# Splitting of simply connected algebraic group

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?

• 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? Aug 14, 2020 at 6:17
• @Arno Fehm: Why is $k'$ contained in $k_v$? I must be missing something basic in algebraic number theory. Aug 14, 2020 at 15:58
• $k'$ 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 $k_v$. Aug 15, 2020 at 4:59