In Prop. 4.1, p. 87 of the article "Ample vector bundles on curves" (Nagoya Math. J. 43 [1971], 73--89), R. Hartshorne states the following:

Let $A$ be an abelian variety [over an alg. closed field $k$, say - my assumption]. Let $X$ be a non-singular subvariety of $A$. Assume that every curve in $X$ generates $A$. Then the normal bundle $N$ to $X$ in $A$ is ample.

The proof he provides is based on a criterion of Gieseker for ampleness and boils down to the following statement (X):

Let $Y$ be a smooth curve over $k$ and let $\phi:Y\to A$ be a $k$-morphism, which is birational onto its image. Suppose that $\phi_*(Y)$ generates $A$. Then the map $\phi^*:H^0(A,\Omega_{A/k})\to H^0(Y,\Omega_{Y/k})$ is injective.

Hartshorne does not spell out a proof of (X) but it is likely that he had the following argument in mind:

"Proof": There is a factorisation $Y\to^{i}{\rm Jac}(Y)\to^{g} A$ of $\phi$, where $i$ is the morphism of $Y$ into the Jacobian ${\rm Jac}(Y)$ of $Y$ given by the choice of some $k$-point on $Y$ and $g$ exists because the Jacobian is an Albanese variety for $Y$. Now $i^*:H^0({\rm Jac}(Y),\Omega_{{\rm Jac}(Y)/k})\to H^0(Y,\Omega_{Y/k})$ is an isomorphism (this is a classical fact) and $g^*:H^0(A,\Omega_{A/k})\to H^0({\rm Jac}(Y),\Omega_{{\rm Jac}(Y)/k})$ is injective because the morphism $g$ is a surjective morphism between abelian varieties and is thus smooth.

(for this argument when $k={\bf C}$, see for instance Prop. 6.3.10 (p. 30) in R. Lazarsfeld, "Positivity in algebraic geometry. II." Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. No. 49. Springer-Verlag, Berlin, 2004.)

Now the "Proof" given above is apparently flawed, because when ${\rm char}(k)>0$, surjective morphisms between abelian varieties are not smooth in general. In fact, I think that (X) is false when ${\rm char}(k)>0$, although I do not have a counterexample handy. This is is also suggested by Serre's discussion of Albanese varieties in Exp. 10 of the "Séminaire Chevalley" (1958-1959) (see proof of Th. 3).

In fact it seems to me that Hartshorne's Prop. 4.1 is likely to be false when ${\rm char}(k)>0.$

My question is: is there a way to give a characteristic free proof of (X) ? Or are there classical counterexamples to (X) ? Is it well-known that Hartshorne's Prop. 4.1 is false when ${\rm char}(k)>0$ ?

I would also like to underline there is no implicit assumption on the characteristic of the base field in Hartshorne's paper. This is clear from the formulation of some corollaries of Prop. 4.1 (for instance Cor. 4.3).

Thank you in advance for your help.

  • $\begingroup$ What if you took $Y=A$ to be an elliptic curve, with $\phi$ the Frobenius? Isn't this a counterexample to (X)? $\endgroup$ Commented Nov 14, 2014 at 14:39
  • $\begingroup$ @Donu Arapura: $\phi$ is birational onto its image (I edited the question 30 minutes ago - you might have read the post just before that - sorry). $\endgroup$ Commented Nov 14, 2014 at 14:43
  • $\begingroup$ OK, I misread it. $\endgroup$ Commented Nov 14, 2014 at 14:44
  • 3
    $\begingroup$ As you suggest, Serre's proof can be turned around to give a counterexample: let $C$ be any smooth curve of genus $g>1$ over a field of positive characteristic and $J(C)$ its Jacobian. Let $f:J(C) \to A$ be any height 1 isogeny. The morphism $f$ induces a separable morphism from $C$ to $f(C)$ since the map $H^0(A, \Omega_A)$ to $H^0(C,\Omega_C)$ is non-zero. Also, $f(C)$ generates $A$ so the (geometric) genus of $f(C)$ is $\geq g$. Since $g>1$, it follows (from the Riemann-Hurwitz formula) that the map from $C$ to $f(C)$ is birational onto its image. $\endgroup$
    – naf
    Commented Nov 15, 2014 at 5:07
  • $\begingroup$ @ulrich. Thank you very much for this interesting comment. That settles it then, as far as (X) is concerned. Here is a possible simplification of your argument: as you write, $C\to f(C)$ must be generically separable because $H^0(A,\Omega_A)\to H^0(C,\Omega_C)$ does not vanish. Now since $C\to f_*(C)$ is also generically inseparable, it must be birational. Note that $f_*(C)$ might be singular, so its geometric genus might not be defined (or is there a way to show that it must be non-singular ?). $\endgroup$ Commented Nov 15, 2014 at 7:31

2 Answers 2


The assertion (X) is false in any characteristic $p > 0$ for any $Y$ with genus at least 2. (It is true and easy for $Y$ of genus 1, and true and easy and uninteresting for $Y$ of genus 0.)

To see this, we may and do choose a subgroup scheme $G \subset J := {\rm{Jac}}(Y)$ of the nonzero infinitesimal Frobenius kernel of $J$ such that $G$ is equal to $\alpha_p$ or $\mu_p$. Note that the map $q:J \rightarrow J/G =: A$ has tangent mapping with kernel equal to the line ${\rm{Lie}}(G) \subset {\rm{Lie}}(J)$ that is a proper subspace (as the genus is larger than 1).

For $y_0 \in Y(k)$, the associated inclusion $i_{y_0}:Y \hookrightarrow J$ defined by $y \mapsto \mathscr{O}(y - y_0)$ carrying $y_0$ to $0$ yields a line ${\rm{Lie}}(i_{y_0})({\rm{T}}_{y_0}(Y)) \subset {\rm{Lie}}(J)$, and it is a classical fact that as we vary through such $y_0$ the resulting lines inside ${\rm{Lie}}(J)$ span this space. In particular, there must be some such $y_0$ (and hence for all but finitely many $y_0$, though we won't use this) for which this line is not contained in the proper subspace ${\rm{Lie}}(G)$. Consequently, the composition $q \circ i_{y_0}:Y \rightarrow J/G$ carrying $y_0$ to 0 is injective between tangent spaces at $y_0$ and $0$. Thus, this map is formally an immersion at $y_0$, so it is separable onto its image (by smoothness considerations for $Y$ and $J/G$), yet it is also purely inseparable onto its image since $Y$ is closed in $J$ and $q$ is a purely inseparable isogeny, so $Y \rightarrow J/G$ is birational onto its image.

The pullback map $\Omega^1(J/G) \rightarrow \Omega^1(Y) = \Omega^1(J)$ is dual to the quotient map ${\rm{Lie}}(J) \rightarrow {\rm{Lie}}(J/G)$ that is not surjective (for dimension reasons, as its kernel ${\rm{Lie}}(G)$ is a nonzero subspace), so this pullback map on global 1-forms is not injective.

  • $\begingroup$ Thank you very much for taking the time to write this answer. If I understand your answer right, your construction is in the same line as what ulrich described in his comment. I would still like to see a proof or a counterexample to Prop. 4.1, though but if nobody provides one in the next few days I will accept your answer. $\endgroup$ Commented Nov 16, 2014 at 13:58
  • $\begingroup$ @DamianRössler: I had not noticed ulrich's comment at the time that I posed my answer; indeed they are the same (and I just edited by answer to clarify the implicit role of inseparability in the proof of the birationality property, which ulrich highlights in his comment but which I don't think requires invoking Riemann-Hurwitz). $\endgroup$
    – user27920
    Commented Nov 16, 2014 at 15:37

Yes, this is classic. A counterexample was found by Serre, and more classes of counterexamples are in Fulton's paper Ample Vector Bundles, Chern Classes, and Numerical Criteria (Inventiones, 1976).

  • $\begingroup$ Thank you for this reference. I had a look at Fulton's paper but I cannot find an argument in there that shows that (X) (in my post) or Prop. 4.1 is wrong in a specific situation. The counterexample of Serre that Fulton mentions is apparently a construction that appears in the very paper by Hartshorne that I am concerned with (in par. 3). It is an example of a curve and a vector bundle on it, which is not ample and yet has only positive quotients. How does all this connect to (X) or Prop. 4.1 ? $\endgroup$ Commented Nov 14, 2014 at 23:29

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