0 (xr)n = xrxr ··· xr = xx ··· xrr ··· r = xnrn = 0 and so we see that xr, rx ∈ I.

Thus we see that I is an ideal.

9 4 Problem 8. Show that x + 1 is reducible in Fp[x] for every prime p.

4 8 Proof. We want to ﬁnd α ∈ Fp such that α = −1 =⇒ α = 1. Thus, the roots will be of order 8. case 1: p = 2 (x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1 = x4 + 1 4 So we have shown that x + 1 is reducible in F2[x].

case 2: p is an odd prime We spilt this into two subcases: p = 1 mod 8 and p2 = 1 mod 8 (notice that 32 = 52 = 72 = 1 mod 8 and so we only need these two cases). case 2a: p = 1 mod 8. × × × So p = 8n + 1 for some n ∈ N. Thus, |Fp | = 8n, so 8 |Fp |. Since Fp is cyclic, it must have an element of order 8 and hence x4 + 1 has a linear factor.

case 2b: p2 = 1 mod 8. 2 × 2 × So p = 8n + 1 for some n ∈ N. Then we know that |Fp2 | = p − 1 = 8n. So 8 |Fp2 | and × 4 Fp2 has an element of order 8. Hence x + 1 has a quadratic factor.

4 Thus we have shown that x + 1 is reducible in Fp[x] for every prime p.