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I have a surjective morphism $\pi: Y \to X$ between smooth projective varieties of the same dimension over some algebraically closed field $k$. Debarre claims in his book "Higher-Dimensional Algebraic Geometry" (1.41) that there is an effective divisor $R$ such that $$ K_Y \equiv \pi^* K_X + R$$ if $K(X) \subset K(X)$ is a separable field extension. In particular, this is always true if $\pi$ is birational. Debarre also claims that in this case the support of $R$ is the exceptional locus.

I search for a reference for this fact because Debarre does not provide one. I'm in particular interested in the case of positive characteristic. In characteristic $0$ this is a well known fact and I would like to confirm that it is still fine in positive characteristic.

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What do you mean by "the exceptional divisor of a birational morphism" ? Do you mean the set-theoretic union of the fibres of dimension $>0$ ? (the term "exceptional divisor" usually only appears in connection with blow-ups) –  Damian Rössler Nov 7 '11 at 11:26
    
I'm sorry. I mean exceptional locus. –  Benjamin Schmidt Nov 7 '11 at 12:02
    
No!!!! 1) Benjamin, you mean the exceptional divisor as you said it. 2) Damian, that's not true. The exceptional divisor is a well defined notion for any birational morphism, not just blow ups. (Then again, every projective birational morphism is a blow-up, at least locally). –  Sándor Kovács Nov 7 '11 at 15:22
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Def Let $f:X\to Y$ be a birational morphism and let $Z\subset Y$ be the smallest (closed) subset such that $f$ is an isomorphism over $Y\setminus Z$. Then $f^{-1}(Z)\subset X$ is called the exceptional locus and its codimension $1$ part the exceptional divisor. Rem The exceptional locus maybe larger than the union of fibres of positive dimension, think of the normalization of a singular curve. Also, the exceptional locus maybe larger than the exceptional divisor. For this, you need to go to dimension $3$ or higher. –  Sándor Kovács Nov 7 '11 at 15:24
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Benjamin, yes, you are right. But it is still true that exceptional divisors are defined in general. Then again, if you look at my answer below you'll see that the statement is true in more generality in which case this distinction is important. I had that in mind when I wrote this comment. Cheers! –  Sándor Kovács Nov 7 '11 at 16:19

3 Answers 3

up vote 9 down vote accepted

EDIT Originally I claimed a more general statement and along the incremental generalizations I reached a statement that was not true. Thanks to Carlos for pointing out this error! So, I thought it would be fair to point out where the error lied. The main issue is that the original proof works for a finite morphism, but not if there are exceptional divisors, because then on the target the localization would not happen at a height $1$ prime. In order to preserve the original proof, here is a fix that divides the statement into two parts.

0

By Stein factorization it is enough to prove the statement for finite or birational morphisms.

1

Thm Let $\pi:Y\to X$ be a separable projective finite morphism between normal varieties of the same dimension. Assume that $K_X$ is a $\mathbb Q$-Cartier divisor (i.e., there exists an $m\in \mathbb N_+$ such that $mK_X$ is a Cartier divisor). Then there exists an effective divisor $R\subset Y$ whose support is contained in the ramification locus of $\pi$ (that is, the complement of the largest open subset of $Y$ on which $\pi$ is smooth) such that $$K_Y\sim \pi^*K_X + R.$$

Proof:
We need to prove that the divisor $\pi^*K_X-K_Y$ is linearly equivalent to an effective divisor supported on the exceptional locus . This can be done by localizing at height $1$ primes, so the question reduces to a question about regular schemes of dimension $1$. One may apply the usual proof of the Hurwitz formula:

Consider the short exact sequence $$0\to \pi^*\Omega_X\to \Omega_Y\to \Omega_{Y/X}\to 0$$ and observe that $\Omega_{Y/X}$ is a torsion sheaf whose support is the ramification locus (this is where you need that the map is separable), which in this case is the same as the exceptional divisor (the smaller parts of the exceptional locus disappear at the localization). This is a finite set, so $\Omega_{X/Y}$ maybe considered as the structure sheaf of a finite subscheme of $Y$. Let $R\subseteq Y$ be this subscheme. Tensoring the above short exact sequence by $\Omega_Y^{-1}$ gives: $$ 0\to \pi^*\Omega_X\otimes \Omega_Y^{-1}\to \mathscr O_Y \to \mathscr O_R \to 0,$$ which shows that $\pi^*\Omega_X\otimes \Omega_Y^{-1}\simeq \mathscr O_Y(-R)$. This implies the needed linear equivalence.

To find the original $R$ all you need to do is to take the divisor that localizes to the $R$ we found in the $1$-dimensional case. Since we're talking about divisors the codimension $2$ ambiguity makes no difference.

2

Thm Let $\pi:Y\to X$ be a projective birational morphism between smooth varieties. Then there exists an effective divisor $R\subset Y$ whose support is the exceptional locus $E\subseteq Y$ of $\pi$ such that $$K_Y\sim \pi^*K_X + R.$$

Proof: Consider (again) the short exact sequence $$0\to \pi^*\Omega_X\to \Omega_Y\to \Omega_{Y/X}\to 0$$ and notice that $\pi$ is an isomorphism on $Y\setminus E$, so similarly $\pi^*\Omega_X\to \Omega_Y$ is an isomorphism there. Now take the determinant of these locally free sheaves and conclude that $\pi^*\omega_X\subseteq\omega_Y$ and $\omega_Y/f^*\omega_X$ is supported on $E$. This implies that the divisor $K_Y-f^*K_X$ is an effective divisor supported on $E$.

We need to prove that the support of $R$ is the entire $E$. For this, first notice that in order to prove the desired statement, using what we have proved already, we may pass to another birational model that dominates $Y$. (The point is that a $\pi$-exceptional divisor will be exceptional for the combined map to $X$ but not for the map to $Y$.) Second, a theorem of Zariski says that every exceptional divisor can be reached by a sequence of blow-ups (see Kollár-Mori98, 2.45). We know how to compute the canonical divisor of a blow-up and we know that the entire exceptional locus is contained in the discrepancy divisor, so the desired statement follows.

3

Comment The reason the second part requires smoothness is that one needs $\Omega_X$ to be locally free so when pulled back it would give the right thing. Otherwise it might pick up torsion or co-torsion. The statement is true in a little bit more general situation, if $X$ has at worst canonical singularities, but that is essentially the definition of those singularities and this statement says that smooth points are canonical, so it is a reasonable condition to use to define singularities.

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I like this proof. –  Benjamin Schmidt Nov 8 '11 at 9:28
    
Let $F_3\to X$ be a contraction of the negative section E on a Hirzebruch surface $F_3=P(O+O(3))$. Then $K_{F_3}=f^*K_X-\frac{1}{3}E$, so R is not effective. –  Carlos Nov 8 '11 at 20:58
    
Carlos, you are absolutely right, I over-generalized it. Obviously one needs a condition on the singularities allowed. –  Sándor Kovács Nov 8 '11 at 23:02
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Benjamin, yes. Well, I didn't try to prove that every variety has canonical singularities, that would be silly, it just turned into that. (So I guess I am saying I did something silly, indeed). What happened was that I was thinking about the finite case. And actually (for the record), I already included in #3 of my correction that we are talking about canonical singularities. Finally, I hadn't realized (like seemingly everyone else who gave an answer, even though you clearly stated) that you want that the support is the entire exceptional divisor. Anyway, I fixed it now. –  Sándor Kovács Nov 9 '11 at 16:16
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Sándor, a dumb question, if you are doing a Stein factorization $Y \to Z \to X$, then I don't think you can guarantee that $Z$ is smooth. Thus you can't reduce to theorem 2 above. However, I think the statement you need basically is the usual transformation rule for discrepancies under finite maps. In particular, even though $Z$ is not smooth (and might have VERY bad), the pair $(Z, -\text{Ramification divisor})$ should be canonical/terminal since $(X, 0)$ was. –  Karl Schwede Nov 10 '11 at 13:56

A down-to-earth approach can be the following.

Assume that the morphism $\pi \colon X \to Y$ is separable; then it is unramified on an open dense subset of $X$ and we have a short exact sequence of tangent sheaves $$0 \to T_X \stackrel{d \pi}{\to} \pi^*T_Y \to N_f \to 0,$$ where $N_f$ is a sheaf supported on the ramification divisor of $\pi$, see [Sernesi, Deformation Theory, page 162].

Dualizing the sequence above we obtain $$0 \to \pi^*\Omega^1_Y \to \Omega^1_X \to \mathcal{E}xt^1(N_f, \mathcal{O}_X) \to 0,$$ where $\mathcal{E}xt^1(N_f, \mathcal{O}_X)$ is again supported on the ramification divisor of $\pi$.

The identity you are looking for (and which is sometimes called Hurwitz formula) now follows by just taking the $n$-th exterior powers, with $n=\dim X = \dim Y$.

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I still don't understand how the fact that the support of $N_f$ is the ramification divisor implies that the support of $\mathcal{E}xt^1(N_f, \mathcal{O}_X)$ is the ramification divisor. But I guess this is something more general? –  Benjamin Schmidt Nov 8 '11 at 9:35
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Well, outside the ramification locus $R$ vthe sheaves $\pi^* \Omega^1_Y$ and $\Omega_X^1$ are isomorphic, so the cokernel has necessarily support contained in $R$. And yes, this is something more general. In fact, by the local fundamental isomorphism, $\mathcal{E}xt^1(\mathcal{O}_R, \mathcal{O}_X)$ is the normal sheaf of $R$ in $X$, which is supported on $R$. –  Francesco Polizzi Nov 8 '11 at 11:01

It is a consequence of general Grothendieck duality, but you are probably looking for a down-to-earth reference. I suggest you to look at

Kleiman, S. L.: The enumerative theory of singularities. in: Real and complex singularities (Proc. Ninth Nordic Summer School/NAVF Sympos. Math., Oslo, 1976), pp. 297–396. Sijthoff and Noordhoff, Alphen aan den Rijn, 1977. (MR 58 #27960)

where it follows from the first fundamental exact sequence for sheaves of differentials.

The formula you look for is (I, 16) on page 303 and it is stated for a surjective separable map between smooth varieties of the same dimension. It is written in terms of sheaves, but they are invertible so the formula follows immediately.

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Dear Leo, which statement exactly is a consequence of Grothendieck duality ? –  Matthieu Romagny Nov 8 '11 at 21:11
    
Dear Matthieu, notice that when $h = g \circ f$ then $h^! = f^! g^!$. In the OP setting this yields $\omega_Y = \pi^*\omega_X \otimes \omega_\pi$. Then, one has to identify $\omega_\pi$ with $R_\pi$. This follows from the equality of Kähler and Dedekind differents. This is discussed with certain detail in books by Kunz, and Lipman's Asterisque monograph is useful, too. –  Leo Alonso Nov 9 '11 at 10:01
    
Dear Leo, muchas gracias para los detalles. –  Matthieu Romagny Nov 11 '11 at 12:26

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