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



Generically finite morphisms
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 $\dim X\le 1$. 


