Let $S/k$ be a smooth variety and $A/S$ be an abelian scheme. Let $Z \hookrightarrow S$ be a reduced closed subscheme of codimension $\geq 2$.

I want to show that in this situation, $H^i_Z(S, A) = 0$ for $i \leq 2$ (étale cohomology).

Note that we have a long exact sequence

$\ldots \to H^i_Z(S, A) \to H^i(S, A) \to H^i(S \setminus Z, A) \to H^{i+1}_Z(S, A) \to \ldots$,

so for $i = 0$, this is equivalent to the injectivity of $H^0(S, A) \to H^0(S \setminus Z, A)$, which is clear since $A/S$ is separated, $S$ is reduced and $S \setminus Z \hookrightarrow S$ is dense. For $i = 1$ this is equivalent to $H^0(S, A) \to H^0(S \setminus Z, A)$ being surjective and $H^1(S, A) \to H^1(S \setminus Z, A)$ being injective. The surjectivity of $H^0(S, A) \to H^0(S \setminus Z, A)$ follows e.g. from [Bosch-Lutkebohmert-Raynaud, Néron models], p. 109, Thm. 1, which states that "Let $S$ be a normal noetherian base scheme, and let $u: Z \to G$ be an $S$-rational map from a smooth $S$-scheme $Z$ to a smooth and separated $S$-group scheme $G$. Then, if $u$ is defined in codimension $\leq 1$, it is defined everywhere." (Edit 2:) See also: [Milne in Cornell-Silverman, Abelian Varieties], Theorem 3.1.

For the injectivity of $H^1(S, A) \to H^1(S \setminus Z, A)$ and for $i = 2$ I don't have an idea.

Edit: For the injectivity of $H^1(S, A) \to H^1(S \setminus Z, A)$ and $i = 2$, one could use the interpretation of $H^1$ as torsors and $H^2$ as gerbes. Does the cited theorem of [Bosch-Lutkebohmert-Raynaud, Néron models] help in this case? What we need is:

  1. If a principal homogeneous space $X/S$ for $A/S$ is trivial over $S \setminus Z$, then it is trivial over $S$.

  2. Any principal homogeneous space $X/(S \setminus Z)$ extends to a principal homogeneous space $X/S$.

  3. If a gerbe $X/S$ for $A/S$ is trivial over $S \setminus Z$, then it is trivial over $S$.

(Another idea would be to use purity, but see there some problems: We would need a /smooth/ pair $(S,Z)$ and finite coefficients.)

  • 2
    $\begingroup$ 1. is true because $S$ is smooth. The trivialisation over $S \backslash Z$ gives a rational map from $S$ to the principal homogenous space and any such map (with $S$ a regular scheme) extends to a morphism. The reason for this is that abelian varieties do not contain rational curves but all positive dimensional fibres of a birational proper morphism $S' \to S$ are covered by rational curves (by a theorem of Murre or Abhyankar (depending on the level of generality)). $\endgroup$
    – naf
    Nov 25, 2011 at 13:03
  • 2
    $\begingroup$ 2. is also true if the order of $X \in H^1(S\backslash Z,A)$ is prime to the characteristic. If this holds then $X$ comes from an element of $H^1(S \backslash Z,A[n])$ for some $n$ prime to the characteristic . As you suggested, we can use purity, actually just the classical Zariski-Nagata purity theorem, to extend this to an element of $H^1(S,A[n])$ and then we get the desired element of $H^1(S,A)$. $\endgroup$
    – naf
    Nov 25, 2011 at 15:27
  • $\begingroup$ Dear ulrich, thank you for your response. By the theorem of Murre, do you mean the article "On a connectedness theorem ..."? And which paper of Abhyankar do you mean? $\endgroup$
    – user19475
    Dec 1, 2011 at 10:12
  • 2
    $\begingroup$ The paper of Murre you refer to is indeed the one I meant. The result of Abhyankar I was referring to is Theorem 1.2 in VI.1 of Kollar's book "Rational curves...". I now see that Abhyankar's theorem actually predates Murre's and it suffices for what I said. $\endgroup$
    – naf
    Dec 1, 2011 at 11:04
  • $\begingroup$ Does anyone have a vague idea how to approach 3? $\endgroup$
    – user19475
    Dec 2, 2011 at 12:44

1 Answer 1


This can be proved for $\mathcal{A} = \mathrm{Pic}_{\mathcal{C}/S}$ using vanishing of cohomology sheaves with supports and values in the multiplicative group and the Leray spectral sequence (modified for cohomology with supports) and vanishing of $R^q\pi_*\mathbf{G}_m$ for $q > 1$, with no restriction on (prime-to-characteristic) torsion.


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