union of infinitely many prime ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:53:26Z http://mathoverflow.net/feeds/question/102546 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102546/union-of-infinitely-many-prime-ideals union of infinitely many prime ideals Yong Hu 2012-07-18T14:35:35Z 2012-08-06T15:23:06Z <p>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.)</p> <p>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? </p> http://mathoverflow.net/questions/102546/union-of-infinitely-many-prime-ideals/102548#102548 Answer by Will Sawin for union of infinitely many prime ideals Will Sawin 2012-07-18T14:49:39Z 2012-08-05T15:32:41Z <p>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 $I$.</p> <p>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 $P \subset \bigcup_i m_i$.</p> http://mathoverflow.net/questions/102546/union-of-infinitely-many-prime-ideals/102602#102602 Answer by Mahdi Majidi-Zolbanin for union of infinitely many prime ideals Mahdi Majidi-Zolbanin 2012-07-19T00:37:57Z 2012-07-19T00:37:57Z <p>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 <em>non maximal</em> prime ideals, because any $x\in\mathfrak{m}$ lies in a prime ideal of height $\leq1$.</p> http://mathoverflow.net/questions/102546/union-of-infinitely-many-prime-ideals/104027#104027 Answer by Neil Epstein for union of infinitely many prime ideals Neil Epstein 2012-08-05T14:42:58Z 2012-08-06T15:23:06Z <p>On a related note, however, there are some important classes of rings $R$ where <em>countable prime avoidance</em> holds. That is, when an ideal $J$ is a subset of a <em>countable</em> 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:</p> <ol> <li><p>$R$ contains an uncountable field, or</p></li> <li><p>$R$ is a complete Noetherian local ring.</p></li> </ol> <p>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.</p> <p>I learned these things from the Hochster-Huneke article some years back.</p> http://mathoverflow.net/questions/102546/union-of-infinitely-many-prime-ideals/104079#104079 Answer by Fred Rohrer for union of infinitely many prime ideals Fred Rohrer 2012-08-06T06:39:05Z 2012-08-06T06:39:05Z <p>As an addition to the answers already given, let me mention the interesting paper <em>Baire's category theorem and prime avoidance in complete local rings</em> by Sharp and Vamos (Arch. Math. 44 (1985), 243-248). Beside other things, it contains the following:</p> <ul> <li><p>A neat proof of Burch's Lemma (cf. the answer by Neil Epstein) on use of Baire's category theorem.</p></li> <li><p>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.</p></li> <li><p>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.</p></li> </ul>