is the tensor product of coordinate rings always a coordinate ring? given an arbitrary field $k$ (in particular, it does not need to be algebraically closed or even perfect) and sets $S ⊆ \mathbb A^m(k)$ and $T ⊆ \mathbb A^n(k)$, is there an isomorphism of coordinate rings
$$k[S × T] \cong k[S] \otimes_k k[T]?$$
from the commutative diagram with the natural arrows
$$
 \require{AMScd}
 \begin{CD}
  k[\mathbb A^m × \mathbb A^n] @= k[\mathbb A^m] \otimes_k k[\mathbb A^n] \\
  @VVV @VVV \\
  k[S × T] @<<< k[S] \otimes_k k[T]
 \end{CD}
$$
we would only need a natural arrow in the other direction in the bottom row to conclude that. for this, it would suffice for a regular functions $h \colon \mathbb A^m × \mathbb A^n → k$ vanishing on $S × T$ to come from a sum of products of regular functions $f \colon \mathbb A^m → k$ and $g \colon \mathbb A^n → k$ vanishing on $S$ and $T$ respectively. is that so? how can i show this?
note that i consider classical affine space, not schemes and $\mathbb A^m (k) = k^m$ and $\mathbb A^n (k) = k^n$ as sets.
(by the way: i don't see that the tensor product of two reduced $k$-algebras of finite type need not be reduced as an obstruction here as, to my knowledge, the counter-examples arise from inseparable extensions, wich are, to my knowledge, not formed by coordinate rings.)
 A: Functors that have a left adjoint preserve $\underleftarrow{\textrm{lim}}$ (inverse limits), in particular finite products. So perhaps this could (be made to) work.
Let $\mathrm{Aff}_k$ be the category of the algebraic sets $S$ living in some $\mathbb A^m(k)$ (any $m$). Following Hartshorne, regular functions $S\to k$ and morphisms $S\to T$ can be defined just like when $k$ is algebraically closed ([H], §I.3).
Let $_k\mathrm{Fta}$ be the category of $k$-algebras of finite type that are coordinate rings. These can be given as pairs $(n,I)$, where $I$ is an ideal of $k[X_1,\cdots\,X_n]$ which any $f$ $\in$ $k[X_1,\cdots\,X_n]$ that vanishes at all common zeroes in $\mathbb A^n(k)$ of the $i$ $\in$ $I$ must belong to. I.e., we need "$I$ $=$ $I(Z(I))$". Morphisms are $k$-algebra homomorphisms. Denote the opposite category (with all arrows reversed) by $_k\mathrm{Fta}^{\circ}$.
Then we have covariant functors $A:$ $\mathrm{Aff}_k$ $\to$ $_k\mathrm{Fta}^{\circ}$ and $Z:$ $_k\mathrm{Fta}^{\circ}$ $\to\mathrm{Aff}_k$ given, for $S$ $\subseteq$ $\mathbb A^m(k)$, by $S$ $\mapsto$ $(m,I(S))$, which is just the $k$-algebra $k[S]$, and by $(n,I)$ $\mapsto$ $\{\vec{\lambda}\in k^n$ $\mid$ $\forall_{i\in I}\,i(\vec{\lambda})=0\}$, which is of course closed.
By corollary I, 3.8 of [H], we have $\mathrm{Aff}_k(Z(n,I),S)$ $\cong$ $_k\mathrm{Fta}^{\circ}(A(Z((n,I))),A(S))$. I think the proof goes through for arbitrary fields $k$. But $A(Z((n,I)))$ $=$ $(n,I)$ because of the restriction we imposed on $I$.
Hence $\mathrm{Aff}_k(Z(n,I),S)$ $\cong$ $_k\mathrm{Fta}^{\circ}((n,I),A(S))$, and these isomorphisms are natural. So $Z\dashv A$, i.e. $Z$ is a left adjoint of $A$, and hence $A$ preserves finite products. Now direct products in $_k\mathrm{Fta}^{\circ}$ are the same thing as direct sums in $_k\mathrm{Fta}$, and those are given, like in the category of all $k$-algebras (for any ring $k$), by $\otimes_k$. So $A(S\times T)$ $\cong$ $A(S)\otimes_kA(T)$.
$A$ also respects any other forms of $\underleftarrow{\textrm{lim}}$ that exist in $\mathrm{Aff}_k$, subject to the fact that all arrows are reversed, so that $A$ will send them to the corresponding $\underrightarrow{\textrm{lim}}$ in $_k\mathrm{Fta}$. So pullbacks become pushouts, etc.
[H] Robin Hartshorne, Algebraic Geometry.
