Let $k$ be a field with separable algebraic closure $k^{\rm s}$ and corresponding absolute Galois group $\varGamma={\rm Gal}(k^{\rm s}/\,k)$ and let $X$ and $Y$ be geometrically connected and geometrically reduced $k$-schemes locally of finite type.

Question. Are there examples where the canonical map $(\,{\rm Pic}\,X^{\rm s})^{\varGamma}\oplus(\,{\rm Pic}\,Y^{\rm s})^{\varGamma}\to{\rm Pic}\,(X^{s}\times_{\,k^{\rm s}}Y^{\rm s})^{\varGamma}$ is an isomorphism but the map ${\rm Pic}\,X^{\rm s}\oplus{\rm Pic}\,Y^{\rm s}\hookrightarrow{\rm Pic}\,(X^{s}\times_{k^{\rm s}}Y^{\rm s})$ is not?

It is known that, if $X$ and $Y$ are smooth, projective, geometrically irreducible and of finite type over $k$, then the cokernel of the latter map is naturally isomorphic to ${\rm Hom}({\rm Pic}_{ Y^{\rm s}}^{\vee},{\rm Pic}_{X^{\rm s}})$, where ${\rm Pic}_{X^{\rm s}}$ is the Picard variety of $X^{\rm s}$ (i.e., the largest reduced subscheme of the identity component of the Picard scheme of $X^{\rm s}$ over $k^{\rm s}$) and ${\rm Pic}_{Y^{\rm s}}^{\vee}$ is the dual of ${\rm Pic}_{Y^{\rm s}}$. Thus, while looking for such examples, one should assume that ${\rm Pic}_{Z}\neq 0$ for $Z=X^{s}$ and $Y^{s}$, in particular that $H^{1}(Z,\mathcal O_{Z})\neq 0$ for such $Z$ (a condition which excludes the rational varieties).

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    $\begingroup$ It seems to me that if $Y$ is a genus $1$ curve without a $k$-point and $X$ is its Jacobian, then there should probably be no $\Gamma$-equivariant homomorphisms $\operatorname{Pic}_{Y^s}^\vee \to \operatorname{Pic}_{X^s}$. This is something you could try. $\endgroup$ Apr 24, 2017 at 17:36
  • $\begingroup$ Dear Dobben de Bruyn. Thanks for the idea. I'll think about this. $\endgroup$ Apr 24, 2017 at 18:27
  • $\begingroup$ @R.vanDobbendeBruyn I don't think this works as there is always a degree $n$ point for some $N$, which gives an $n$-fold covering $Y \to X$ and hence a homomorphism (but not isomorphism) of picard groups. $\endgroup$
    – Will Sawin
    Apr 24, 2017 at 20:46
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    $\begingroup$ I think an elliptic curve over $k$ and its nontrivial quadratic twist provide an example. $\endgroup$
    – Will Sawin
    Apr 24, 2017 at 20:47
  • $\begingroup$ @WillSawin: ah very good, my example indeed doesn't seem to work. I do not immediately see why yours does, but is seems plausible. $\endgroup$ Apr 25, 2017 at 0:24

1 Answer 1


Take $X$ a non-CM elliptic curve over $k$ (over finite fields, a similar argument involving ordinary curves will work) and $Y$ its quadratic twist over some quadratic extension of $k$.

$X$ and $Y$ are smooth, projective, geometrically irreducible and of finite type over $k$, so as you note there is an exact sequence $$0 \to ( \operatorname{Pic} X^s \oplus \operatorname{Pic} Y^s \to \operatorname{Pic} (X^s \times Y^s) \to \operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)) \to 0$$

I have written the Picard varieties as $ \operatorname{Pic}^0$ to distinguish them from Picard groups.

This implies, by the left exactness of $G \mapsto G^\Gamma$, an exact sequence.

$$0 \to (\operatorname{Pic} X^s)^\Gamma \oplus (\operatorname{Pic} Y^s)^\Gamma \to (\operatorname{Pic} (X^s \times Y^s))^\Gamma \to (\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)))^\Gamma$$

So it suffices to check that $\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)) \neq 0$ but $(\operatorname{Hom} ( \operatorname{Pic}^0(Y^{\vee s}), \operatorname{Pic}^0(X^s)))^\Gamma=0$, so that the second exact sequence becomes

$$0 \to \operatorname{Pic} X^s)^\Gamma \oplus (\operatorname{Pic} Y^s)^\Gamma \to (\operatorname{Pic} (X^s \times Y^s))^\Gamma \to 0.$$

To do this, observe that, because $X$ and $Y$ are elliptic curves, they are their own $\operatorname{Pic}^0$, and are also self-dual, so we are just working with $\operatorname{Hom}(Y^s, X^s)$. Because $Y^s$ and $X^s$ are isomorphic, and are a non-CM elliptic curve $\operatorname{Hom}(Y^s, X^s) = \mathbb Z$. The quadratic twisting means exactly that the Galois action on this is by the quadratic character of the corresponding field extension, and in particular the invariants are trivial, as desired.

  • $\begingroup$ Dear Will, thanks for your effort. I'll read your answer carefully later today. Right now I'm wondering if your argument can be extended to a higher-dimensional setting, i.e., instead of an elliptic curve take any self-dual abelian variety without "non-trivial" endomorphisms and take a twist of it more general than a quadratic one. Also, can you see a case where there will be non-trivial endomorphisms but still no Galois equivariant ones? $\endgroup$ Apr 25, 2017 at 13:28
  • $\begingroup$ @CristianD.Gonzalez-Aviles Yes, it's actually very common. One can take a twsit by any quadratic character except one that appears already inside the Galois action on the endomorphism ring. For number fields, say, there will be infinitely many such characters. $\endgroup$
    – Will Sawin
    Apr 25, 2017 at 14:05

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