Finite maps between algebraic varieties are closed

Question

In the proof given by the book of Shafarevich "Basic algebraic geometry 1", section 5.3 of chapter 1, that every finite map between algebraic varieties $f:X \longrightarrow Y$ (where we assume $f(X)$ is dense in $Y$ and finite means that $k[X]$ is an integral extension of $k[Y]$), he uses that the restriction of $f$ to a closed set $Z$, $f':Z \longrightarrow \overline{f(Z)}$ is also finite. Why is this so?

Thanks in advance

Answer 1

Algebraically, this is the fact that if $\varphi: R \to S$ is an integral extension, then $R/\varphi^{-1}(I) \to S/I$ is again integral. This is more or less immediate to prove -- if $S$ is finitely-generated as an $R$-module, then so is $S/I$, by the same generators. And the same is true whether you think of $S/I$ as an $R$- or $R/\varphi^{-1}(I)$-module. (Here $R = k[Y], S = k[X]$ and $I = I(Z)$ in $X$.)

Geometrically, this is essentially the fact that the following maps are all finite:

• closed embeddings,
• compositions of finite maps,
• restrictions of finite maps $X \to Y$ to closed subsets of $Y$.

So $Z \hookrightarrow X \to Y$ is finite; and so is the "restriction" of $Z \to Y$ to $\overline{f(Z)}$.