# Galois invariant line bundles on a product of varieties

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).

• 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. Apr 24, 2017 at 17:36
• Dear Dobben de Bruyn. Thanks for the idea. I'll think about this. Apr 24, 2017 at 18:27
• @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. Apr 24, 2017 at 20:46
• I think an elliptic curve over $k$ and its nontrivial quadratic twist provide an example. Apr 24, 2017 at 20:47
• @WillSawin: ah very good, my example indeed doesn't seem to work. I do not immediately see why yours does, but is seems plausible. Apr 25, 2017 at 0:24

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.

• 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? Apr 25, 2017 at 13:28
• @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. Apr 25, 2017 at 14:05