# union of infinitely many prime ideals

Consider a noetherian ring $R$ and a collection $m_i$, $i\in I$ of maximal ideals of $R$. Let $P$ be a prime ideal of $R$. It is well-known that if the collection is finite (i.e. the index set $I$ is finite), then $P\subseteq \cup_{i\in I}m_i$ if and only if $P$ is contained in one of the $m_i$. (This is true under weaker assumptions.)

Now my question is what about the case of an infinite collection. Does the same hold (under suitable assumptions, e.g. $R$ is normal integral, etc.)? Or can anyone give a counter-example?

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I fixed your notation so that there wasn't any equivocation. I changed $P_i$ to $m_i$, changed the ideal $I$ to $P$, and kept the set $I$ as $I$.

There is indeed a counterexample. Take $R=\mathbb C[x,y]$, the prime ideal $P=(x,y)$ is not contained in any of the ideals $m_{a,b}=(x-a,y-b)$ for $(a,b)$ not both $0$, since $P$ is also a maximal ideal. Yet every element of $P$ is contained in at least one of the $m_i$, because there is no polynomial function on $\mathbb A^2$ that vanishes on one point but not any other point. Thus $P \subset \bigcup_i m_i$.

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[edited only to fix the typo "prime ideal ideal"] –  Noam D. Elkies Aug 5 '12 at 15:32
That seems like an ideal ideal reason to edit. –  Will Sawin Aug 5 '12 at 22:34

On a related note, however, there are some important classes of rings $R$ where countable prime avoidance holds. That is, when an ideal $J$ is a subset of a countable union of prime ideals in such a ring, it has to be contained in a single element of the collection. In particular, $R$ satisfies countable prime avoidance if either:

1. $R$ contains an uncountable field, or

2. $R$ is a complete Noetherian local ring.

Case 1 was mentioned in an article by Hochster and Huneke in the Michigan Math. Journal in 2000. Case 2 was proved by Lindsay Burch in 1972. Both arguments are fairly straightforward.

I learned these things from the Hochster-Huneke article some years back.

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It seems "enumerate" does not work well, so I removed it. Yet merely numering the lines (or just starting them with 1.) has a similar effect, which I thus did. –  quid Aug 5 '12 at 16:15

As an addition to the answers already given, let me mention the interesting paper Baire's category theorem and prime avoidance in complete local rings by Sharp and Vamos (Arch. Math. 44 (1985), 243-248). Beside other things, it contains the following:

• A neat proof of Burch's Lemma (cf. the answer by Neil Epstein) on use of Baire's category theorem.

• An example showing that in Burch's Lemma the hypothesis of completeness cannot be omitted; this is essentially the same as the example given by Mahdi Majidi-Zolbanin.

• A noetherian local ring with uncountable residue field has countable avoidance (i.e., if an ideal is contained in the union of a countable family of ideals (not necessarily prime!) then it is contained in one of these ideals); this can be viewed as a generalisation of a special case of the Hochster-Huneke result mentioned by Neil.

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Actually, the countable avoidance result (without assuming any of the ideals are prime), at least when the ring contains an uncountable field, also appears in the Hochster-Huneke paper. Perhaps they weren't aware of the Sharp-V\'amos result, or maybe it was just that their result had slightly different hypotheses. Avoidance (prime, countable, or otherwise) wasn't the main point of their paper anyway; it was about tight closure theory and the localization problem. In any case, thank you for the interesting reference! –  Neil Epstein Aug 8 '12 at 13:38

As another counterexample, take any noetherian local ring $(R,\mathfrak{m})$ of dimension $>1$, such that $\mathfrak{m}\not\in\mathrm{Ass}(R)$. Then $\mathfrak{m}$ is subset of the union of all non maximal prime ideals, because any $x\in\mathfrak{m}$ lies in a prime ideal of height $\leq1$.

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Nice example ! But why do you assume $\mathfrak{m}$ is not an associated prime ? –  Qing Liu Sep 3 '12 at 22:22
That is a good point. I was being "too" cautious when I wrote the answer, but the assumption $\mathfrak{m}\not\in\mathrm{Ass}(R)$ is not needed. –  Mahdi Majidi-Zolbanin Sep 5 '12 at 4:05