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If $Q$ is an ideal whose radical is a maximal ideal, then $Q$ is a primary ideal.

I wanted to prove this using these facts:

  • $Q$ is primary if and only if every zero divisor in $R/Q$ is nilpotent.
  • Lattice isomorphism theorem: $R/J \cong (R/I)/(J/I), \ I \subset J$.
  • $R/J$ is an integral domain if and only if $J/I$ is prime.
  • If $R/\text{rad } Q$ is an integral domain, then every zero divisor in $R/Q$ is nilpotent. Proof: When $I \subset J$, there is a surjective hom $R/I \to R/J$ given by $\bar{x} = x + I \mapsto x + J = \bar{x}$. So if $\overline{xy} = 0$ in $R/Q$ with $\bar{x}, \bar{y} \neq 0$, then $\overline{xy} = 0$ in $R/\text{rad } Q$, which means $\bar{x} = 0$, wlog, since we're in an integral domain. But $\bar{x} = 0 $ in $R/ \text{rad } Q$ iff $x \in \text{rad } Q$ iff $x^n \in Q$ for some $n \geq 1$, which means then that $\bar{x}^n = 0 $ in $R / Q$.

To complete the proof we need to show that $(\text{rad } Q) / Q $ is prime. But by the correspondence of ideals of $R$ that $\supset Q$ and ideals of $R/Q$, $\text{ rad } Q $ is maximal, hence prime and so then must $(\text{rad } Q)/ Q$ be prime. But isn't $(\text{rad } Q) / Q$ also maximal and hence $R/\text{rad } Q $ is a field as well? But I haven't seen any mention of this.

user26857
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Daniel Donnelly
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1 Answers1

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Your fourth bullet point asserts that if $\text{rad}(Q)$ is prime, then $Q$ is primary. Since maximal ideals ar prime, that would prove the title question, miraculously over-proving and generalizing it.

Unfortunately, the fourth bullet is incorrect. An ideal can have a prime radical and be nonprimary.

The problem is that you can't conclude $\bar{x}=0$ in $R/\text{rad }{Q}$. It could be nonzero, while $\bar{y}$ is zero. In the counterexample given at the link, $\overline{xy}$ is zero mod $(xy, y^2)$, but $\bar x$ is not zero mod $(y)=\text{rad}(xy,y^2)$

Take a look at other solutions to this problem on the site.

But isn't $\text{rad}(Q) / Q$ also maximal and hence $R/\text{rad } Q $ is a field as well?

Yes, if $\text{rad}(Q)$ is maximal in $R$, $\text{rad}(Q)/Q$ is maximal in $R/Q$ and $R/\text{rad}(Q)$ is a field. The second fact follows immediately from the hypothesis, and it does not really have anything to do with the first fact, as you suggest with that boldface line.

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rschwieb
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