Let $H/k$ be a genus $2$ curve. Consider the function field of $H$ given by $k(H)$. Take the base extension of $H$ to $k(H)$, namely $\hat H:=H\otimes_k k(H)$. Consider the Jacobian $J$ of the curve $H$.

I want to know if $\text{Pic}^0(\hat H)\cong J(k(H))$

This is because I want to work with divisor classes having $k(H)$-rational points in their support.

Is well known that $\text{Pic}^0(H)\cong J(k)$ when $k$ is perfect. When is not perfect, when do I have this isomorphism?

In this paper it seems to be defined for Jacobians of genus $2$ curves according to Propositions 3.1,3.2 and 3.3


But I am not sure, can anybody help me to understand?.

  • $\begingroup$ Why is being perfect relevant here? In the cited paper, they make no such assumptions on the base field $k$. $\endgroup$ – Daniel Loughran Oct 27 '17 at 17:33
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    $\begingroup$ Thanks for the question to clarify. Is not relevant, the question is in which situation, if $V/K$ is an algebraic curve where $K$ is its function field over some base field $F$, the isomorphism $Pic^0(V)\cong J(K)$ holds $\endgroup$ – Eduardo R. Duarte Oct 27 '17 at 17:43
  • $\begingroup$ What is your definition of the Jacobian? (I can think of multiple possible definitions, not all of which are always equivalent.) $\endgroup$ – R. van Dobben de Bruyn Oct 27 '17 at 18:45
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    $\begingroup$ Yes. For a smooth geometrically connected proper curve $X$ over a field $F$ (such as your $\widehat{H}$ over $k(H)$) there is an obstruction in ${\rm{Br}}(F)$ for the surjectivity of the map ${\rm{Pic}}^0(X) \to J(F)$. So for finite or separably closed $F$ one has surjectivity, and usually not otherwise (many perfect $F$!). But if $X(F)$ is non-empty then the obstruction vanishes for any $F$, so one wins (any $x_0 \in X(F)$ can be used to reformulate the moduli problem defining $J$ in more tangible terms). Since $\widehat{H}(k(H))$ is obviously non-empty by design, you get what you want. $\endgroup$ – nfdc23 Oct 27 '17 at 19:07
  • $\begingroup$ Hello, thanks, for the reply, do you have an example for a non surjective map? i mean, when the Brauer group obstructs? If you could reformulate this comment as an answer I could close this question. $\endgroup$ – Eduardo R. Duarte Oct 28 '17 at 19:20

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