Krull dimension and zero divisors of $k[x,y,z]/(x^ay,x^bz)$

Question

I found the primary decomposition of $(0)$ in the ring $k[x,y,z]/(x^ay,x^bz)$, where $a\geq b \geq 1$, $k$ is alg. closed, to be $(x^b) \cap (x^a,z) \cap (y,z)$ (is this correct?). Now I am now looking for the zero divisors of the ring, and I understand that they can be related to the primary decomposition somehow, but I don't know how. On the dimension question I don't even know where to start, so a nudge in the right direction would be great.

Answer 1

The primary decomposition is correct.

The set of zero divisors is the union of the associated primes, that is, $(x)\cup(x,z)\cup (y,z)=(x,z)\cup (y,z)$.

For the Krull dimension let's look at the prime ideals in $k[x,y,z]$ minimal over $(x^ay,x^bz)$ and to the maximal chains of prime ideals they provide, that is, find their coheight. The minimal primes are $(x)$ and $(y,z)$. The first one gives rise to a chain of prime ideals of maximal length two. (Think what's going on in the factor ring $k[x,y,z]/(x)\simeq k[y,z]$ whose dimension is $2$.) The second provides a chain of length one, so the Krull dimension of your ring is $2$.