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Let $G$ be a linear algebraic group over an algebraically closed field, and let $H$ be a closed subgroup. Identify all groups with their closed points. General theory guarantees the structure of a quasiprojective variety on the space of cosets $G/H$ in the quotient topology.

This is done by constructing a rational representation $\pi: G \rightarrow \textrm{GL}(V)$, together with an element $0 \neq v \in V$, such that $H$ is the stabilizer in $G$ of the line $[v]$ through $v$, and $\mathfrak h$ is the stabilizer in $\mathfrak g$ the same line. The representation induces an action of $G$ on $\mathbb{P}(V)$, and $G/H$ is given its variety structure via the resulting bijection with the orbit $G[v]$.

Let $G = \mathbb{G}_m$, and let $H = \{ \pm 1 \}$. The quotient $G/H$ is a one dimensional algebraic group consisting of semisimple elements, so it should be isomorphic to $\mathbb{G}_m$. I'm wondering how one can construct an explicit isomorphism of $G/H$ with $\mathbb{G}_m$.

D_S
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  • From the functor of points view (see e.g. Milne's book) it's perhaps easier, because the map $x^2: {\mathbb G}_m\to {\mathbb G}_m$ has kernel $H$ and is surjective with respect to the fpqc-topology (for any $R$ and $\alpha\in R^{\times}$, over the faithfully flat $R$-extension $R[t]/(t^2-\alpha)$ we have the solution $x:=t$ to $x^2=\alpha$). – Hanno Nov 06 '17 at 07:00

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Assume we aren't in characteristic two. Let $\lambda$ be a generator of $X(G)$. Then $2\lambda$ is a generator of $X(G/H)$.

Define an isomorphism of abelian groups $\phi:X(\mathbb{G}_m) \rightarrow X(G/H)$ by $\phi(\chi) = 2\lambda$, where $\chi$ is a generator of $X(\mathbb{G}_m)$. The morphism of algebraic groups $G \rightarrow \mathbb{G}_m, x \mapsto x^2$ induces a morphism of algebraic groups $f: G/H \rightarrow \mathbb{G}_m$.

We have $\phi(2\lambda) \circ f(xH) =\chi \circ f(xH) = x^2 = 2\lambda(xH)$. This shows that $f$ is the morphism of tori corresponding to $\phi$. Since $\phi$ is an isomorphism, the equivalence of categories between tori and free abelian groups tells us that $f$ is an isomorphism as well.

This is the weird thing: the inverse of $f$ does not look like a morphism of varieties. At least to me, it does not appear to be continuous. But it must be. Explicitly, one chooses for each $x \in \mathbb{G}_m$ a square root $\sqrt{x}$, and defines $f^{-1}(x) = \sqrt{x}H$.

D_S
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  • The ambiguity in the choice of square root is determined upto $\pm 1$ and one inability to make a coherent choice along a loop is what causes the problem but here one has absorbed this ambiguity by taking the quotient. – random123 Nov 06 '17 at 07:39