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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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After some weeks, I think the following works fine. Divide it into some cases, $1)a=b=0, 2)a\not=0, b=0$ or one is divided by the other, 3) non of the above. Then, use the fact that, for a an ideal $I$ of $R$ which has a minimal primary decomposition, $I=\bigcap_{i=1}^n \mathfrak{q}_i$, we have $Z(R/I)=\bigcup_{i=1}^n \mathfrak{p}_i$ where $\mathfrak{p}_i \in \text{Ass}(I)$ and $Z(.)$ stands for the set of zero-divisors in $R.$ – Ehsan M. Kermani Jun 25 '11 at 14:13
Is this a solution to the exercise, or just few remarks? – user26857 Jun 18 '14 at 17:12
@user26857, as I recall, they're just some important points I used in my proof. It was very long time ago though! – Ehsan M. Kermani Jul 3 '14 at 22:33
It's hard to believe that these trivial remarks can lead to a solution. The exercise is not trivial at all. – user26857 Jul 12 '14 at 8:27
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. – Ehsan M. Kermani Jul 12 '14 at 16:57
up vote 3 down vote accepted

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. – user26857 Jun 18 '14 at 21:54

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