MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #5

JOHN VOIGHT

Problem 1. Let A be a commutative ring. Let M be a module, and N a submodule. Let N = Q_{1} ∩ ⋯ ∩ Q_{r} be a primary decomposition of N. Let \overline{Q_{i}} = Q_{i}/N. Show that 0 = \overline{Q_{1}} ∩ ⋯ ∩ \overline{Q_{r}} is a primary decomposition of 0 in M/N. State and prove the converse.

Solution. Set (or module) theoretically, we indeed have 0 = ⋂_{i} \overline{Q_{i}} if and only if N = ⋂_{i} Q_{i}.

Now Q ⊂ M is primary if and only if Q ≠ M and a_{M/Q} is injective or nilpotent. Therefore \overline{Q} = Q/N ⊂ M/N is primary if and only if Q/N ≠ M/N and a_{(M/N)/(Q/N)} is injective or nilpotent. For any Q ⊃ N, we have (M/N)/(Q/N) ≅ M/Q, so Q ⊂ M is primary (contaning N) if and only if \overline{Q} ⊂ M/N is primary.

The converse is also now proven: Let 0 = ⋂_{i} \overline{Q_{i}} be a primary decomposition for 0 in M/N. Let Q_{i} be the inverse image of \overline{Q_{i}} in the surjection M → M/N. Then N = ⋂_{i} \overline{Q_{i}} is a primary decomposition for N.

Problem 2. Let 𝔭 be a prime ideal, and 𝔞, 𝔟 ideals of A. If 𝔞𝔟 ⊂ 𝔭, show that 𝔞 ⊂ 𝔭 or 𝔟 ⊂ 𝔭.

Solution. If 𝔞 ⊄ 𝔭 and 𝔟 ⊄ 𝔭, then there exists an a ∈ 𝔞 \ 𝔭 and b ∈ 𝔟 \ 𝔭. Then ab ∈ 𝔞𝔟 ⊂ 𝔭, so since 𝔭 is prime, we have a ∈ 𝔭 or b ∈ 𝔭, a contradiction.

Problem 3. Let 𝔮 be a primary ideal. Let 𝔞, 𝔟 be ideals, and assume 𝔞𝔟 ⊂ 𝔮. Assume that 𝔟 is finitely generated. Show that 𝔞 ⊂ 𝔮 or there exists some positive integer n such that 𝔟^{n} ⊂ 𝔮.

Solution. If 𝔞 ⊂ 𝔮, we are done. Otherwise, let a ∈ 𝔞 \ 𝔮. Let b_{1}, …, b_{r} generate 𝔟 as an ideal. Then for each i, ab_{i} ∈ 𝔮 but a ∉ 𝔮 so there exists an integer n_{i} such that b_{i}^{n_{i}} ∈ 𝔮. Let n = r max_{i} n_{i}. Then 𝔟^{n} is generated by products b = b_{1}^{m_{1}} … b_{r}^{m_{r}} with ∑_{i} m_{i} = n; by the pigeonhole principle i such that e_{i} ≥ n_{i}, so b ∈ 𝔮 for all b ∈ 𝔟^{n}.

Problem 4. Let A be noetherian, and let 𝔮 be a 𝔭-primary ideal. Show that there exists some n ≥ 1 such that 𝔭^{n} ⊂ 𝔮.

Date: March 6, 2003.

X: 1–8, *.

Solution. Since A is noetherian, the ideal 𝔭 is finitely generated; let p_{1}, …, p_{r} generate 𝔭. We have by definition that

𝔭 = {a ∈ A : a^{n} ∈ 𝔮 for some n ≥ 1}.

Let n_{i} be such that p_{i}^{n_{i}} ∈ 𝔮, and let n = r max_{i} n_{i}. Then as in the previous exercise we see that 𝔭^{rn} ⊂ 𝔮.

Problem 5. Let A be an arbitrary commutative ring and let S be a multiplicative subset. Let 𝔭 be a prime ideal and let 𝔮 be a 𝔭-primary ideal. Then 𝔭 intersects S if and only if 𝔮 intersects S. Furthermore, if 𝔮 does not intersect S, then S^{−1}𝔮 is S^{−1}𝔭-primary in S^{−1}A.

Solution. If f ∈ 𝔭 ∩ S, then there exists an n such that f^{n} ∈ 𝔮 since 𝔮 is 𝔭-primary and S is multiplicatively closed. The reverse inclusion follows immediately from 𝔮 ⊂ 𝔭.

To the second statement, we know that S^{−1}𝔮 ≠ A, since S ∩ 𝔮 = ∅. We now see that

S^{−1}A/S^{−1}𝔮 ≅ S^{−1}(A/𝔮),

since the map S^{−1}A → S^{−1}(A/𝔮) clearly has kernel S^{−1}𝔮. Therefore a_{S^{−1}A/S^{−1}𝔮} = a_{S^{−1}(A/𝔮)}, we know that if a_{A/𝔮} is injective or nilpotent then a_{S^{−1}(A/𝔮)} is injective (since localization is flat) or nilpotent, respectively.

Finally, we see that a ∈ A has a^{n} ∈ 𝔮 for some n ≥ 1 if and only if a/1 ∈ S^{−1}A has (a/1)^{n} ∈ S^{−1}𝔮 if and only if for all s ∈ S, (a/s)^{n} ∈ S^{−1}𝔮, so S^{−1}𝔮 is S^{−1}𝔭primary. (Or note that S^{−1}𝔭 = S^{−1}\sqrt{𝔮} = \sqrt{S^{−1}𝔮}.)

Problem 6. If 𝔞 is an ideal of A, show there is a bijection between the prime ideals of A which do not intersect S and the prime ideals of S^{−1}A, given by 𝔭 ↦ S^{−1}𝔭 and S^{−1}𝔭 ↦ S^{−1}𝔭 ∩ A = 𝔭.

Prove a similar statement for primary ideals instead of prime ideals.

Solution (sketch). Let 𝔭 ⊂ A be a prime ideal. We show that 𝔭 ∩ S = ∅ if and only if S^{−1}𝔭 is a prime ideal of S^{−1}A. From the previous exercise, we have S^{−1}(A/𝔭) ≅ S^{−1}A/S^{−1}𝔭. Since A/𝔭 is a domain, S^{−1}(A/𝔭) is a domain if and only if S ∩ 𝔭 = ∅ if and only if S^{−1}𝔭 is prime.

Also from the previous exercise, we see that S^{−1}𝔮 is primary if and only if S^{−1}𝔭 is primary, and by the correspondence of primes we have a similar bijection for 𝔭-primary ideals.

Problem 7. Let 𝔞 = 𝔮_{1} ∩ ⋯ ∩ 𝔮_{r} be a reduced primary decomposition of an ideal. Assume that 𝔮_{1}, …, 𝔮_{i} do not intersect S, but that 𝔮_{j} intersects S for j > i. Show that

S^{−1}𝔞 = S^{−1}𝔮_{1} ∩ ⋯ ∩ S^{−1}𝔮_{i}

is a reduced primary decomposition of S^{−1}𝔞.

Solution. Note that for j > i, since 𝔮_{j} ∩ S = ∅, we have S^{−1}𝔮_{j} = S^{−1}A, so already S^{−1}𝔞 = ⋂_{j=1}^{i} S^{−1}𝔮_{j} is a primary decomposition of S^{−1}𝔞.

The bijection of the previous exercise implies that the primes S^{−1}𝔭_{i} are distinct whenever 𝔭_{i} are distinct, for j ≤ i. If the decomposition is not reduced, then we have

S^{−1}𝔮_{k} ⊂ ⋂_{j≠k} S^{−1}𝔮_{j}

for some 1 ≤ i ≤ k; as the bijection preserves inclusions, we have

S^{−1}𝔮_{k} ∩ A ⊂ 𝔮 ⋂_{j≠k} S^{−1}𝔮_{j} ∩ A = ⋂_{j≠k} 𝔮_{j}

so the original decomposition is not reduced, a contradiction.

Problem 8. Let A be a local ring. Show that any idempotent ≠ 0 in A is necessarily the unit element.

Solution. Note that e^{2} = e if and only if (1 − e)^{2} = 1 − e. We cannot have both e and 1 − e in the maximal ideal 𝔪, since then e + (1 − e) = 1 ∈ 𝔪. Therefore, say, e ∉ 𝔪, so since A is local, e is a unit. But e(1 − e) = 0, so 1 − e = 0, i.e. e = 1.

Problem *. Find an example of an ideal I in an integral domain A for which there is a prime in Ass(A/I) that is not in Ass(A).

Solution. Take I = p^{2}ℤ and A = ℤ. Then Ass(ℤ) = {(0)} since the annihilator of any x ∈ ℤ is just 0. However, Ass(ℤ/p^{2}ℤ) = {(0), (p)}, since this is the set of all primes of ℤ/p^{2}ℤ and (0) is the annihilator of 1 and (p) is the annihilator of the element p ∈ ℤ/p^{2}ℤ.