One of the first theorems one encounters in the study of commutative algebra is that if $I$ is an ideal of a ring $A$ not contained in any of the prime ideals $P_1,\cdots,P_n$, then $I$ is not contained in $\cup_{i=1}^n P_i$. This is proposition 1.11(i) from Atiyah-MacDonald and exercise 1.6 from Matsumura's commutative ring theory.

Now consider the following situation: Let $(R,m)$ be a regular local ring of dimension $n>1$. Let $P_1,\cdots,P_r$ be its minimal prime ideals. Since $\operatorname{dim} R=n>1$, we can not have $m=m^2$, since otherwise we would get $m=0$ by Nakayama's Lemma and $R$ would be a field.

Question: why is it true that there exists an $x \in m$ which is not inside $P_1\cup \cdots \cup P_r \cup m^2?$

Remark 1: We can not apply proposition 1.11(i) from Atiyah-MacDonald, since $m^2$ might not be a prime.

Remark 2: If $m \subset P_1 \cup \cdots \cup P_r \cup m^2$, then $m \subset P_1\cup \cdots \cup P_r + m^2 \Rightarrow m=P_1\cup \cdots \cup P_r + m^2$. Now if $P_1 \cup \cdots \cup P_r$ were an ideal, we could apply NAK and get $m=P_1\cup \cdots \cup P_r$ and we would be done. But $P_1 \cup \cdots \cup P_r$ might not be an ideal. And so what we get is $m = <P_1 \cup \cdots \cup P_r>$. But now proposition 1.11(i) is not applicable.

Remark 3: In the argument of remark 2, i did not use neither the fact that $R$ is regular, nor the fact that the $P_i$ are minimal.

Reference: Matsumura's Commutative Ring Theory, proof of Theorem 14.3.


1 Answer 1


Are you aware of the more general theorem that has the same statement and conclusion, except that it allows at most two of the $P_i$ to be nonprime? This statement would allow you to draw the conclusion you want.

Here is one place it appears. Pretty sure it appears in something like Kaplansky's commutative ring book. I think the proof is similar to the proof of the theorem you cited, it is just "more careful."

I see it also appears on page 90 of Eisenbud's Commutative algebra with a view...


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