2
$\begingroup$

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?

$\endgroup$
1
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ – Asvin Apr 22 at 17:50
  • $\begingroup$ @Asvin Tensor it with some line bundle of high degree $\endgroup$ – Giulio Bresciani Apr 22 at 17:51
  • $\begingroup$ Oh right, of course. I thought of that and convinced myself it couldn't work... $\endgroup$ – Asvin Apr 22 at 17:55
2
$\begingroup$

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

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.