Trace map from finite surjective morphism of normal irreducible varieties I am trying to understand why if $f: Y \rightarrow X$ is a finite surjective morphism of normal (irreducible) complex varieties, there is a trace-map $f_{\ast} \mathcal{O}_Y \rightarrow \mathcal{O}_X$ splitting the natural inclusion in the other direction. 
What I understand so far is (1) that the finiteness condition ensures that $f_{\ast} \mathcal{O}_Y$ is integral over $\mathcal{O}_X$, so $K(Y)$ is algebraic over $K(X)$ and (2) that there is a trace map from the algebraic closure $K(Y)$ to $K(X)$, which restricts since the extension is integral.
However these steps feel a little shakey to me, and also I haven't used normality. Is it just required for the extension of fields to be Galois (I guess it guarantees it will be a normal extension?) I don't quite understand this part. 
For reference, I am trying to understand the proof of the Injectivity Lemma 4.1.14 in Lazarsfeld. 
Thanks for your help.
 A: As $Y\to X$ is finite, $K(X)\hookrightarrow K(Y)$ is finite extension and therefore there is a trace map $Tr:K(Y)\to K(X)$ which comes from the trace of multiplication by an element $a\in K(Y)$ acting as a $K(X)$-linear endomorphism of $K(Y)$ viewed as a finite-dimensional $K(X)$-vector space. The goal is to show that this gives to a morphism of sheaves $f_*\mathcal{O}_Y\to \mathcal{O}_X$, which we can do by showing that this is true on each affine open subset $\operatorname{Spec} A = U\subset X$.
Since normal + irreducible implies integral, we can take $A$ to be a normal integral domain, and then $f^{-1}(U)=\operatorname{Spec} B$ for $B$ a normal integral domain which is a finite (thus intergal) $A$-module. As the characteristic polynomial of $b$ as an endomorphism of $K(Y)$ over $K(X)$ is also a power of it's minimal polynomial (think about $K(X)\subset K(X)(b)\subset K(Y)$) and has the trace as a coefficient, all we need to do is to show that this minimal polynomial lives in $A[t] \subset K(X)[t]$. The following lemma proves this:
Lemma: Let $R$ be an integrally closed domain and $F=Frac(R)$. Let $F\subset K$ be a finite field extension. Then for any $\alpha\in K$ integral over $R$, the minimal polynomial $m(x)$ of $\alpha$ is actually in $R[x]$.
Proof: As $\alpha$ is integral, there's a monic polynomial $f\in R[x]$ with $f(\alpha)=0$. But by the inclusion $R[x]\subset F[x]$, $f$ is also a polynomial in $F$ which vanishes on $\alpha$. So it's divisible by $m(x)$, and we may write $f=gm$. As all roots of $m$ are roots of $f$, all roots of $m$ are integral over $R$. As the coefficients of $m$ are the elementary symmetric polynomials in these roots, the coefficients of $m$ are again integral over $R$. As $R$ is integrally closed, these coefficients are actually in $R$, and thus we've shown that $m(x)\in R[x]$. $\blacksquare$
As a result of this lemma, we get that the trace of any element $b\in B$ actually lands in $A$ and therefore gives us a morphism $f_*\mathcal{O}_Y\to \mathcal{O}_X$. After averaging by dividing by $\deg K(Y)/K(X)$, we get that the composite $\mathcal{O}_X\to f_*\mathcal{O}_Y \to \mathcal{O}_X$ is the identity, demonstrating a splitting as required in the question.
