I am reading Shaferevich's Basic Algebraic Geometry Book 1. In page 50 it proved that regular maps between quasiprojective varieties are continuous. I do not understand the proof:
Suppose $f:X\rightarrow Y$ is a regular map between quasiprojective varieties. $Z\subset Y$ is a closed subset of $Y$. Want to show that $f^{-1}(Z)$ is closed in $X$. Now for any point $x\in X$ there are neighbourhoods $U$ of $x$ and $V$ of $f(X)$ such that $f(U)\subset V\subset\mathbb A^m$ nad the map $f:U\rightarrow V$ is regular. Cover $X$ with such $U$'s. We can assume that $U$ is an affine variety. Since closedness is a local property it is enough to check that $f^{-1}(Z)\cap U=f^{-1}(Z\cap V)$ is closed in $U$. Since $Z\cap V$ is closed in $V$, it is defined by equations $g_1=\cdots=g_m=0$, where $g_i$ are regular functions on $V$. But then $f^{-1}(Z\cap V)$ is defined by the equation $f^*(g_1)=\cdots =f^*(g_m)=0$.
I do not understatnd the following things:
(i) Why $Z\cap V$ is closed in $V$ implies it is defined by equations $g_1=\cdots=g_m=0$, where $g_i$ are regular functions on $V$.
(ii) Where do we use the fact that "We can assume that $U$ is an affine variety".
Please help me. Thank you.
This is an application of Hilbert's Nullstellensatz (or of the usual corollaries) and Hilbert's Basissatz. Because $V$ is affine it can be embedded as a closed subset into some affine space $V\subset \mathbb{A}^m$. By Hilberts Nullstellensatz it is given as the vanishing set of some (radical) ideal $I\subset [X_1,\ldots,X_m]$. Similarly, $V$ is the vanishing set of some (radical) ideal $J\supset I$ and by the Basissatz it is finitely generated, say $J=(g_1,\ldots,g_n)$. (I think your implied statement that the number of generators and the dimension of the affine space coincides is a typo.) Now, the ring of regular functions on $V$ is isomorphic to (or defined as) the quotient $k[X_1,\ldots,X_m]/I$, and you can consider the $g_i$ as elements in it.