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Here is an another problem in Commutative Rings by Kaplansky, p. 103, no. 15.

Let $R$ be a Noetherian UFD. Let $(a,b) \not= R$ where $a,b \in R.$ Prove that any maximal prime of $(a,b)$ has grade of at most $2.$

Note: By a maximal prime of $I=(a,b)$, I assumed a maximal prime ideal $\mathfrak{p} \in \text{Ass}(I).$ In other words, an embedded prime ideal associated with $I.$

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    Dear @user26857,$I$wish I could remember it's proof. That was for 2 years ago when I took a grad commutative algebra course. I'm no longer working in that direction (no longer in pure math of course) and retrieving them all is a pain. So please don't want me to write a proof for it.2014-07-12

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Since $R$ is noetherian, a maximal prime of $(a,b)$ is nothing but a maximal prime in the set $\operatorname{Ass}_R(R/(a,b))$. We can localize and reduce the problem to the following:

Let $(R,\mathfrak m)$ be a local noetherian UFD, and $a,b\in R\setminus\{0\}$ such that $(a,b)\ne R$. If $\mathfrak m\in\operatorname{Ass}_R(R/(a,b))$, then $\operatorname{depth}R\le 2$.

We have an exact sequence of $R$-modules $0\to R/(a)\cap (b)\to R/(a)\oplus R/(b)\to R/(a,b)\to0.$ Since $R$ is a UFD, $(a)\cap (b)$ is a principal ideal (generated by $ab/\gcd(a,b)$), and thus the projective dimension of $R/(a)\cap (b)$ is one. Now we get $\operatorname{pd}_R(R/(a,b))\le 2$. Applying the Auslander-Buchsbaum formula we obtain $\operatorname{depth}R=\operatorname{pd}_R(R/(a,b))\le 2$.

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    In Kaplansky's book this exercise appears before the Auslander-Buchsbaum formula. It would be nice to have a proof which doesn't use it.2014-06-18