Galois stable elements of the Picard group of a curve and the rational divisors

Let $$C$$ be a (smooth,proper) curve over a field $$k$$. Let $$\operatorname{Div}_C(k)$$ be the free abelian group generated by the closed points of $$C/k$$ and $$k(C)^\times$$ be the group of rational functions. We have an injective map: $$i \colon \operatorname{Div}_C(k)/k(C)^\times \to \left(\operatorname{Div}_C(\overline{k})/\overline{k}(C)^\times\right)^{\operatorname{Gal}(\overline{k}/k)}$$

When is this map surjective? I think this is definitely true if the Brauer group $$H^2(k,\mathbb G_m) = 0$$, as it follows from the four-term exact sequence $$0 \to \overline{k}^\times \to \overline{k}(C)^\times \to \operatorname{Div}_C(\overline{k}) \to \operatorname{Pic}_C(\overline{k})\to 0.$$

Q1. Are there other situations where this is true? Also, what are some examples where the map is not injective?

Q2. In particular, I saw a comment somewhere that if $$k$$ is a number field and $$C$$ has a point locally for every completion $$k_v$$, then the map is indeed surjective. Is this true and if so, how does one see it?

Q3. Is the map surjective if $$C$$ has a rational point?

2 Answers

It is always injective, and the cokernel is the group of Brauer classes split by $$k(C)$$, see On the period-index probem in light of the section conjecture by J. Stix. Equivalently, those are the classes of Brauer-Severi varieties $$P$$ with a morphism $$C\to P$$.

If $$k$$ is a number field and $$C$$ has local points everywhere, it cannot map to a non-trivial Brauer-Severi variety (a Brauer-Severi variety with local points everywhere is trivial). Also, if C has a rational point it cannot map to a non-trivial Brauer-Severi variety.

• What if our galois stable picard element has degree 0 and has no global sections? How do we construct the corresponding Brauer-Severi variety in that case? Apr 22, 2021 at 17:50
• @Asvin Tensor it with some line bundle of high degree Apr 22, 2021 at 17:51
• Oh right, of course. I thought of that and convinced myself it couldn't work... Apr 22, 2021 at 17:55

A partial (I have a few questions at the end) proof to Q.2,3 communicated to me by Ananth Shankar:

Suppose $$\mathscr L$$ is a line bundle corresponding to a Galois stable element in $$Pic_C(\overline{k})$$ and assume also that it is very ample.

Let $$f: C \to \mathbb P(H^0(C))$$ be the induced map on global sections defined over $$\overline{k}$$. This map is in fact Galois stable by assumption and therefore we can descent to: $$f: C \to B$$ where $$B$$ is, a priori, only a twist of projective space over $$k$$. However, if $$C$$ has a k-rational point, we can push it forward to get a rational point on $$B$$ which forces the twist to be trivial and then the pull back of a divisor gives a rational divisor on $$C$$ in the same class as $$\mathscr L$$.

Similarly, if $$C$$ has a $$k_v$$ rational point for every completion $$v$$ of k, then $$B$$ is the trivial twist for every $$k_v$$ and by the Hasse local-global theorem for twists of projective space, this forces $$B$$ to be projective space over $$k$$ and we conclude as before.

Q: What if $$\mathscr L$$ is not very ample? Is the cokernel of the map $$i$$ torsion-free? If so, we can deal with $$\mathscr L$$ not of degree $$0$$. Finally, what if $$\mathscr L$$ is degree 0