I would like to know if it is always possible to find a one-dimensional ideal in a local commutative ring... actually I am interested in finding a curve through a point on a scheme (locally). If the ring is of finite dimension it should be obvious, but does anybody know about the situation in more general rings?
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$\begingroup$ I assume you mean "a local commutative ring in which not every prime ideal is maximal;" otherwise, Artinian local rings provide a finite-dimensional counterexample. $\endgroup$– Charles StaatsFeb 2, 2012 at 16:49
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$\begingroup$ Also see mathoverflow.net/questions/75320/… $\endgroup$– ParsaFeb 4, 2012 at 21:09
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2 Answers
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Given your comment about curves I suppose by a one dimensional ideal you mean an ideal such that the ring mod this ideal is one-dimensional.
The answer is this: if you assume your ring to be noetherian yes, if not no.
Let $(A,m)$ be a local ring.
Case 1: $A$ is noetherian. Take the set of prime ideals other than $\mathfrak m$. By the noetherian assumption this set has a (possibly more than one) maximal element. Call that $\mathfrak p$. Then the quotient ring $A/ \mathfrak p$ is a local domain that has exactly two prime ideals, so it has dimension one.
Case 2: $A$ is not necessarily noetherian. Karl Schwede constructs an example in this paper that has a single infinite chain of prime ideals $(0)\subset \mathfrak p_1\subset \mathfrak p_2\subset \dots \subset \mathfrak m\subset A$. He does this to exhibit the scheme $(\mathrm{Spec} A)\setminus\mathfrak m$ as a scheme without a closed point, but it also works here. Since all the prime ideals of $A$ appear in that chain and there are infinitely many "above" any of them except $\mathfrak m$, this ring does not have a prime ideal "of dimension one". In fact taking any prime other than $\mathfrak m$ gives you an infinite dimensional quotient ring. This implies that if the quotient ring by any ideal is finite dimensional, then it has to be 0 dimensional: Take the quotient, if it has finite dimension then the only prime ideal it can have is the image of $\mathfrak m$ and hence its 0-dimensional.
answered Feb 1, 2012 at 21:07
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The above answer says there is a ring such that for any nonmaximal prime ideal there is a nonmaximal prime ideal containing it. In fact valuation ring with infinite products of integers group as valuation group is an example. A paper about spectral space of commutative rings written by Hochster tells us there is also ring with spectral space which order just inverse the spectral space of a given ring. So there also exists a ring such that for any nonminimal prime ideal there exists another nonminimal prime ideal which is contained in it. That is there is a ring which has no prime with height 1.
