Why is this ring not Cohen-Macaulay?

commutative-algebra cohen-macaulay

Solution 1:

Certainly $(x^3y)^2y^4 = x^6y^6 = x^4(xy^3)^2$ is zero when you mod out by $x^4$. The key thing to see is that $(x^3y)^2$ is not inside $(x^4)$. Why is this?

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