How to define morphisms using coordinates in scheme theory? It is very usual in scheme theory to define morphisms between schemes using coordinates. For example: one could say that the morphism of rings
\begin{align*}
B^n\times B^m &\to B^{nm}\\
(x_1,\dotsc,x_n,y_1,\dotsc,y_m)&\mapsto (x_iy_j)_{ij}
\end{align*}
induces a morphism of schemes $\mathbb{A}_A^n\times_{\operatorname{Spec}A}\mathbb{A}_A^m\to \mathbb{A}_A^{nm}$. I think this comes from the functor of points somewhat, but I don't know precisely how. I would be happy if someone could explain that in detail.
 A: You find a brief description of the "functor of points" here:
https://math.stackexchange.com/posts/3977439/edit
Let $k$ be any field and let $B:=k[x_1,..,x_n]$ be the polynomial ring in $n$ variables. Let $I \subseteq B$ be an ideal with $A:=B/I$ and let $X:=Spec(A)$ and let $S:=Spec(k)$.
Let
R0. $X(k)$ be the set of $n$-tuples $(a_1,..,a_n)\in k^n$ with $f(a_1,..,a_n)=0$ for all polynomials $f \in I$.
R1. Let $Hom_{k-alg}(A,k)$  be the set of all maps of $k$-algebras $\phi: A \rightarrow k$.
Lemma. There is a one-to-one correspondence of sets $X(k) \cong Hom_{k-alg}(A,k)$.
If $a:=(a_1,..,a_n)\in X(k)$ we define a map $\phi_a: A\rightarrow k$ by letting $\phi_a(x_i):=a_i$. Why is the map $\phi_a$ well defined? To give a map of $k$-algebras $\psi: k[x_1,..,x_n]\rightarrow k$ is equivalent to specify how $\psi$ acts on the variables $x_j$. Hence we must define
$$\phi_a((f(x_1,..,x_n)):=f(a_1,..,a_n) \in k.$$
If $g(x_1,..,x_n)\in I$ it follows $\phi_a(g)=0$ by definition, hence we get a well defined map
$$\phi_a: A \rightarrow k$$
of $k$-algebras.
Conversely if $\phi: A\rightarrow k$ is a map of $k$-algebras it follows $\phi(x_i)=a_i \in k$ for all $i=1,..,n$. Since $\phi$ is well define it follows $$\phi(g(x_1,..,x_n))=g(a_1,..,a_n)=0$$
for all $g\in I$, hence it follows
$$ g(a_1,..,a_n)=0$$
for all $g\in I$ and it follows $a:=(a_1,..,a_n)\in X(k)$ by definition. You may prove that his is a one-to-one correspondence of sets. QED.
I include the following discussion in the post:
Question: "Dear @hm2020, I understand that morphisms of k-algebras $A \rightarrow k$ are determined by n-tuples $(a_1,…,a_n)\in k^n$ which satisfy $f(a_1,..,a_n)=0$ for all $f \in I$. What I don't understand is why, if $X$ and $Y$ are $A$-schemes, it suffices to define morphisms $X(B)\rightarrow Y(B)$ for all $A$-algebras $B$ in order to determine a morphism $X\rightarrow Y$. (I understand that it suffices to give $X(S)\rightarrow Y(S)$ for all $A$-schemes $S$ by the way.)"
Answer: This is "deeper" in a sense. The functor of points $h_X$ is a "sheaf in the zariski topology". If a functor $F$ is a sheaf in the zariski topology and if $F$ has an open cover of representable functors, it follows $F$ itself is representable. You need to investigate the "embedding"
$$Sch(S)⊆Funct(Sch(S),Sets)$$
and what it means for a functor $F$ to have an "open cover of representable sub-functors". This leads naturally to other grothendieck topologies (the etale topology etc), and it is a good exercise to do this construction. A scheme $X$ has an open cover of affine schemes $U_i:=Spec(A_i)$, and similarly the functor of points $h_X$ of $X$ has an open cover $h_{U_i}$ of the functor of points of the affine schemes $U_i$. To define maps of functors $h_X \rightarrow h_Y$ you can do this on an "open affine cover" and prove that this "glues" to a global map. A good exercise is to prove that the projective space bundle functor (or grassmannian functor) is representable. You do this by proving it is a sheaf in the zariski topology and by giving an open cover of representable sub-functors.  This is essentially Proposition II.7.12 in Hartshornes book. There are books where this is written out in complete details but I do not have a good reference. I believe you also find articles on this on the arXiv preprint server. It is not extremeley "deep" - it is a systematic use of the Yoneda lemma and the functor of points.
