# maximal ideal properly contains union of its square with the union of minimal prime ideals

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.

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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.