Primary ideals are those whose quotient has irreducible prime spectrum.

Question

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.

Answer 1

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.