Commutative algebra, Krull dimension. Eisenbud, Exercise 10.6

Question

I need help to solve this exercise. If anyone can help, thanks in advance!

Let $R=k[x,y,s,t]/(xs-yt)$ and $S=R/(x,y)=k[s,t]$. Let $P=(s,t)\subset R$. Show that $\operatorname{codim} P=1$ but $\operatorname{codim} PS=2$.

Answer 1

First note that $xs-yt$ is an irreducible polynomial in $k[x,y,s,t]$.
If the height of $P$ in $R$ is greater than $1$, then there is a prime ideal $Q$ such that $(xs-yt)\subsetneq Q\subsetneq P$, and thus the height of $P$ in $k[x,y,s,t]$ is at least $3$, a contradiction.
Since $PS=(s,t)k[s,t]$ we get the height of $PS$ equals $2$.

Answer 2

Did you think about the geometry? $R$ corresponds to a cone (a three dimensional cone in $4$-space). The ideals $P$ and $Q := (x,y)$ each cut out a plane in $4$-space; these two planes meet in a single point --- the cone point.

Each plane has codimension one in the cone; this corresponds to the fact that each of $P$ and $Q$ has codimension one as prime ideals.

But $PS = P (R/Q)$ has codimension two, because $S/PS = R/(P,Q)$, corresponding to the intersection of the two planes, which is the single cone point. (And a point has codimension two in the plane that $S$ corresponds to.)