Consider the ring $R = \mathbb Z[x_1,...,x_n]$ and let $\text{Spec}(R) = \{q \subset R : q \text{ a prime ideal}\}$ be the set of prime ideals of $R$. For $I \subset R$ an ideal, define $V(I) = \{q \in \text{Spec}(R) : I \subset q\}$ to be the set of prime ideals containing $I$. The $V(I)$'s are the closed sets of $\text{Spec}(R)$ in the Zariski topology.
For $f, g \in R \setminus \{0\}$ prime elements, I'm asked to compute $\dim V(I)$, where $I = (f,g)$ and the dimension is defined as here and is linked to here in the sense that $\dim V(I) = \dim (R/\text{rad}(I))$.
If $R$ were a finitely generated $K$-algebra for a field $K$, it would be easier. So I've tried to consider the quotient $\mathbb Z/p[x_1,...,x_n]$ for a prime $p$ in $\mathbb Z$.
However, I am not sure on whether $(f)$ and $(g)$ would still be prime in this quotient, but maybe I can choose $p$ so it is always the case. Even if they are, $I$ is not necessarily prime so this is also a problem.
So I wanted to quotient first by $(f)$ and then by $(g)$ to have the quotient by $(f,g)$ but again, we don't know if $(\bar{g})$ is prime in $\mathbb Z/p[x_1,...,x_n]/(f)$...
Has anyone got an idea ?