# 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.

• Normal affine variety is about the coordinate ring being integrally closed, with $Y=V( u^2-t, (u+1)v-1)$ then $O_Y(Y) = K[t,u,v]/(u^2-t,(u+1)v-1)= K[t^{1/2},\frac1{t^{1/2}+1}]$ is a localization of the PID $K[t^{1/2}]$ thus it is integrally closed and $(t,u,v)\to t$ is a surjective morphism to $X=\Bbb{A}^1_K,O_X(X)=K[t]$ but $Tr_{K(Y)/K(X)} (v) = \frac1{t^{1/2}+1}+\frac1{-t^{1/2}+1}=\frac2{1-t}$ is not in $O_Y(Y)$. – reuns Oct 20 at 7:19
• Thank you for this example. I am having trouble seeing how I am using integral closure though. I guess I am using the Gauss lemma? Because an element of $f_{\ast} \mathcal{O}_y$ satisfies a monic polynomial with coefficients in $\mathcal{O}_X$ but this may not be the minimal polynomial, so I have to know that its factors are also integral? – hedgehog enthusiast Oct 20 at 14:57
• But this seems to only require integral closure of $\mathcal{O}_X$! – hedgehog enthusiast Oct 20 at 15:06
• @reuns this is not a finite morphism. If it were, $K[Y]$ would be a finite $K[X]$-module, but for any finite collection of elements of $K[Y]$, when written in lowest terms there's a maximum power $n$ of $t^{1/2}+1$ in the denominator, so $(\frac{1}{t^{1/2}+1})^{n+1}$ is not in this span, contradiction. – KReiser Oct 21 at 0:19
• In KReiser's answer : $\alpha$ is integral over $R$ because $R[\alpha]$ is a finitely generated $R$-module because this is in the definition of finite morphism. Thus its minimal polynomial is in $R$'s integral closure (in $Frac(R)$) which is $=R$ because this is in the definition of normal variety. – reuns Oct 21 at 13:57

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.