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In chapter I, section 5.3 of Shafarevich Basic Algebraic Geometry 1, he defines a finite morphism between affine varieties as a dominant morphism $f: X \rightarrow Y$, such that the inclusion $f^{*}:k[Y] \rightarrow k[X]$ defines a integral extension ($k[X]$ is the coordinate ring of $X$).

He then proves that a finite morphism is surjective, and as a consequence he proves that a finite morphism is closed. The proof is as follows. He first observes that it is enough to check that $f(Z)$ is closed for every irreducible $Z \subset X$. Then he considers the restriction $f$ to $Z$, which is a finite morphism between affine varieties between $Z$ and the closure of $f(Z)$. The conclusion follows from the surjectivity.

Now I understand that both $Z$ and the closure of $f(Z)$ are affine varieties, but I can't see why the restriction of $f$ is a finite morphism. Specifically I can't see why the inclusion between the coordinate rings $k[\bar{f(Z)}] \subset k[Z]$ is an integral extension. Any suggestions?

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Any element $a\in k[Z]$ is the restriction of some element $b\in k[X]$, and this $b$ is integral over $k[Y]$. Now if $g$ is a monic polynomial with coefficients in $k[Y]$ such that $g(b)=0$, then $\tilde{g}(a)=0$, where $\tilde{g}$ is obtained by restricting each coefficient to an element of $k[\overline{f(Z)}]$. Indeed, $\tilde{g}(a)$ is just the image of $g(b)$ under the restriction map $k[X]\to k[Z]$. Thus $a$ is integral over $k[\overline{f(Z)}]$. Since $a\in k[Z]$ was arbitrary, $k[Z]$ is integral over $k[\overline{f(Z)}]$.

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