Trying to prove that $\mathbb{G}_m^2$ is isomorphic to $D_2$ (algebraic groups)

Question

Let $k$ be an algebraically closed field. I have learned that $\mathbb{G}_m^2$ is isomorphic as algebraic groups to $D_2$ (invertible diagonal matrix $2$ by $2$), but I am struggling to provide all the details due to perhaps my lack of understanding of some basic materials. I have no doubt that this is indeed the case, but I would appreciate if someone could provide me with a more detailed explanation. Thank you.

Answer 1

The isomorphism just sends $(x,y)$ to the $2$ by $2$ matrix that has $x,y$ on the diagonal and $0$ in the other corners.

This is clearly natural in the $k$-algebra $R$, and is clearly a pointwise isomorphism, hence a natural isomorphism.

You can also see it in terms of representing algebras : both $\mathbb{G}_m^2$ and $D_2$ are representable by $k$-algebras and as such are isomorphic if and only if their representing algebras are isomorphic.

$\mathbb{G}_m^2$ is represented by $k[X,X^{-1},Y,Y^{-1}]$ while $D_2$ is represented by $k[X,Y,Z,T, (XT-YZ)^{-1}]/(Y,Z)\simeq k[X,T,(XT)^{-1}] \simeq k[X,T,X^{-1},T^{-1}]$ and therefore they are isomorphic.