Let $B$ a ring of Krull dimension $1$ and suppose that $\text{Spec }B$ is irreducible. And let $f:X\to\text{Spec}(B)$ a $B$ scheme with the following properties:
- $X$ is projective, regular and integral
- $f$ is a flat morphism of relative dimension $1$ (this implies that $X$ has dimension $2$).
Now suppose that we have an embedding $\tau:B\hookrightarrow F$ in a field $F$. Then we have the base change $f_\tau:X_\tau:=X\times_B \text{Spec } F\to \text{Spec }F$ (through $\tau$). The morphism $f_\tau$ is again flat, but what about its relative dimension? It should be again $1$, so $X_\tau$ has also dimension $1$, but I am not able to prove it. Can you help me?
Edit: In other words I'm asking a particular case of the following claim: "the relative dimension is stable under base change".