If $A$ is a finitely generated $k$-algebra, is $A$ is determined by to isomorphism by its functor of points $\textrm{Hom}_{\textrm{$k$-alg}}(B,A)$ on finitely generated $k$-algebras $B$?
Let $X$ be a scheme over a field $k$. Then the contravariant functor
$$\textrm{Hom}_{\textrm{$k$-sch}}(-,X): \textrm{$k$-sch} \rightarrow \textrm{set}$$
completely determines $X$ up to isomorphism of $k$-schemes. If $X = \textrm{Spec } A$ is affine, I believe that the functor is completely determined by its values on affine $k$-schemes, that is spectra of $k$-algebras. Thus $X$, or $A$, is completely determined up to isomorphism by the covariant functor
$$\textrm{Hom}_{\textrm{$k$-alg}}(A,-) \rightarrow \textrm{set}$$
Suppose moreover that $X$ is of finite type over $k$, i.e. $A$ is finitely generated as a $k$-algebra. Can we say that $A$ is determined by the values of that covariant functor, only allowing finitely generated $k$-algebras?
If $B$ is an arbitrary $k$-algebra, then $B$ is the direct limit of its finitely generated $k$-subalgebras. I was thinking we could do something like say that:
$$\textrm{Hom}_{\textrm{$k$-alg}}(A,B) = \textrm{Hom}_{\textrm{$k$-alg}}(A,\varinjlim B_0) = \varinjlim \textrm{Hom}_{\textrm{$k$-alg}}(A,B_0)$$
where $B_0$ runs through the finitely generated $k$-subalgebras of $B$. I know that you can take direct limits in and out in the category of $k$-vector spaces if $A$ is finite dimensional, but since we are in the category of $k$-algebras and $A$ is probably not finite dimensional, this argument doesn't immediately carry over.