Let $k$ be an infinite field, $S=k[x_{11},x_{12},x_{21},x_{22}]/(x_{11}x_{22}-x_{12}x_{21})$. Let $p=(x_{11},x_{12})$, $q=(x_{21},x_{22})$. Show that
(i) $p$ and $q$ are prime ideals in $S$ and maximal Cohen-Macaulay $S$-modules.
(ii) $p^j$ and $q^j$ are not Cohen-Macaulay for $j\geq 2$.
This question is from Bruns-Herzog, Cohen-Macaulay Rings, Exercise 6.4.17 (b). I am not sure here we say an ideal $p$ is Cohen-Macaulay, means $p$ as module is Cohen-Macaulay or its quotient ring $S/p$ is Cohen-Macaulay.
Typically, when an ideal $I$ is called Cohen-Macaulay, it is the quotient ring $R/I$ that is intended to be Cohen-Macaulay. However, in this case, $\dim S = 3$ but $S/p \cong k[x_{21}, x_{22}]$ $\implies \dim S/p = 2$, so $S/p$ is not maximal Cohen-Macaulay as an $S$-module (although $S/p$ is CM, either as an $S$-module or as a ring, and similarly $S/p^j$ is not CM for $j \ge 2$).
Note that any $S$-ideal $I$, viewed as an $S$-module, has trivial annihilator (since $S$ is a domain), so $\dim I := \dim S/\text{ann}_S(I) = \dim S$. Finally, the following fact (proven using the Depth Lemma) suffices to finish the problem:
Proposition: Let $R$ be a Noetherian local ring, and $I$ an $R$-ideal with $\text{depth}(R) > \text{depth}(R/I)$. Then viewing $I$ as an $R$-module, $\text{depth}(I) = \text{depth}(R/I)+1$.