# Finite morphism is stable under base change.

In the scheme theory, there is a notion 'finite morphism'.

I find the fact that a finite morphism is stable under base change in the Wikipedia. (Link: https://en.wikipedia.org/wiki/Finite_morphism)

But I cannot prove or find why.

How can I prove it?

Proposition. If $f\colon X \to S$ is a finite morphism, then the morphism $f' \colon X \times_S S' \to S'$ is finite for all base extensions $g\colon S' \to S$.
Proof. Cover $S$ with open affines $V_i = \operatorname{Spec} B_i$; then, $g^{-1}(V_i)$ is an open cover for $S'$. Now cover these $g^{-1}(V_i)$ with open affines $V'_{ij} = \operatorname{Spec} B'_{ij}$. We claim that the preimages $$f^{\prime-1}(V'_{ij}) = X \times_S V'_{ij} \cong f^{-1}(V_i) \times_{V_i} V'_{ij}$$ of the $V'_{ij}$ are affine, with coordinate rings $A'_{ij}$ that are finitely generated as modules over $B'_{ij}$. Note that the isomorphism above is by the construction of the fiber product; see [Hartshorne, Thm. 3.3, Step 7].
First, affinity of $f^{\prime-1}(V'_{ij})$ follows since $$f^{\prime-1}(V'_{ij}) \cong f^{-1}(V_i) \times_{V_i} V'_{ij} \cong \operatorname{Spec}(A_i \otimes_{B_i} B'_{ij})$$
Now defining $A'_{ij} := A_i \otimes_{B_i} B'_{ij}$, we want to show that $A'_{ij}$ is finitely generated as a module over $B'_{ij}$. Since $A_i$ is finitely generated as a module over $B_i$, we have surjections $$B_i^{\oplus n_i} \twoheadrightarrow A_i$$ for each $i$. Now applying $-\otimes_{B_i} B'_{ij}$, by the right-exactness of the tensor product, we have surjections $$B_{ij}^{\prime\oplus n_i} \twoheadrightarrow A_i \otimes_{B_i} B'_{ij} =: A'_{ij}$$ for each $i,j$. Thus, $A'_{ij}$ is a $B'_{ij}$-algebra, which is finitely generated as a module over $B'_{ij}$. $\blacksquare$