We say $G$ is an affine group scheme if it is a representatble functor from $F:\text{Alg}_k\to \text{Grp}$. We say it is represented by $A$ if
$F(R)\cong \text{Hom}_k(A,R)$.
I can't work out how to use this definition.
Say we take $\Bbb{G}_m$ and represent it by $A=k[x,y]/(xy-1)$. Taking $R=k[a,b,c]$ we have $\Bbb{G}_m(R) = \{x\in R: x\text{ is invertible }\}$ so then if $a,b,c$ are transcendental, this is just the units of $k$ and hence $F(R)$ is trivial, and thus this would imply $\text{Hom}(A,R)$ is trivial?
Is this correct? It seems reasonable, since the homomorphism would be required to take $x$ and $y$ to units, which must be elements of $k$.