2
$\begingroup$

My question concerns the notion of a generically finite morphism

$f: X \rightarrow Y$ of "nice" schemes, say integral and noetherian.

I want to define $f$ gen. finite if the generic fibre is finite.

Can I characterize this property somehow by the relation of the dimensions of $X$ and $Y$?

For example, can I conclude that if the morphism is gen.finite, then $dim(X) \le dim(Y)$? Does also the converse hold? Or what are related criteria for a morphism to be gen.finite?

Thanks

$\endgroup$
1
  • 2
    $\begingroup$ Better assume $f$ dominant; otherwise it is (trivially) generically finite but you can say nothing about $\dim(X)$. $\endgroup$ Sep 2, 2011 at 11:15

2 Answers 2

6
$\begingroup$

If $f$ is locally of finite type EDIT: and dominant, then your inequality holds. First replace $Y$ by the Zariski closure of $f(X)$ and we are reduced to the case when $f$ is dominant. For all $x\in X$ and $y=f(x)$, we have the dimension formula $$ \dim O_{X,x}+ \dim (\overline{\lbrace x \rbrace}\cap X_y)\le \dim O_{Y, y}+\dim X_{\eta}=\dim O_{Y,y}$$ where $\eta$ is the generic point of $Y$.

If $f$ is not necessarily of finite type, you have to define the notion of "generic finite". Do you mean the generic fiber is a finite set or is finite over $k(Y)$ ? But in anyway, I don't think it is true in general.

Add A reference for the above dimension formula is EGA, Proposition IV.5.6.5.

Full answer to the question

Let $f : X\to Y$ be a dominant morphism of integral Noetherian schemes. Suppose that the generic fiber of $f$ is finite as a scheme over $k(Y)$ ($f$ not necessarily of finite type). Then $\dim X\le \dim Y$.

Proof. 1) One can suppose $\dim Y<\infty$ and $X, Y$ are affine.

2) The finiteness hypothesis implies that $k(X)$ is a finite extension of $k(Y)$ (algebraic extension will be enough).

3) write $X=\mathrm{Spec} B$ and $Y=\mathrm{Spec} A$ and let $d\ge 1$ be a positive integer. Let $$P_0 \subset P_1 \subset ... \subset P_d$$ be a strictly increasing chain of prime ideals of $B$. As $B$ is Noetherian, there exists a finite subset $S$ of $B$ which contains a familly of generators of $P_i$ for all $i\le d$. Let $C$ be the sub-$A$-algebra of $B$ generated by $S$. Let $Q_i=P_i\cap C$ and let us show that $Q_i\ne Q_{i+1}$. Otherwise $P_{i+1}\cap C\subseteq P_i$. As $C$ contains a set of generators of $P_{i+1}$, this would imply that $P_{i+1}=P_i$. Contradiction. So $d\le \dim C$.

4) By construction $C$ is finitely generated over $A$. Moreover, if $K=k(Y)$, then $C\otimes_A K$ is a sub-$K$-algebra of the algebraic extension $k(X)/K$, so it is a field. This implies that $\mathrm{Spec} C\to Y$ is dominant, of finite type, and generically finite. By the previous result, $\dim C\le \dim Y$. Hence $d\le \dim Y$ and $\dim X\le \dim Y$.

Remark Without the finiteness hypothesis on $k(X)/k(Y)$, one can still say something. Suppose for instance that $\dim Y=1$ and the generic fiber of $X\to Y$ is a single point (not necessarily a finite scheme). One can show $\dim X\le 1$ as follows: first we can suppose $X, Y$ are affine and $Y=\mathrm{Spec} A$ is local. Let $h$ be a non-zero non-invertible element of $Y$. Then $D(h)$ consists in the generic point of $Y$. So $D(h)$ in $X$ is just the generic point of $X$. So $O(X)_h$ is a field. By a result of Artin-Tate (see Ulrich Görtz & Torsten Wehorn, Algebraic geometry I, Corollary B.62), $O(X)$ is semi-local of dimension $\le 1$. So $\dim X\le 1$.

$\endgroup$
2
  • 3
    $\begingroup$ As I said in a comment to the question, you should assume $f$ dominant. $\endgroup$ Sep 2, 2011 at 11:22
  • $\begingroup$ Oops, I implicitly thought $f$ is generically finite over its image. I will correct. $\endgroup$
    – Qing Liu
    Sep 2, 2011 at 12:12
4
$\begingroup$

The converse is true if $X$ and $Y$ are of finite type over a field. It is false in general: take $Y=\mathrm{Spec}(R)$ where $R$ is a discrete valuation ring with quotient field $K$, and $X=\mathbb{A}^1_K$ (which is of finite type over $Y$ because $K=R[t^{-1}]$ if $t$ is a uniformizer).

$\endgroup$
3
  • $\begingroup$ If $f$ is surjective locally of finite type and $Y$ is universally catenary, then the converse is true: for any $y\in Y$, take $x\in X_y$ closed. Then $\dim O_{X,x}=\dim O_{Y,y}+\dim X_\eta$. $\endgroup$
    – Qing Liu
    Sep 2, 2011 at 12:18
  • $\begingroup$ OK, so assumptions for the converse would be: $f$ of finite type, $Y$ universally catenary, and $f(X)$ contains a closed point of $Y$. $\endgroup$ Sep 2, 2011 at 14:06
  • $\begingroup$ The assumption on $f(X)$ is not sufficient in general. Let $R$ and $K$ be as above. Consider the morphism $f: X=\mathbb A^2_K\to Y=\mathbb A^1_R$ which consists in the projection to $\mathbb A^1_K$ composed with the canonical inclusion. Then $\dim X=\dim Y$ and there are plenty of closed points of $Y$ in $f(X)$. $\endgroup$
    – Qing Liu
    Sep 2, 2011 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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