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
 A: 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 $\dim X\le 1$. 
A: 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).
