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.

  • 1
    $\begingroup$ 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 $O_Y(Y)$. $\endgroup$ – reuns Oct 20 at 7:19
  • $\begingroup$ 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? $\endgroup$ – hedgehog enthusiast Oct 20 at 14:57
  • $\begingroup$ But this seems to only require integral closure of $\mathcal{O}_X$! $\endgroup$ – hedgehog enthusiast Oct 20 at 15:06
  • $\begingroup$ @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. $\endgroup$ – KReiser Oct 21 at 0:19
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


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