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

Let $R$ (always commutative with $1$) be a Noetherian UFD ring. 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 the 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
What you say here is nonsense. –  user26857 Oct 17 '12 at 20:30
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