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Let $k$ be a field, let $C\subset\mathbb{P}^2_k$ be a smooth plane cubic.

Suppose $C$ does not admit $k$-rational points, and for every degree-$3$ closed point $P$ in $C$, the Galois closure of $k(P)/k$ has Galois group $S_3$.

Is it necessarily true that $\mathrm{Pic}^0(C)=\{1\}$?

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    $\begingroup$ Perhaps the "generic" curve of genus with two inequivalent embeddings in $\mathbb{P}^2$ will give a counterexample. $\endgroup$
    – Kapil
    Commented Aug 14, 2020 at 4:03
  • $\begingroup$ Just curious: are there curves known that satisfy your hypotheses? $\endgroup$
    – R.P.
    Commented Aug 14, 2020 at 10:54
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    $\begingroup$ @RP_ I think any cubic that degenerates to a triangle with monodromy $S_3$ will be an example, the degree 3 closed point specialize to a length 3 subscheme on the triangle, with one point on each of the conjugate components or the conjugate nodes. $\endgroup$
    – user39380
    Commented Aug 14, 2020 at 11:22
  • $\begingroup$ @Kapil Thanks! This sounds a very plausible example $\endgroup$
    – user39380
    Commented Aug 14, 2020 at 11:23

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