What is the Krull dimension of $A[x,y,z]/\langle xy,xz \rangle$

Question

What is the Krull dimension of $B=A[x,y,z]/\langle xy,xz \rangle$, given that $A$ is a Noetherian commutative ring?

Answer 1

It's obvious that $$\dim B=\sup\{\dim B/P:P\text{ minimal prime}\}.$$

Let $P$ be a minimal prime of $B$. Then $P=\mathfrak p/(XY,XZ)$ with $\mathfrak p$ prime in $A[X,Y,Z]$ minimal over $(XY,XZ)$. Then $\mathfrak p$ is minimal over $(X)$ or $\mathfrak p$ is minimal over $(Y,Z)$. Conversely, every minimal prime over $(X)$, respectively $(Y,Z)$ gives rise to a minimal prime of $B$.

Then $$\dim B=\sup(\dim A[X,Y,Z]/(X),\dim A[X,Y,Z]/(Y,Z))=\sup(\dim A[Y,Z],\dim A[X])=\sup(\dim A+2,\dim A+1)=\dim A+2.$$

Answer 2

The dimension is precisely $\dim A+2$. We have $\dim B\geq\dim B/x=\dim A+2$. On the other hand, since any prime ideal of $B$ must either contain $x$ or both $y,z$, one can check that the above inequality is actually an equality.