2

I'm currently trying to figure out primary ideals and I think I proved the following geometric statement, but I'm not sure if I did it right and would like to have some feedback.

Lemma: An ideal $\mathfrak{a}$ of a commutative ring $A$ is primary if and only if $\text{Spec}(A/\mathfrak{a})$ is irreducible.

Proof: First note that $\text{Spec}(A/\mathfrak{a})$ is irreducible if and only if $\sqrt{(0)} \subset A / \mathfrak{a}$ is a prime ideal. Now suppose $\mathfrak{a}$ is primary, and $\bar x \cdot \bar y \in\sqrt{0} \subset A / \mathfrak{a}$. Then $x^n\cdot y^n = (xy)^n \in \mathfrak{a}$ for some $n \in \mathbb{N}$, so either $x^n \in \mathfrak{a}$ or $(y^n)^m \in \mathfrak{a}$ for some $m \in \mathbb{N}$. I.e. either $x \in \sqrt{(0)}$ or $y \in \sqrt{(0)}$.

Conversely, suppose $x \cdot y \in \mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $\mathfrak{a}$, i.e. $x^n \in \mathfrak{a}$ or $y^n \in \mathfrak{a}$.

I'm still not 100% sure that I'm done here, because I only have $x^n \in \mathfrak{a}$, not $x\in \mathfrak{a}$. Am I missing something, or is the claim wrong? If so, I would like to see a counterexample.

red_trumpet
  • 8,515
  • 1
    Conversely, suppose $x \cdot y \in \mathfrak{a}$. Then either $x$ or $y$ is nilpotent modulo $\mathfrak{a}$, i.e. $x^n \in \mathfrak{a}$ or $y^n \in \mathfrak{a}$. . Are you aware that this condition ("$xy\in I$ implies $x^n\in I$ or $y^n\in I$") is strictly weaker than being primary? You seem to be using it as if you thought it was equivalent. That's what catches my eye. The converse proof seems to fall short in this manner. I guess that's what you're getting at in the last sentence. – rschwieb Jan 29 '19 at 18:21
  • 1
    Yeah exactly. Do you know a counterexample of an ideal where $xy \in I \Rightarrow (x^n \in I$ or $y^n \in I)$, but I is not primary? – red_trumpet Jan 29 '19 at 18:30
  • Of course I do! :) – rschwieb Jan 29 '19 at 18:31
  • I am not handy at all with the topology on Spec. It feels to me like it should somehow be very straightforward to say that the spectrum is irreducible since $A/\mathfrak{a}$ only has one minimal prime., – rschwieb Jan 29 '19 at 18:37
  • Thanks! As far as I understand up to now, the spectrum is irreducible if and only if $\mathfrak{a}$ satisfies the weaker condition above, as shown by my proof. So in particular if $\mathfrak{a}$ is primary, this holds. – red_trumpet Jan 29 '19 at 18:48
  • I'm a little unsure about how to follow up the approach you took. If ${0}=\sqrt{\mathfrak a}/\mathfrak a$ is prime, that would be equivalent to $\sqrt{\mathfrak{a}}$ being prime in $A$, but there are nonprimary ideals with prime radicals. This is a major problem, no? – rschwieb Jan 29 '19 at 18:54
  • What I mean is that $\sqrt{(0)} \subset A/\mathfrak{a}$ is prime if and only if $xy \in \mathfrak{a} \Rightarrow (x^n \in \mathfrak{a}$ or $y^n \in \mathfrak{a})$. This should be equivalent to saying that $\sqrt{\mathfrak{a}}$ is prime. You counterexample shows, that there are ideal which are not primary, but still satisfy this condition. But if $\mathfrak{a}$ is primary, then in particular $\sqrt{\mathfrak{a}}$ is prime, or $A/\mathfrak{a}$ has irreducible spectrum. – red_trumpet Jan 29 '19 at 19:02
  • $nil(R/I)=\sqrt{I}/I$, and by prime correspondence if the former is prime in $R/I$, $\sqrt{I}$ is prime in $R$. it seems you disagree with this? I'm having trouble seeing what's wrong with it. For sure I believe the lemma you cited at the start of the proof. I'm just not entirely sure it goes back all the way to prove the necessity that $\mathfrak{a}$ is primary. – rschwieb Jan 29 '19 at 19:56
  • I don't disagree with your first sentence. And as you said, prime correspondence shows that $nil(R/I)$ is prime if and only if $\sqrt{I}$ is prime. But this proves the lemma wrong, because there are non-primary ideals with prime radical. – red_trumpet Jan 29 '19 at 20:16
  • Yes! Now I guess we're on the same page. Where did the lemma in the yellow box appear for you? – rschwieb Jan 29 '19 at 20:58
  • Just thought of it trying to make sense of primary ideals, especially in the context of algebraic geometry. – red_trumpet Jan 29 '19 at 22:17

1 Answers1

3

The lemma I wrote above is wrong. The proof actually shows

The quotient $A/\mathfrak{a}$ has irreducible prime spectrum if and only if $\mathfrak{a}$ satisfies: $xy \in \mathfrak{a} \implies (x^n \in \mathfrak{a} \text{ or }y^n \in \mathfrak{a})$.

Clearly, any primary ideal satisfies this, so quotients of primary ideals have irreducible prime spectrum, but this condition is strictly weaker. Here is a counterexample.

A similar characterisation of primary ideals which can be found here would be:

$\mathfrak{a}$ is primary if and only if all zerodivisors of $A/\mathfrak{a}$ are nilpotent.

red_trumpet
  • 8,515