Let $K$ be an algebraically closed field. Let $G_m$ be the multiplicative subgroup of $K$. In Lectures on Linear Algebraic Groups by Tamás Szamuely (you can easily find these notes online) it is said that $G_m$ is an affine algebraic group, actually isomorphic to the affine hyperbola $($i.e $V(f)$ where $f=x_1x_2-1 \in K[x_1,x_2])$. I have two questions regarding this matter:
Why is $G_m$ an affine variety? I cannot understand $G_m$ as the zero-locus in a certain affine n-space over $K$ of a finite family of polynomials of n indeterminates over $K$ such that it generates a prime ideal of $K[x1,...,x_n]$.
I tried to verify that the affine hyperbola is actually an affine algebraic group. I would appreciate if someone could verify the correctness of the following argument which I gave:
The set $G=\{(u,u^{-1})$ $|$ $u\in K^*\}$ is a subgroup of the direct product $G_m\times G_m$. Also, $G\subset K\times K$ (as the affine hyperbola) is an affine variety since $G=V(f)$ where $f=x_1x_2-1 \in K[x_1,x_2]$. It remains to show that both multiplication $m:G \times G \longrightarrow G$ and inversion $i:G \longrightarrow G$ are regular maps. To do so, we must be able to write $m=(f_1,f_2)$ and $i = (f_3,f_4)$ for polynomials $f_1,f_2 \in K[x_1,x_2,x_3,x_4]$ and $f_3,f_4 \in K[x_1,x_2]$ $($here we make the natural identification $K^2 \times K^2 = K^4)$. Let $P :=(u,u^{-1},v,v^{-1}) \in G \times G$. Then for $m$ we have $m(P) = (uv,u^{-1}v^{-1}) = (f_1(P),f_2(P))$ where $f_1 = x_1x_3 $ and $f_2 =x_2x_4$. Similarly, let $Q:=(u,u^{-1}) \in G$. Then $i(Q)=(u^{-1},u)=(f_3(Q),f_4(Q))$ where $f_3 = x_2$ and $f_4=x_1$. This shows that $G$ is an affine algebraic group.
A priori, if we write $\mathbb G_m := \mathbb A^1\setminus \{0\}$, then it is a quasi-affine variety (i.e. a Zariski open subset of an affine variety). However, the projection from the affine hyperbola is an isomorphism (in the category of quasi-affine varieties), and so $\mathbb G_m$ is isomorphic to an affine variety.
By an absolutely standard abuse of language, we say that $\mathbb G_m$ is itself an affine variety.