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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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put on hold as off-topic by Sanath Devalapurkar, user26857, anorton, Michael Albanese, Claude Leibovici Jul 10 at 4:15

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Sanath Devalapurkar, user26857, anorton, Michael Albanese, Claude Leibovici
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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 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 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 yesterday
    
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 18 hours ago
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