Primary ideals are those whose quotient has irreducible prime spectrum. [duplicate]

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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.

marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 30 at 20:28

• 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 at 18:21
• 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 at 18:30
• – rschwieb Jan 29 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 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 at 18:48

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