Hartshorne Exercise II.3.7.
A morphism $f:X \ra Y$, with Y irreducible, is generically finite if $f^{-1}(\eta)$ is a finite set, where $\eta$ is the generic point of Y. A morphism $f:X \ra Y$ is dominant if $f(X)$ is dense in $Y$. Now let $f:X \ra Y$ be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset $U\subseteq Y$ such that the induced morphismf $f^{-1}(U)\ra U$ is finite. [Hint: First show that the function field of $X$ is a finite field extension of the function field of $Y$.]
Remark. Let us first make the following remark. Suppose $X$ and $Y$ are varieties over a field $k$ and $f:X\ra Y$ is any morphism. Given an open affine $V=\spec B\subset Y$, an affine open $U=\spec A\subset X$ such that $f^{-1}(V)\cap U \ne\varnothing$ can be covered with standard opens $\{g\ne 0\}$ with $g\in A$, giving a morphism of affines $f:\{g\ne 0\}\ra V $ that makes $A[\frac 1g]$ into a finite type $B$-algebra. We can cover $f^{-1}(V)$ with such standard affine opens, and because $X$ is quasicompact, finitely many will do. That is to say that $f$ is automatically a finite type morphism.
Let $f:X\ra Y$ be a dominant, generically finite morphism of varieties. Let $V=\spec{B}$ be any affine open, necessarily a neighborhood of the generic point $\eta_Y\in Y$. Let $U=\spec{A}\subset X$ be any affine open and denote $L=\op{Frac}(A)=\Bbb K(X)$, $K=\op{Frac}(B)=\Bbb K(Y)$. Due to the dominance of $f$, there is an induced map of function fields $\Bbb K(Y)\ra \Bbb K(X)$, which I claim is finite extension. The generic fiber $X_{\eta_Y}$ is covered by affine sets of the form $U_{\eta_Y}=\spec{A\otimes_BK}$ which is a finite set by assumption. As $A$ is a finite type $B$-algebra, $A\otimes_BK$ is a finite type $K$-algebra. By NNL, it follows that $U_{\eta_Y}\tra\Bbb A_K^d$ is a finite morphism for some $d\in\N$, so we must have $d=0$, hence $A\otimes_BK$ is finite over $K$. Since $A\otimes_BK$ is just a localization of $A$, it is a domain, so it follows that $A\otimes_BK$ is field and is equal to $L$.
Now, if $X=\spec{A}$ and $Y=\spec{B}$ are affine, then the homomorphism $\phi:B\ra A$ gives $A\cong B[a_1,…,a_n]$ as a finite type B algebra. As $L/K$ is finite, each $a_i$ is algebraic over K, and thus integral over $B[\frac{1}{b_i}]$ for some $b_i\in B$. It follows that $A[\frac{1}{\phi(b)}]$ is integral over $B[\frac{1}{b}]$ for some $b\in B$, which is the product of many elements. Since integral + finite type = finite and $f^{-1}(\{b\ne 0\})=\{\phi(b)\ne 0\}$, the affine case is done. For the general case, take $V$ and $U$ as before. Shrinking $U$ if necessary we may assume $U\subset f^{-1}(V)$ so that $U\ra V$ is induced by the ring homomorphism $\phi:B\ra A$. Then cover $f^{-1}(V)$ with standard affine opens, finitely many of which will do: $W_1,…,W_r$. Each $W_j=\spec{A_j}$ gives a morphism of affine varieties that lands in $V$. Obviously it remains generically finite, and dominance follows from $f(\eta_X)=\eta_Y$. By the earlier remark, $f$ is already finite type, at least with the stronger assertion that the preimage of any affine open $V$ is covered by affine opens whose rings of regular functions are all finite as $\OO_Y(V)$-modules. However, there are examples of finite type, quasifinite, and dominant (even surjective) morphisms with this property (e.g., the affine line with a doubled origin mapping to $\A{k}{1}$); a finite morphism is so strong because the preimage of any affine open must be affine. Clearly the generically finite hypothesis is crucial. Indeed, we saw that $(W_j)_{\eta_Y}$ must be the point ${\eta_X}$, hence $(W_j-U\cap W_j)\cap (W_j)_{\eta_Y}=\varnothing$. Writing $Z_j=W_j-U\cap W_j=\spec{A_j/I_j}$ as a closed subset of $W_j$, the corresponding scheme theoretic intersection with the generic fiber is,
$$Z_j\times_{W_j}(W_j)_{\eta_Y}=\spec{A_j/I_j\otimes_{A_j}(A_j\otimes_BK})=\spec{A_j/I_j\otimes_BK}=\varnothing.$$
Thus, there is some nonzero element $h_j\in B$ whose image in $A_j/I_j$ is zero; that is, the ring homomorphism corresponding to
$$W_j\ra V$$
sends $h_j$ inside $I_j\subset A_j$, and we thus have $f^{-1}(\{h_j\ne 0\})\cap W_j\subset W_j\cap U$ (because any point in $W_j$ that is sent to a point in $V$ which does not contain $h_j$ cannot live in $Z_j$). Taking $h:=h_1\cdot…\cdot h_r\in B$, since the $W_j$’s form an open cover for $f^{-1}(V)$, we conclude that $f^{-1}(\{h\ne 0\})\subset U$. Thus $f$ restricts to
$$f^{-1}(\{h\ne 0\})=\{\phi(h)\ne0\}\ra\{h\ne 0\}.$$
By our previous work, it follows that $\{h\ne0\}\subset Y$ is the desired dense open subset on which $f$ restricts to be finite.
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