Isomorphism between $\mathbb{G}_m\times SL(n)$ and $GL(n)$

Question

In the wikipedia page for reductive groups under the isogenies section it says that for any field $k$ (You can assume the field to be non archimedean of $0$ characteristic here) one has an isomorphism $(\mathbb{G}_m\times SL(n))/\mu_n\cong GL(n)$ where $\mu_n$ is the scheme of $n$-th roots of unity.

What is this isomorphism explicitly? Does the same hold for the symplectic group, meaning $(\mathbb{G}_m\times Sp(2n))/\{\pm I_{2n}\}\cong GSp(2n)$?

Answer 1

Yes, this holds quite generally.

Let us assume that $G$ is an algebraic group i.e., a finite type group scheme over some field $k$. There is a natural map

$$p\colon G^\mathrm{der}\times Z(G)\to G,\qquad (a,z)\mapsto az.$$

Here $G^\mathrm{der}$ is the derived subgroup scheme over $G$, and $Z(G)$ the center. This is a group homomorphism as can be easily checked. Moreover, it's clear that

$$\ker (p)=\{(a,z)\in G^\mathrm{der}\times Z(G):a=z^{-1}\}\cong G^\mathrm{der}\cap Z(G)=: K,$$ where this isomorphism is via diagonal embedding.

So, one sees that if $p$ is surjective, then we have the following isomorphism

$$G\cong (G^\mathrm{der}\times Z(G))/K,$$ where again $K$ is embedded diagonally. Your desired isomorphism then follows from the following.

Theorem (see [Milne, §19.D]): Let $G$ be a reductive group. Then, $p$ is surjective.

Of course, one must be careful in intepreting this isomorphism when $k$ is note algebraically closed. Indeed, as $K$ is central in $G^\mathrm{der}\times Z(G)$, one has an exact sequence of groups

$$1\to K(k)\to G^\mathrm{der}(k)\times Z(G)(k)\to G(k)\to H^1(k,K).$$

In particular, there is no reason that $G^\mathrm{der}(k)\times Z(G)(k)\to G(k)$ is surjective or, said differently, that $G(k)$ is isomorphic to $(G^\mathrm{der}(k)\times Z(G)(k))/K(k)$ via the obvious map.

Example: When $G=\mathrm{GL}_n$, the isomorphism given by $p$ is given by taking a diagonal matrix $zI_n$ and an element $g$ of $\mathrm{SL}_n$ to $zI_ng$.

To show this surjective in the sense of algebraic groups, one is free to show that the map $p$ is surjective on $\overline{k}$-points. But then, for an arbitrary $h$ in $\mathrm{GL}_n(\overline{k})$ one has that $h=zI_n g$ where $z$ is an $n^\text{th}$-root of $\det(h)$ and $g=z^{-1}I_n g$.

Such a $z$ exists as $\overline{k}$ is algebraically closed, but over $k$ itself one may not have such a decomposition. Indeed, if one can decompose $h=zI_n g$ as above, then we see that $\det(h)=z^n$ and thus $\det(h)$ is an $n$-th root in $k$. In fact, it's not hard to see from this that over $k$ the cokernel of $p$ can be precisely identified with $k^\times/(k^\times)^n$. On the other hand, it's trivial to see that $K=\mu_n\cdot I_n\cong \mu_n$, and observe that $k^\times/(k^\times)^n$ is precisely $H^1(k,\mu_n)$ by Kummer theory.

This exaclty corresponds to the general picture we discussed above.

References:

[Milne] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.