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Is there an example of a finite dimensional local ring $(A,m)$ (maybe Noetherian, preferably not too far away from a ring that would arise when studying algebraic varieties) with a prime ideal $P \subset A$ that cannot be extended to a maximal chain of primes?

As user26857 points out, a cleaner way to ask what I want is:

For a Noetherian local ring $A$, is it true that for every prime $P \subset A$ we have $\operatorname{ht}P+\dim A/P=\dim A$? (Or at least for a finite dimensional ring $A$, not necessarily Noetherian.)

I know of an example when the locality is dropped: $\operatorname{Spec} k[x]_{(x)}[t]$ and the prime $P = (tx - 1)$. I don't see how to profitably localize this example to get an example of what I want. For varieties I know that this cannot happen.

user26857
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Elle Najt
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  • @user26857 I mean that there is no chain of primes $P_0 \subset P_1 \ldots P = P_k \subset P_{k+1} \subset \ldots ... P_n$ which is of the maximal length guaranteed by the Krull dimension of the ring being $n$. – Elle Najt Nov 25 '15 at 12:37
  • Maybe you want to ask if for any prime ideal $p$ the following holds: $ht p+\dim A/p=\dim A$. – user26857 Nov 25 '15 at 12:39
  • @user26857 Yes, the truth of that formula is what I am asking about. – Elle Najt Nov 25 '15 at 12:41

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There are prime ideals $\mathfrak p$ in noetherian local rings $A$ such that $\operatorname{ht}\mathfrak p+\dim A/\mathfrak p<\dim A$.

For instance, let $A=K[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, and $\mathfrak p=(y,z)$. Then $\operatorname{ht}\mathfrak p=0$ (since $\mathfrak p$ is minimal), and $\dim A/\mathfrak p=1$ (since $A/\mathfrak p\simeq K[X]_{(X)}$). On the other side, $\dim A=2$ (since $(x)\subset(x,y)\subset(x,y,z)$ is a chain of prime ideals).

If the local ring $A$ is Cohen-Macaulay, then for any prime $\mathfrak p$ we have $\operatorname{ht}\mathfrak p+\dim A/\mathfrak p=\dim A$.

user26857
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  • Thanks. So this kind of phenomenon will occur at any intersection between components of different dimension on a reducible variety? It seems to me that the point is that this example works already for $K[X,Y,Z] / (XY,XZ)$ because of the pathologies of codimension for reducible varieties, and localizing at the intersection of the line and the plane doesn't change any dimension computations. – Elle Najt Nov 25 '15 at 12:59