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I have been reading Deligne-Milne's Tannakian Categories and got to the point at the end of part 3 where they discuss what went wrong with Saavedra's definition. Motivated by their counterexample, they ask the following question:

Let $C$ be a $k$-linear rigid abelian tensor category over a field $k$. Let us further assume that $End (\underline{1})=k$ and there exists a fiber functor with values in a field extension $k'\supseteq k$. Does it follow that $C$ is Tannakian?

What is the current status of this question? A reference to a solution or to a modern discussion would be very useful.

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    $\begingroup$ Would be invaluable if you can explain the point of this story $\endgroup$
    – Alexander Chervov
    Commented Sep 13, 2012 at 19:49
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    $\begingroup$ There's an error in Saavedra's thesis where he claims to show that any two fiber functors on a Tannakian category are locally isomorphic. In fact, with his definition it's not true --- it is necessary to add the condition that for a Tannakian category over a field k, the endomorphism ring of the identity object is k. With this extra condition, it was expected (but not proved) in 1982 that Saavedra's claim is correct. This was proved (not easily) by Deligne. $\endgroup$
    – anon
    Commented Sep 13, 2012 at 20:49
  • $\begingroup$ MO 3446/tannakian-formalism mathoverflow.net/questions/3446/tannakian-formalism $\endgroup$
    – Alexander Chervov
    Commented Sep 14, 2012 at 6:41

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The answer is yes; in fact, this is mentioned in a footnote to the definition of Tannakian category in Milne's corrected version of the Deligne-Milne paper, on page 32. This is proved in Deligne's Catégories tannakiennes, Theorem 1.12.

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