# Converse of Principal Ideal Theorem

$(A, \mathfrak{m})$ a Noetherian local ring, $a\in\mathfrak{m}$ a zero divisor. Then is it true that $\mbox{dim}\ A/(a) = \mbox{dim}\ A$ ?

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No. Let A be the ring k[x,y,z]/(xz,yz) localized at (x,y,z). Then the dimension of A is 2, but the dimension of A/(x) is one. Note that xz = 0 in A.

The geometric picture is this: In 3-space, take the union of the horizontal z=0 plane with the vertical line x = y = 0, and look at the local ring at the origin. Then restricting to x=0, we get the union of the y-axis and the z-axis.

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Such pictures are good for the memory. – Martin Brandenburg May 12 '10 at 0:05
@Charles Staats: What is the most easiest way to see that dimA=2? Thanks a lot! – TmobiusX May 13 '10 at 7:02

Necessary and sufficient conditions for your question to have a positive answer are that $A$ be $\textit{unmixed}$, that is, every associated prime is a minimal prime, and $\textit{equidimensional}$, that is, $\mathrm{dim} (A/\mathfrak{p}) = \mathrm{dim}(A)$ for every minimal prime $\mathfrak p$. Charles Staats' example is not equidimensional. The one-dimensional ring $A=k[x,y]/(x^2,xy)$ is not unmixed, and $A/(y)$ has dimension zero. Sufficiency follows from the fact that the set of zerodivisors of $A$ is equal to the union of the associated primes.

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