Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.