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Let $k$ be a number field and let $G$ be a connected semisimple, simply connected algebraic group defined over $k$. Let $k'$ be a finite Galois extension over which $G$ splits. By the Chebotarev Density Theorem, there are an infinite number of places $v$ of $k$ such that $v$ splits completely in $k'$. The paper I'm reading states without proof that over all such $v$, $G \times_k k_v$ is split. Why is this true?

Is the simply connected or even semisimple hypothesis necessary?

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    $\begingroup$ If $v$ splits completely in $k'$, then $k'$ is contained in $k_v$, hence if $G\times_kk'$ is split, so is $G\times_kk_v$. (I assume $k_v$ denotes completion.) Am I missing something? $\endgroup$
    – Arno Fehm
    Aug 14, 2020 at 6:17
  • $\begingroup$ @Arno Fehm: Why is $k'$ contained in $k_v$? I must be missing something basic in algebraic number theory. $\endgroup$
    – Mehta
    Aug 14, 2020 at 15:58
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    $\begingroup$ $k'$ embeds into $k'\otimes_k k_v$. Since $v$ splits completely in $k'$, we have $k'\otimes_k k_v=k_v\times\dots\times k_v$. Thus $k'$ embeds into $k_v$. $\endgroup$ Aug 15, 2020 at 4:59

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