# Generically finite morphisms

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

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Better assume $f$ dominant; otherwise it is (trivially) generically finite but you can say nothing about $\dim(X)$. – Laurent Moret-Bailly Sep 2 '11 at 11:15

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.

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 $\dim X\le 1$.

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As I said in a comment to the question, you should assume $f$ dominant. – Laurent Moret-Bailly Sep 2 '11 at 11:22
Oops, I implicitly thought $f$ is generically finite over its image. I will correct. – Qing Liu Sep 2 '11 at 12:12

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).

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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 $\dim O_{X,x}=\dim O_{Y,y}+\dim X_\eta$. – Qing Liu Sep 2 '11 at 12:18
OK, so assumptions for the converse would be: $f$ of finite type, $Y$ universally catenary, and $f(X)$ contains a closed point of $Y$. – Laurent Moret-Bailly Sep 2 '11 at 14:06
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 $f(X)$. – Qing Liu Sep 2 '11 at 22:17