Checking a ring is not Cohen-Macaulay

Question

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".

Answer 1

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.

Answer 2

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.