Matsumura Example 2 (16.E) I am reading example 2 (16.E) of Matsumura's Commutative Algebra where he gives an example of a non-CM ring. 
Let $A = k[x,y]$ and $B = k[x^2, xy,y^2, x^3, x^2y, xy^2,y^3]$. Then $A,B$ have the same fraction field and $A$ is integral over $B$. Let $\mathfrak{m} = (x,y) \cap B$. Now it is clear to me that $x^4 \notin x^3B$ and $x^4\mathfrak{m} \subseteq x^3B$; whence $\mathfrak{m}\in \operatorname{Ass}_B (B/x^3B)$. However how does this imply that $B_{\mathfrak{m}}$ is not Cohen-Macaulay?
 A: Since $A$ is integral over $B$, $\dim A = \dim B=2$. In fact $\dim B_m = 2$, since $\operatorname{ht}(m) = \operatorname{ht}(x,y)=2$ (*). Hence, to show that $B_m$ is not CM, it is enough to show that its depth is less than $2$. 
In a CM local ring $(R,n)$ we have that a sequence $x_1,\dots,x_r$ is an $R$-sequence if and only if it is part of a system of parameters. Suppose that $B_m$ is CM. Now $x^3$ is regular in $B_m$ so it is part of a system of parameters. Since $\dim B_m =2$, there exists some $\omega \in mB_m$ such that $x^3,\omega$ is a system of parameters. But since $B_m$ is CM, $x^3,\omega$ is also a $B_m$-sequence. Hence $\omega$ is $B_m/(x^3B_m)$-regular. 
But $mB_m \in \operatorname{Ass}_{B_m}[B_m/(x^3B_m)]$, and so $\omega$ is a zero divisor on $B_m/(x^3B_m)$, contradiction.
(*, credit to user26857) By Corollary 5.9 in Atiyah-MacDonald (AM) or exercise 9.8 in Matsumura's Commutative Ring Theory we have that $2 = \operatorname{ht}_A(x,y) \le \operatorname{ht}_B(m)$ since $A$ is integral over $B$. But $\dim B = 2$ and so $\operatorname{ht} (m) = 2$. 
