9
$\begingroup$

I have a question concerning one of the exercises of Hartshorne, Ch. II. Namely: Exercise 3.7 about gerneically finite morphisms.

A morphism $f: X \rightarrow Y$ with Y irreducible and $\eta$ generic point of Y, is called generically finite, if $f^{-1}(\eta)$ is a finite set. Now let $f: X \rightarrow Y$ be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset $U \subseteq Y$ sth. $f^{-1}(U) \rightarrow U$ is finite.

As a hint, one should prove first that the function field of $X$ is a finite field extension of the function field of $Y$.

I have started with the affine case: $X = \operatorname{Spec}\ A$, $Y = \operatorname{Spec}\ B$ with function fields $K$ resp. $L$. The assertion that $K$ now is a finite field extension of $L$ follows from Noether's Normalization theorem and Zariski's Lemma.

But how can I proceed from there? I would be grateful for an idea to help me go on. Thank you!

$\endgroup$
1
$\begingroup$

Here's a diagram of the algebraic setup. All the arrows are inclusions. $$ \require{AMScd} \begin{CD} A @>>> A \otimes_B L @>>> K\\ @AAA @AAA\\ B @>>>L \end{CD} $$ Since $K$ is finite over $L$ you know that algebra generators for $A$ over $B$ satisfy monic polynomials with coefficients in $L$. If you localize $B$ at a well-chosen element $f$ you can upgrade this statement!

$\endgroup$
  • $\begingroup$ Thank you. So after localization at $f$ I get an integral extension which means since $A_f$ is still finitely generated over $B_f$ that I got my finite morphism on the open, dense subset $U = \text{Spec}\ A_f$. $\endgroup$ – Chris Jun 1 '15 at 21:24
  • $\begingroup$ localized at f, which is the multiplication of the denominators of all the L coefficients of all the algebra generators for A over B. Then all the algebra generators for $A_f$ over $B_f$ are integral over $B_f$. So $A_f$ is integral over $B_f$. $\endgroup$ – Shuhang Dec 21 '17 at 23:07
8
$\begingroup$

I want to post a message to say, after applying Hoot's answer, you're still 'not done yet' (this problem is quite technical).

Let's recall the situation. You can reduce to $f:X\rightarrow Y=\operatorname{Spec} B$ dominant morphism of integral schemes of finite type (we no longer need the generically finite assumption) inducing a finite extension $K(Y)\rightarrow K(X)$. This means that $X$ is covered by $U_i=\operatorname{Spec} A_i$ open such that $B\rightarrow A_i$ is finitely generated, and $A_i$ is generated by elements algebraic over $B$ as Hoot said. Note that in fact $B\hookrightarrow A_i$ is mono ($\forall i$) since $U_i$ is dense in $X$ since $X$ irred., and $f$ is dominant, so $U_i$ is dense in $Y$. Let $x_{ij}$ generate $A_i$ over $B$. Then $x_{ij}$ satisfies an equation of algebraic dependence over $B$ with leading coefficient $b_{ij}$. Put $U=\{b_{ij}\}_{i,j}$; then $W:=\operatorname{Spec} B[U^{-1}]$ is an open affine subset of $Y$ and again since $U_i\rightarrow Y$ is dense, $f^{-1}(W)\cap U_i\ne\emptyset\forall i$. Moreover, now $B[U^{-1}]\hookrightarrow A_i[U^{-1}]$ has $A_i[U^{-1}]$ finitely generated by integral elements; hence is finite over $B[U^{-1}]$.

We have reduced to $f : X\rightarrow Y=\operatorname{Spec} B$ with $U_i=\operatorname{Spec} A_i$ covering $X$ and $B\hookrightarrow A_i$ finite. But we need to find a nonempty open subset $V\subset Y$ such that $f:f^{-1}(V)\rightarrow V$ is finite (in particular, affine). Put $W:=\bigcap U_i$ (nonempty open since $X$ irred.). Let $\mathfrak a_i\subset A_i$ be such that $U_i\supset V(\mathfrak a_i)=U_i-W$. $B\subset A_i$ are integral domains, and $K(A_i)$ is algebraic over $K(B)$. It is then a fact that every nonzero ideal of $A_i$ intersects $B$ nontrivially. $\mathfrak a_i$ is nonzero since $W\ne\emptyset$. Hence $\exists f_i\in B\subset A_i$ such that $D(f_i)\subset W\subset U_i$. Put $U:=\{f_i\}_i$. Then $V:=\operatorname{Spec} B[U^{-1}]$ is nonempty affine open in $Y$ and so $\emptyset\ne f^{-1}(V)\subset W$. Note that $f^{-1}(V)$ coincides with $\bigcap_i X_{f_i}$, where $f_i$ is the image of $f_i$ in $\Gamma(X,\mathscr O_X)$ ($B$ injects into global sxns since $f$ dominant). Moreover $f^{-1}(V)\subset W$ hence coincides with $\bigcap_i X_{f_i}\cap U_j=\bigcap_i D(f_i)\cap U_j\forall j=\operatorname{Spec} A_j[U^{-1}]$, where as always we identify $B$ with its image as appropriate. Hence $f:f^{-1}(V)\rightarrow V$ is our desired finite morphism.

One last thing: here is a different (probably functionally equivalent) way to prove that $K(X)$ is a finite extension of $K(Y)$. Reduce to $f:\operatorname{Spec} A=X\rightarrow Y=\operatorname{Spec} B$ dominant generically finite morphism of integral affine schemes of finite type. Identifying $B$ with its monomorphic image we have $B\subset A$ domains with only finitely many primes of $A$ lying over $(0)\in\operatorname{Spec} B$. Replacing $B$ by $K(B)$ we have $A$ finitely generated over a field hence Noetherian and Jacobson (by the Nullstellensatz). A Noetherian ring $A$ is Jacobson iff $\forall\mathfrak p\in\operatorname{Spec} A$ of dimension one, $R/\mathfrak p$ has infinitely many ideals. $A$ has only finitely many ideals after replacing $B$ by $K(B)$, so all primes of $A$ must be maximal. Thus $A$ is Artinian (and Noetherian) hence finite over $K(B)$. Therefore $K(A)=K(X)$ is a finite extension of $K(B)=K(Y)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.