Math 593 Assignment # 2 Solutions

Bryden R. Cais

September 27, 2002

1.

2. Let A be a commutative ring with unit and let I, J ⊂ A be ideals with I + J = A.

(a) Since I + J = A and 1 ∈ A, there exist i ∈ I and j ∈ J such that i + j = 1. Now let a, b ∈ A and

consider

x = aj + bi ∈ A.

We have x ≡ bi ≡ b mod J since i ≡ 1 mod J. Similarly, x ≡ a mod I.

(b) By part (a), we have a surjective map

A ⟶ A/I ⊕ A/J

given by

a ↦ (a mod I, a mod J).

This is obviously a homomorphism. An element a of A is in the kernel of this map if and only if a ≡ 0 mod I and a ≡ 0 mod J, that is, a ∈ I and a ∈ J. In other words, we have an exact sequence

0 ⟶ I ∩ J ⟶ A ⟶ A/I ⊕ A/J ⟶ 0.

Therefore,

A/(I ∩ J) ≃ A/I ⊕ A/J,

and if I ∩ J = 0, then A ≃ A/I ⊕ A/J.

Remark 1. This is just a special case of the Chinese Remainder Theorem.

3. (a) From class, we know that ℤ is a PID, so that every ideal is of the form nℤ. If 0 ≠ n = xy is composite (with 1 < x, y < n) then x, y ∉ (n) since n |̸x, y but xy = n ∈ (n). Therefore (n) is not prime. Conversely, if p ∈ ℤ is a prime and xy ∈ (p) then p|xy so that p|x or p|y; that is, x ∈ (p) or y ∈ (p). In short, (p) is prime. Since ℤ is an integral domain, by part d, (0) is prime.

(b) Let M ⊂ R be a maximal ideal and suppose that ab ∈ M and a ∉ M. Then M + (a) is an ideal of R properly containing M. It follows that we can write

1 = m + ax

for some m ∈ M and x ∈ R. Therefore, b = mb + abx ∈ M since ab ∈ M. It follows that M is prime.

Remark 2. Of course, we could just observe that R/M is a field and hence an integral domain, and then apply part (e).

(c) The ideal (x) is prime since x is irreducible. Moreover, (x) ⊊ (x, y) ⊊ ℚ[x, y] so that (x) is not maximal.

Remark 3. If you want to use overkill, observe that (x) is the kernel of the homomorphism

φ : ℚ[x, y] ⟶ ℚ[y]

given by

p(x, y) ↦ p(0, y).

We therefore have an exact sequence

0 ⟶ (x) ⟶ ℚ[x, y] ⟶ ℚ[y] ⟶ 0,

and since ℚ[y] is an integral domain, it follows that ℚ[x, y]/(x) ≃ ℚ[y] is an integral domain and hence, by part (e), that (x) is prime.

(d) This follows from part (e) since R/(0) ≃ R.

(e) We have that R/P is an integral domain if and only if for any x, y ∈ R/P, xy = 0 implies x = 0 or y = 0. Let x̃, ỹ be any lifts of x, y to R. Then R/P is an integral domain if and only if x̃ỹ ∈ P implies x̃ ∈ P or ỹ ∈ P, i.e. if and only if P is prime.

4. Since parts (a) and (b) follow from (c), we prove only (c). Let I ⊂ R be an ideal and let \sqrt{I} be the radical of I. Let x, y ∈ \sqrt{I}. Then there exist m, n such that x^{m}, y^{n} ∈ I. Set N = m + n and consider

(x + y)^{N} = ∑_{i=0}^{N} \binom{N}{i}x^{N−i}y^{i},

(by the binomial theorem, which holds in any commutative ring). Observe that for any 0 ≤ i ≤ N, either N − i ≥ m or i ≥ n. Therefore, every summand is in I (since I is a two sided ideal of R) and hence (x + y)^{N} ∈ I. It follows that x + y ∈ \sqrt{I}.

Remark 4. Notice that we don’t need R to be commutative for this to hold. All we need is that xy = cyx for come c in the center of R when considered as a group under multiplication.

Now let r ∈ R, x ∈ \sqrt{I}. Then for some n, we have x^{n} ∈ I and hence, since R is commutative,

(rx)^{n} = r^{n}x^{n} ∈ I.

It follows that rx ∈ \sqrt{I} and hence that \sqrt{I} is an ideal.

Let f ∈ R/\sqrt{I} and suppose that f^{n} = 0. Let f̃ be any lift of f to R. Then f̃^{n} ∈ \sqrt{I}, and hence there exists some m such that f̃^{nm} ∈ I. It follows that f̃ ∈ \sqrt{I} and hence that f = 0. Therefore R/\sqrt{I} is reduced.