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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.

amWhy
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Gabriel
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  • Yes, functor of points will work. You interpret the $x_i$ and $y_i$ as taking values in an arbitrary $A$-algebra. Then what you've written down is a natural transformation between the functors of points, which for affine schemes (such as these) corresponds to a unique map on coordinate rings by the Yoneda lemma. – Qiaochu Yuan Jan 14 '21 at 09:11
  • @QiaochuYuan Unfortunately I know nothing about the functor of points beyond its very definition. Would you mind giving me just a little more detail (or a reference which explains this)? For example, how would we work with such morphism (as it's defined by "abstract nonsense")? – Gabriel Jan 14 '21 at 09:13
  • Your example above appears to be badly confused. Firstly, your proposed morphism is not a morphism of rings. (Try the case when $n = m = 1$; additivity is not satisfied.) Secondly, what is $A$, and how is it related to $B$? Affine $n$-space over a ring $A$ corresponds to the spectrum of the polynomial ring $A[X_{1}, \ldots, X_{n}]$, and not $A^{n}$. This leads me to another point: the product in the category of rings does not correspond to the fiber product in the category of schemes. – Alex Wertheim Jan 14 '21 at 09:21
  • Anyway, Qiaochu gives great advice above, as always. I really recommend working out exercises 3.2O and 3.2P in Ravi Vakil's "Foundations of Algebraic Geometry"; it provides great practice in understanding how the philosophy of schemes is compatible with the classical language of coordinates. When those examples are clear, the functor of points perspective will be much easier to follow. Again, Vakil is a great resource here: sections 6.3 and 6.6 of the aforementioned text should provide a nice introduction to the essential ideas. – Alex Wertheim Jan 14 '21 at 09:21
  • @AlexWertheim my example is basically how U. Gortz and T. Wedhorn construct the Segre embedding in the section 4.14 of their book on algebraic geometry. (In this case they assume that $n,m\geq 2$ but even then this isn't a morphism of $A$-algebras... So I now understand even less how this morphism is constructed. By the way, I thought that the product worked since the fibered product corresponds to the product on the level of the functor of points.) I will surely try to do the exercises you said. – Gabriel Jan 14 '21 at 09:34
  • See here, for instance (in fact, I would lean towards this being a duplicate of that question). You don't need to know anything about the functor of points to make sense of this if you don't want to. – KReiser Jan 14 '21 at 09:35
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    The map you've written above describes the Segre embedding on coordinates, which is different from the ring morphism which induces it. In fact, Gortz and Wedhorn point this out, since they specifically describe it as "the morphism of $R$-schemes which is given on $A$-valued points by...". (In fact, once you've mastered this philosophy of coordinates, it would be a great exercise to go back and try and write down the map on polynomial rings which induces your proposed map on affine spaces.) – Alex Wertheim Jan 14 '21 at 09:45
  • @KReiser If I understood correctly Qiaochu Yuan's answer there, the map I mentioned in the post is the spectrum of the morphism $A[z_{ij}]{ij}\to A[x_1,\dotsc,x_n,y_1,\dotsc,y_m]$ which sends $z{ij}$ to $x_i y_j$? – Gabriel Jan 14 '21 at 14:58
  • @AlexWertheim I did the exercises and read the sections you suggested. Indeed, exercises 3.2O and 3.2P are basically what Qiaochu Yuan did in the answer cited by KReiser, and it clarified my mind. Nevertheless, what I had in mind was more in the lines of sections 6.3 and 6.6 (which indeed talk about the functor of points). I understood that a map of schemes $X\to Y$ is determined by the maps $X(Z)\to Y(Z)$, where $Z$ ranges over all schemes (this is Yoneda's lemma) but I still don't understand why, when $X$ and $Y$ are $A$-schemes, it suffices to consider the spectrum of $A$-algebras. – Gabriel Jan 14 '21 at 15:01
  • Vakil says something about using the ringed space definition of a morphism to prove that things are well-defined, then proving that it is equivalent to the definition using affine covers, and then throwing the ringed space definition out like yesterday's news. In other words, even if you write down a map in coordinates there is still a lot more going on in the background. No idea if this helps but I've always kept it in mind. – Tabes Bridges Jan 14 '21 at 19:35

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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.

hm2020
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  • Dear @hm2020, I understand that morphisms of $k$-algebras $A\to k$ are determined by $n$-tuples $(a_1,\dotsc,a_n)\in k^n$ which satisfy $f(a_1,\dotsc,a_n)=0$ for all $f\in I$. What I don't understand is why, if $X$ and $Y$ are $A$-schemes, it suffices to define morphisms $X(B)\to Y(B)$ for all $A$-algebras $B$ in order to determine a morphism $X\to Y$. (I understand that it suffices to give $X(S)\to Y(S)$ for all $A$-schemes $S$ by the way.) – Gabriel Jan 15 '21 at 18:35