4
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.