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Let $A$ be a central simple algebra of degree $n$ over $k$, $\dim_kA=n^2$, let $K/k$ be a cyclic galois extension of degree $n$. Suppose $A\times_kK\cong M_n(K)$, does this imply that $A$ is a cyclic algebra?

So the question is in the definition, $A$ is cyclic if it splits after some cyclic extension $K$ contained in $A$, I am not sure if one can drop this?

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  • $\begingroup$ I am trying to see why every CSA over number field is cyclic. Is there a modern exposition of the fact?(I could only find the original proof in the 1930s) $\endgroup$
    – user39380
    Commented Oct 16, 2016 at 3:51
  • $\begingroup$ If you are willing to accept the reciprocity exact sequence from class field theory, then you can simply write down a cyclic algebra with the same invariants as your CSA. Of course, cyclicity may be a step in the proof of the reciprocity exact sequence . . . Anyway, I think this result is usually attributed to Albert-Brauer-Hasse-Noether. $\endgroup$
    – Jason Starr
    Commented Oct 16, 2016 at 12:58
  • $\begingroup$ @JasonStarr Could you breifly mention which is the reciprocity exact sequence? Why is cyclicity a step in establishing the sequence? Thank you! $\endgroup$
    – user39380
    Commented Oct 16, 2016 at 16:50
  • $\begingroup$ I might be using the wrong name. The exact sequence I meant is $0\to \text{Br}(K) \to \oplus_{v} \text{Br}(K_v) \xrightarrow{\text{inv}} \mathbb{Q}/\mathbb{Z} \to 0,$ together with the identification $\text{inv}:\text{Br}(K_v) \xrightarrow{\cong} \mathbb{Q}/\mathbb{Z}$ for nonarchimedean places and $\text{inv}:\text{Br}(K_v)\xrightarrow{\cong} (1/2)\mathbb{Z}/\mathbb{Z}\subset \mathbb{Q}/\mathbb{Z}$ for real places. $\endgroup$
    – Jason Starr
    Commented Oct 16, 2016 at 16:58
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    $\begingroup$ @JasonStarr I found Gille and Szamuely's book section 4.7 contains answers to my questions, thank you! $\endgroup$
    – user39380
    Commented Oct 19, 2016 at 20:36

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