Compute $\dim k[x,y, z]/(xz, yz)$ How do I compute the Krull dimension $\dim R=\dim k[x,y, z]/(xz, yz)$?
The variety $V$ defined by $k[V]=R$ is the $z$-axis together with the $(x, y)$-plane. How can I find the height of chains of irreducible algebraic sets here? Is it easier to look at chains of prime ideals instead?
 A: Here is a more 'hands on' approach.
$(\bar{z})\subset (\bar{z}, \bar{x})\subset (\bar{z}, \bar{x}, \bar{y})$ is a chain of prime ideals in $R$. So Krull $\dim R\geq 2$.
For any $p$ a prime ideal in $R$ such that $\bar{z}\notin p$, either $\bar{x}\in p_0$ or $\bar{y}\in p_0.$
Let $\phi: k[x, y, z]\to R$.
Case (1) Assume there is a chain $p_0\subset p_1\subset p_2\subset p_3$ of distinct prime ideals in $R$ such that $\bar{z}\notin p_0$.
If both $\bar{x}, \bar{y}\in p_0$ then $(\bar{x}, \bar{y})\subset p_0.$ So the quotient map $k[x, y, z]\to R/p_0$ factors through $\psi: k[x, y, z]/(x, y)=k[z]\to R/p_0$.
Since $k[z]$ is a PID, $$\psi^{-1}(p_0)\subset \psi^{-1}(p_1)\subset 
 \psi^{-1}(p_2)\subset \psi^{-1}(p_3)$$ is a chain of prime ideals in $k[z]$, which implies $\psi^{-1}(p_1)=
 \psi^{-1}(p_2)= \psi^{-1}(p_3)$. In particular $p_1=p_2=p_3$, a contradiction.
On the other hand if $x\in p_0$ and $y\notin p_0$, then  since in $R$, $\bar{y}\bar{z}=0$. $\bar{y}\bar{z}\in p_0$ implies $\bar{z}\in p_0$, again a contradiction to the assumption that  $\bar{z}\notin p_0$.
Case (2) So by above reasoning we can start with a chain $p_0\subset p_1\subset p_2\subset p_3$ in $R$ such that $\bar{z}\in p_0$.
So the quotient map $k[x, y, z]\to R/p_0$ factors through $\psi: k[x, y, z]/(z)=k[x, y]\to R/p_0$.
Since $k[x, y]$ is of Krull dimension 2, and $$\psi^{-1}(p_0)\subset \psi^{-1}(p_1)\subset 
 \psi^{-1}(p_2)\subset \psi^{-1}(p_3)$$ is a chain of prime ideals in $k[x, y]$, which implies $\psi^{-1}(p_2)= \psi^{-1}(p_3)$. In particular $p_2=p_3$, a contradiction.
A: Here's a quick proof if you're willing to take it on faith that $\dim k[x_1,\dots,x_n]=n$ for $n\in\mathbb{N}$; let me know if you would like a reference for this fact. In particular, $\dim k[x,y,z]=3$. Now, for an arbitrary integral domain $S$ and a non-zero ideal $I\leqslant S$, we have $1+\dim S/I\leqslant \dim S$. (Why?)

 Suppose $Q_0<\dots<Q_n$ is a strictly ascending chain of prime ideals of $S/I$. Recall that the prime ideals of $S/I$ correspond with the prime ideals of $S$ that contain $I$, so we can pull the $Q_i$ back along the projection map $S\to S/I$ to obtain a chain $$I\leqslant P_0<\dots<P_n$$ of prime ideals of $S$. Since $I\neq \langle 0\rangle$, we have in particular that $P_0\neq\langle 0\rangle$, so that $$\langle 0\rangle<P_0<\dots<P_n$$ is a strictly ascending chain of ideals of $S$. Furthermore, since $S$ is a domain, $\langle 0\rangle$ is prime, so this chain forces that $1+n\leqslant \dim S$, as desired.

Taking $S=k[x,y,z]$ and $I=\langle xz,yz\rangle$ thus tells us that $\dim R\leqslant 2$. Conversely, the prime ideals of $R$ correspond to the prime ideals of $S$ containing $I$. In particular, since $I$ is contained in the ideal $\langle z\rangle<S$, the strictly ascending chain of prime ideals $$\langle z\rangle<\langle z,x\rangle<\langle z,x,y\rangle$$ of $S$ pushes forward to a strictly ascending chain of prime ideals of $R$, whence $\dim R\geqslant 2$. So we get the desired result that $\dim R=2$.
