Let $k[x,y,z]$ ring of polynomials in $3$ indeterminants and consider an arbitrary non-constant polynomial $f \in k[x,y,z] \backslash k$.
How can I concretely make use of Krull's principal ideal theorem to show that $\dim k[x,y,z]/(f) =2$ holds?
I tried to argue in following way: Since $f$ is not constant and especially not zero Krull's principal ideal theorems says:
For every prime ideal $\mathfrak {p}$ that is minimal with property $f \in \mathfrak {p}$ holds $ht({\mathfrak {p}}) = 1$ (because $f \neq 0$ and $k[x,y,z]$ integral domain). This gives the estimation $\dim k[x,y,z]/(f) \le2$.
How can I conclude with PIT that also $\dim k[x,y,z]/(f) \ge 2$ holds?