$A$ a CohenMacaulay ring (not necessarily local), $M$ a CohenMacaulay $A$module. Then does it necessarily follow that $\mbox{ann}(M)$ is heightunmixed?
This is true if $A$ is local but fails in general.
First, a counterexample. Let $A=\mathbb Z[X]$ and $M= A/p\oplus A/q$ with $p=(2)$, $q=(3,X)$. Since any maximal ideal $m$ of $A$ can not contain both $2$ and $3$, when you localize at $m$ only one of the summands can survive at most, so $M_m$ will be CM. The annihilator is $p\cap q$ which is mixed.
Now assume $A$ is local and $I = \text{ann} (M)$ . We need to show that all the associated prime of $I$ have same height. Any associated prime $p$ of $I$ would be an associated prime of $M$, as $M$ is a faithful $A/I$ module. But then $\dim A/p = \dim M$ (see BrunsHerzog, Theorem 2.1.2). Now, as $A$ is also CM, $\text{height}(p) = \dim A  \dim A/p =\dim A\dim M$. QED

$\begingroup$ Thanks! I was aware of the local result, but spent an inordinate amount of time today trying to prove the nonlocal version. Your example puts an end to that. Greatly appreciated! $\endgroup$ – ashpool Nov 21 '10 at 1:45