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Suppose $Q$ is a $P$-primary ideal and $Q'$ is a $P'$-primary ideal such that $P$ and $P'$ are comaximal in the Noetherian ring $R$. Show that $Q$ and $Q'$ are comaximal.

Proof. Since $Q$ is a $P$-primary ideal and $Q'$ is a $P'$-primary ideal, then $P=\sqrt{Q}$ and $P'=\sqrt{Q'}$. Since $P$ and $P'$ are comaximal, then $P+P'=R$.

I also have a proposition that says if $R$ is a Noetherian ring, then for any ideal $I$ some positive power of $(\sqrt{I})$ is contained in $I$. That is, for positive $p$, $(\sqrt{I})^p \subset I$.

I want to show $Q+Q'=R$. Here's what I have so far:

$Q+Q'\subset\sqrt{Q}+\sqrt{Q'}=P+P'=R$

As for the other direction I'm having a bit of trouble. I have:

$R=P+P'=\sqrt{Q}+\sqrt{Q'}$ I want this to be a subset of $(\sqrt{Q} )^q +(\sqrt{Q'})^{q'} \subset Q+Q'$ and then I would be done.

I'm studying the section about Dedekind domains right now in class so I don't know if that helps?

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  • $\begingroup$ You don't need $R$ noetherian here. $\endgroup$ – user26857 Apr 1 '16 at 13:19
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You know you can write $1=x+y$ where some power of $x$ lies in $Q$ and some power of $y$ lies in $Q'$. Pick $N$ large enough so that $(x+y)^N$ is a sum of terms of $Q$ and $Q'$, and conclude that $1\in Q+Q'$.

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  • $\begingroup$ Note that the only hypothesis we used is that two ideals had comaximal radicals to obtain they were comaximal. In particular, noetherianity and primality are irrelevant. $\endgroup$ – Pedro Tamaroff♦ Apr 1 '16 at 11:12

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