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Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$ of height $d$, and suppose $\mathfrak{p}_1, \ldots, \mathfrak{p}_s$ are prime ideals of height $i - 1 < d$. It's quite clear that each $\mathfrak{p}_j$ is a proper subset of $\mathfrak{m}$, but why is $\bigcup_j \mathfrak{p}_j$ also a proper subset of $\mathfrak{m}$? In the case $s = 2$ this can be reduced to the fact that the union of two ideals is an ideal if and only if one contains the other, but this doesn't generalise for $s > 2$.

Background. This claim appears in my lecturer's proof that $A$ has an $\mathfrak{m}$-primary ideal generated by $d$ elements. I think the proof can be rewritten to avoid using this claim, but the claim on its own seems interesting enough.

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up vote 5 down vote accepted

If the union of primes equals $\mathfrak{m}$, then $\mathfrak{m}$ is contained in their union and by using the prime avoidance lemma it must be contained in one of these primes, contradiction.

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