MODERN ALGEBRA (MATH 210) MIDTERM SOLUTIONS

POKMAN CHEUNG

This solution set was written by Pokman; Ravi has added some comments and hopefully hasn’t introduced too many errors!

1. Show that a ring in which all ideals are finitely generated cannot have an infinite sequence of ideals

I_{1} ⊊ I_{2} ⊊ I_{3} ⊊ ….

Conversely, show that if a ring has no infinite sequence of ideals, then all ideals are finitely generated.

Solution. Suppose all ideals of a ring R are finitely generated, and we have an increasing sequence of ideals, I_{1} ⊂ I_{2} ⊂ I_{3} ⊂ …. The union

I = ∪_{i≥1}I_{i}

is also an ideal. (Not true for any union of ideals. The sequence being increasing is important.) By assumption, I = (a_{1}, …, a_{n}) is finitely generated. Since the sequence {I_{k}}_{k≥1} is increasing, all a_{i} are contained in I_{N} whenever N is big enough, which implies I = (a_{1}, …, a_{n}) ⊂ I_{N} ⊂ I ⇒ I_{N} = I for all sufficiently large N. Hence, the increasing sequence of ideals stabilizes after a finite number of steps.

Conversely, suppose R satisfies the ascending chain condition on its ideals, and let I be an ideal of R. Choose a sequence of ideals inductively as follows: Pick some a_{1} ∈ I and let I_{1} = (a_{1}). Suppose I_{n} = (a_{1}, …, a_{n}) ⊂ I has been chosen. If I_{n} ⊊ I, pick some a_{n+1} ∈ I − I_{n}, and let I_{n+1} = I_{n} + (a_{n+1}) ⊂ I. Note that I_{n} ⊊ I_{n+1}. By assumption, this procedure must terminate after a finite number of steps, which happens only when I = I_{n} = (a_{1}, …, a_{n}). Hence any ideal is finitely generated.

(You can’t start with: let (a_{1}, a_{2}, …,) be a generating set. Do you see why? How do you know there is a countable generating set? Another common mistake is talk about “the” generating set.)

2.

(a) Show that if n ≠ 4, the only normal subgroups of S_{n} are {e}, A_{n}, and S_{n}.

(b) Describe all group homomorphisms from S_{7} → S_{5}.

Solution. (a) The cases n = 2, 3 are easily checked. Consider n ≥ 5. Let N be a normal subgroup of S_{n}. Then N ∩A_{n} is a normal subgroup of A_{n}. Since A_{n} is simple, N ∩A_{n} must equal {e} or A_{n}. In the latter case, N ⊃ A_{n}. As |S_{n} : A_{n}| = 2, the only subgroups of S_{n} containing A_{n} are A_{n} and S_{n}, and both of them are normal.

Date: Thursday, November 7, 2002.

Suppose N ∩ A_{n} = {e}. Assume N ≠ {e} and let σ, τ ∈ N − {e}. Since σ, τ ∉ A_{n}, they are odd permutations. Since the product of two odd permutations is even, A_{n} consists of all even permutations and N ∩ A_{n} = {e}, we have σ^{2} = e = στ, which implies σ = τ. Hence N contains at most one nontrivial element. [This is a proof of problem 2.95(i) in Rotman.] However, since N is normal, whenever N contains an element, N contains all its conjugates, and the conjugacy class of any nontrivial element of S_{n} consists of more than one elements (true for any n > 2). Thus N cannot contain exactly one nontrivial element. We conclude N = {e}.

(b) Let φ : S_{7} → S_{5} be a homomorphism. By (a), ker φ = {e}, A_{7} or S_{7}. The first case is impossible since |S_{7}| > |S_{5}|. If ker φ = A_{7}, we must have im φ = {e, τ}, where τ ∈ S_{5} is of order 2. Conversely, for any such τ, since S_{7}/A_{7} ≅ {e, τ}, there exists a homomorphism with A_{7} as the kernel and {e, τ} as the image. Finally, we have ker φ = S_{7} if and only if φ maps all elements of S_{7} to e ∈ S_{5}.

3. Prove that every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal.

Solution. Let (p) ≠ (0) be prime and (a) ⊋ (p). Since p ∈ (a), we have p = ab for some b ∈ R. Then ab ∈ (p) but a ∉ (p) implies b ∈ (p), or b = cp for some c ∈ R. Hence we have p = ab = acp ⇒ 1 = ac, since R is a domain and p ≠ 0. Therefore, a is a unit and (a) is the whole ring. This proves (p) is maximal.

(Many peope never explicitly used the fact that (p) ≠ (0). It is necessary — do you see where?)

4.

(a) Note that ω = \frac{−1+\sqrt{−3}}{2} is a cube root of 1, and ω^{2} + ω + 1 = 0. Prove that the subset {x + yω ∈ ℤ[ω] : x + y is divisible by 3 } ⊂ ℤ[ω] is an ideal. Is it prime? .

(b) Describe the set of integers of the form a^{2} − ab + b^{2} (a, b integers).

Solution. (a) Let I be the subset defined in the question. It is straightforward to directly check that I is an ideal. Instead, observe that the following are equivalent:

(i) 3|a + b,  (ii) 3|N(a + bω),  (iii) a + bω ∈ (1 − ω).

(i) ⇔ (ii): since N(a + bω) = a^{2} − ab + b^{2} = (a + b)^{2} − 3ab. (iii) ⇒ (ii): since N(1 − ω) = 3. (i) ⇒ (iii): since a + bω = a + b − b(1 − ω) = (1 − ω) [\frac{a+b}{3}(1 − ω̄) − b]. Hence by (iii), I = (1 − ω) is an ideal. Also, (ii) implies I is prime: 3|N(αβ) = N(α)N(β) ⇒ 3|N(α) or 3|N(β).

(b) We first determine when a prime integer p can be so expressed. If p = a^{2} −ab+b^{2}, by reducing to mod 3, it is easy to see that we must have p ≡ 0, 1(mod 3). Conversely, if p = 3, take a = 1, b = −1. If 3|p − 1, by Theorem 2.78 in Rotman, there exists τ ∈ 𝔽_{p}^{×} of order 3. Hence we have (in 𝔽_{p}) (τ − 1)(τ^{2} + τ + 1) = τ^{3} − 1 = 0 but τ − 1 ≠ 0, thus τ^{2} + τ + 1 = 0. This means there exist t ∈ ℤ such that t^{2} + t + 1 = pk for some k ∈ ℤ.

In particular, we may require 1 < t ≤ p − 1, so that

0 < pk = t^{2} + t + 1 = (t + \frac{1}{2})^{2} + \frac{3}{4} ≤ (p − \frac{1}{2})^{2} + \frac{3}{4} = p^{2} − p + 1 < p^{2},

and therefore 0 < k < p. In ℤ[ω], we have

p | pk = t^{2} + t + 1 = (t − ω)(t − ω̄).

Since ℤ[ω] is a UFD, some (nonunit) irreducible factor α of p must divide, say, t − ω. Then, N(α) divdes both N(p) = p^{2} and N(t − ω) = pk. Since 0 < k < p and N(α) ≠ 1, we conclude N(α) = p. Let α = a + bω. We then have p = a^{2} − ab + b^{2}. Therefore, p can be expressed in the desired form if and only if p ≡ 0, 1(mod 3).

Alternatively, here is a little more concise proof, paralleling our proof of which numbers are expressible as the sum of two squares: If p ≠ 3,

p = a^{2} − ab + b^{2} = (a + bω)(a + bω̄)
⇔ (p) ⊂ ℤ[ω] is not prime
⇔ ℤ[ω]/(p) ≅ ℤ[x]/(x^{2} + x + 1, p) ≅ 𝔽_{p}[x]/(x^{2} + x + 1) has zerodivisors
⇔ x^{2} + x + 1 is reducible over 𝔽_{p}
⇔ τ^{3} = 1 for some τ ∈ 𝔽_{p}^{×} − {1} (since p ≠ 3)
⇔ 3 | p − 1 (since 𝔽_{p}^{×} is cyclic of order p − 1)

Let S = {q a prime integer : q ≡ 2(mod 3)}. Suppose n = a^{2} − ab + b^{2} = (a + bω)(a + bω̄) and q ∈ S divides n. Since q stays irreducible in ℤ[ω], we must have, say, q|a + bω, which implies q^{2} = N(q)|N(a + bω) = n. Then, if we let n = q^{2}n′, a + bω = q(a′ + b′ω), we have n′ = (a′ + b′ω)(a′ + b′ω̄). By the above argument, if q|n′, then q^{2}|n′. Therefore, every q ∈ S divides n exactly an even number of times. Conversely, suppose

n = ∏_{p_{i}∉S} p_{i}^{k_{i}} ∏_{q_{j}∈S} q_{j}^{2ℓ_{j}}

is the prime factorization in ℤ. Since for any p_{i} ∉ S, we have p_{i} = N(α_{i}) for some α_{i} ∈ ℤ[ω], we have

n = ∏_{i} N(α_{i})^{k_{i}} ∏_{j} N(q_{j})^{ℓ_{j}} = N (∏_{i} α_{i}^{k_{i}} ∏_{j} q_{j}^{ℓ_{j}}) = a^{2} − ab + b^{2}

for some a, b ∈ ℤ.

5. Find (with proof) an ideal I of ℤ[i] whose quotient F is a field of 9 elements. Is there an ideal whose quotient is a field of 25 elements?

Solution. Since 3 is irreducible in ℤ[i] (a^{2}+b^{2} = 3 has no integer solution) and ℤ[i] is a UFD,

(3) ⊂ ℤ[i] is a prime ideal, and thus a maximal ideal by problem 3. Hence, F = ℤ[i]/(3) is a field. It is easy to see that the nine elements {a + bi : 0 ≤ a, b ≤ 2} of ℤ[i] map to distinct elements, and all the elements of F. Therefore, F is a field of nine elements. [Alternatively, ℤ[i]/(3) ≅ ℤ[x]/(3, x^{2} + 1) ≅ 𝔽_{3}[x]/(x^{2} + 1) is a degree 2 field extension over 𝔽_{3}, since x^{2} + 1 is irreducible over 𝔽_{3}, and hence is a field of 3^{2} = 9 elements.]

(A common mistake was to miscount the number of elements of ℤ[i]/(2), by abusing the division algorithm; there are 4, not 9!.)

For the second part, here are two solutions.

First solution. Suppose I is an ideal such that R = ℤ[i]/I is a field of 25 elements. Then every element of R has order dividing 25, so in particular 25 ∈ I. As I is prime, 5 ∈ I. However, ℤ/(5) has 25 elements (by the same method as above), so I = (5). But then (5) isn’t prime: 5 = (2 + i)(2 − i).

Second solution, using what we know now. Suppose we have a ring homomorphism φ : ℤ[i] → 𝔽_{25}, where 𝔽_{25} is the field of 25 elements. If φ ≠ 0, φ(1) = 1 (since φ(1)^{2} = φ(1)), and thus

0 = φ(i^{2} + 1) = φ(i)^{2} + 1 = (φ(i) − 2)(φ(i) − 3).

The last equality holds since char = 5. Hence, φ(i) = 2 or 3, and for any a, b ∈ ℤ, φ(a + bi) = a + bφ(i) ∈ {0, 1, 2, 3, 4} = 𝔽_{5} ⊂ 𝔽_{25}. Therefore, there doesn’t exist any surjective homomorphism mapping ℤ[i] onto a field of 25 elements. In other words, there doesn’t exist any ideal I ⊂ ℤ[i] such that ℤ[i]/I ≅ 𝔽_{25}.

By a slight extension of these arguments, you can show that if there is an ideal whose quotient is a field of q elements, then q is prime not 3 modulo 4, or the square of a prime that is 3 modulo 4.

6. How many abelian groups are there of order 288?

Solution. 288 = 32 × 9. In the direct sum decomposition into primary factors of an abelian group of order 288, the orders of the 2-primary ones can be:

{2, 2, 2, 2, 2}, {2, 2, 2, 4}, {2, 2, 8}, {2, 4, 4}, {2, 16}, {4, 8}, {32}.

Those of the 3-primary ones can be:

{3, 3}, {9}.

Hence there are 7 × 2 = 14 nonisomorphic abelian groups of order 288.

(Here is is most useful to use the version of the Fundamental Theorem of Abelian groups that separates the group into p-parts.)