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I asked this basic question in MSE and got a comment "This belongs to Mathoverflow", so I ask my question here.

Let $G$ be a linear algebraic group over a field $k$, and $H\subset G$ be a $k$-subgroup. Let $K/k$ be a field extension. I need an isomorphism between $G_K/H_K$ and $(G/H)_K$.

In more detail, let $G$ be a linear algebraic group over a field $k$ (of arbitrary characteristic). Let $H\subset G$ be an algebraic $k$-subgroup. Set $X=G/H$, it exists (because $G$ is linear), and it is a $k$-variety (a reduced scheme of finite type over $k$, not necessarily connected). It has a base point $x_0=1\cdot H\subset X(k)$. The group $G$ acts on $X$ by $$(g',gH)\mapsto g'gH,$$ and the scheme-theoretic stabilizer of $x_0$ in $G$ is $H$.

Now let $K/k$ be a field extension (not necessarily finite). The base change $G_K:=G\times_k K$ acts on $X_K$, and the scheme-theoretic stabilizer of $x_0\in X(k)\subset X(K)=X_K(K)$ in $G_K$ is $H_K$. Consider the map (morphism) $$\lambda\colon G_K\to X_K,\quad g\mapsto g\cdot x_0\,.$$ Clearly this map is constant on the orbits of $H_K$ acting on the right on $G_K$, hence it induces a morphism $$\bar\lambda\colon G_K/H_K\to X_K\,,$$ see Borel, Linear Algebraic Groups, 2nd edition, Section 6.

Question. How can I prove that $\bar\lambda$ is an isomorphism of $K$-varieties?

I would be grateful for a reference or proof.

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    $\begingroup$ Is not the map $\overline{\lambda}$ a dominant map? Thus the image contains an open subset of the variety $(G/H)_K$ and by homogeneity is all of the variety (for the same reason, it is an open map). $\endgroup$
    – Venkataramana
    Commented Jun 24, 2017 at 15:04
  • $\begingroup$ @Venkataramana: Exactly! The differential of $\bar\lambda$ is bijective at $x_0$ and by homogeneity at any point, and the result follows. $\endgroup$
    – Mikhail Borovoi
    Commented Jun 24, 2017 at 15:29
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    $\begingroup$ Tangent space arguments don't work when $H$ is not smooth. For a robust theory of smooth affine groups in characteristic $p>0$ we need to confront non-separable surjective homomorphisms and geometrically transitive actions whose scheme-theoretic stabilizer at any point is not smooth (so the orbit map is not a submersion). Hence, one needs a positive answer (including a good existence result for $G/H$!) even when $H$ isn't smooth. Chevalley's "stable line" trick to construct $G/H$ works for non-smooth $H$ by considering scheme-theoretic normalizers, and flatness + descent theory do the rest. $\endgroup$
    – nfdc23
    Commented Jun 24, 2017 at 16:05

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