# Checking a ring is not Cohen-Macaulay

Mathematics Asked on December 6, 2020

While reading a book, I found an example that said that the ring $$K[w,x,y,z]/(wy,wz,xy,xz)$$ is not Cohen-Macaulay. In order to check this, it is stated to take the quotient by the ideal generated by the non-zero divisor $$w-y$$, that gives the ring $$K[x,y,z]/(y^2,yz,xy,xz)$$. The book assures that all the elements of its maximal ideal are zero divisors.

I couldn’t follow through this last part: I don’t see why this ring would be local (do I need to localize in a maximal ideal?)

You can find this section of the pdf file by looking up "25. Regular local rings Corollary 25.1".

Observe that $$(wy, wz, xy, xz) = (w, x) cap (y, z),$$ hence the height of $$(wy, wz, xy, xz)$$ is $$2.$$ We have therefore that $$R = k[w, x, y, z] / (wy, wz, xy, xz)$$ has dimension $$2.$$ On the other hand, the element $$w - y$$ is neither a zero divisor nor a unit, hence the sequence $$(w - y)$$ is $$R$$-regular. Going modulo $$(w - y),$$ we have that $$R / (w - y) cong k[x, y, z] / (y^2, yz, xy, xz),$$ and each of the generators of the unique homogeneous maximal ideal $$mathfrak m = (x, y, z)$$ is a zero divisor. Consequently, there are no regular sequences in $$R / (w - y),$$ hence it follows that $$operatorname{depth} R = 1 < 2 = dim R$$ so that $$R$$ is not Cohen-Macaulay.

Answered by Carlo on December 6, 2020

If it were local, all its quotients would be local, but factoring out by the ideal generated by $$y$$ and $$z$$, you get $$K[x]$$, which isn't local.

Secondly, the minimal primes of the second quotient are pretty clearly the ideals $$(x,y)$$ and $$(y,z)$$, quotienting by them we get $$K[z]$$ and $$K[x]$$ respectively, so it becomes clear that among the maximal ideals there are $$(x,y,p(z))$$ and $$(p(x), y, z)$$, where the $$p$$ represents an irreducible polynomial.

It isn't clear to me how $$x-1$$ would be a zero divisor in $$(x-1,y,z)$$ in the second quotient.

That doesn't completely rule out that the example isn't Cohen-Macaulay, though, only that the hint appears to have problems. Unfortunately, I'm unfamiliar with Cohen-Macaulay rings and can't tell you straightaway. Perhaps if you provide more info about the work this appeared in.

Answered by rschwieb on December 6, 2020