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Let $G$ be a group scheme over a field $F$, and let $f:G\to G$ be a homomorphism. Written in my notes, I have the following statement:

To check surjectivity (on $F$-rational points), it suffices to show that the induced morphism $G_K\to G_K$ is surjective, for some finite extension $K/F$.

I would like a reference for this fact.

Remarks

  • As such, this statement may be (very) wrong. Among the hypotheses that I have on $G$: it is a connected reductive algebraic group. However, the above is stated in this level of generality because I want to see what breaks down if we don't assume enough.
  • Moreover, one might need further hypotheses on the morphism $f$.
  • 1
    This fails in the very simple situation when $F=\mathbb R$, $G$ is the torus $\mathbb G_{m, F}$, $f$ is the square map $x\mapsto x^2$, and $K=\mathbb C$.2012-05-14
  • 2
    To elaborate a little more: let $H$ be the kernel of $f$. Suppose $K/F$ is Galois of group $\Gamma$. Then the exact sequence $$ 1\to H(K)\to G(K)\to G(K) \to 1$$ gives rise to an exact sequence of pointed sets $$ 1\to H(F)\to G(F)\to G(F) \to H^1(\Gamma, H(K)).$$ So to get surjectivity on $G(F)$, it would be enough to have $H^1(\Gamma, H(K))=$ one point.2012-05-14
  • 0
    @QiL Ok, I see. However, in my situation, the kernel is a unitary group, and so the first cohomology group is *not* trivial. It thus seems that my question is more context-dependent than I first thought. Have you ever encountered a situation where a nice criterion like the one above exists?2012-05-14
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    No, sorry, I have no idea.2012-05-14

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