How is the multiplicative group an algebraic variety?

Question

According to various places, we define an algebraic group as a group that is also an algebraic variety (along with some compatibility conditions). Many places also list some examples, one of which is the multiplicative group $\mathbb{G}_m$ of a field $k$.

But $\mathbb{G}_m = \mathbb{A}^1\setminus\{0\}$, which is not Zariski-closed in $\mathbb{A}^1$ (i.e. is not a variety). So how exactly is it an algebraic group? I'm sure there's something really obvious that I've missed, but I'm new to this area and have yet to really get to grasps with such basics.

Answer 1

A different way to view it is as the variety corresponding to the algebra $k[x,y]/(xy-1)$, so it is a closed subvariety of $\mathbb{A}^2$ (you should check that this all works out).

Answer 2

To view $G_m$ as an affine variety, one needs to consider it inside $k^2$, namely as the zeros of $yx=0$ and $y=0$ in $k[x,y]$. The affine coordinate ring is $k[x,y]/(xy−1)=(k[x])[x^{−1}]=k[x,x^{−1}]$.