necessary and sufficient condition for Normalization of Integral schemes Let $f:Y \longrightarrow X$ be a morphism of integral schemes. I was wondering if the following is true?
$f$ is the normalization morphism $\Leftrightarrow$ $Y$ is normal and $f$ is birational and integral.
I know that the direct sense ''$\Rightarrow$" hold, but I don't really see it for the other implication, can anyone clarify if it does hold? or how?
Thanks.
 A: This result is true. Let $Q(R)$ denote the fraction field of a domain $R$.
We'd like to show that $f:Y\to X$ is the same as the normalization. It suffices to do this locally on $Y$ if we can do it in a compatible manner. Since $f$ is affine by definition of an integral morphism, we can cover $Y$ by open affines which are the preimage of open affines in $X$, say $V=\operatorname{Spec} B \to \operatorname{Spec} A=U$  where $U\subset X$ and $V\subset Y$ are open affines. Since birational morphisms are dominant, we have that the generic point of $V$ maps to the generic point of $U$ and thus an injective map on fraction fields $Q(A)\to Q(B)$ which by birationality is an isomorphism. Further, $A\subset Q(A)$ lands in $B\subset Q(B)$ under this morphism, and so the integral closure of $A$ is contained in the integral closure of $B$. But $B$ is integrally closed and an integral extension of $A$ by the assumption that $f$ is integral, and this is exactly the manner in which we construct the normalization, so we're done.
This result can be shown in more generality - the best version I'm aware of is here at Stacks. The idea of the proof is the same, but in our case the assumption that $X,Y$ are integral saves a fair amount of technical hassle.
