In Vakil's FOAG, the projective space $\Bbb{P}^n_A$ is defined to be $\operatorname{Proj} A[x_0, x_1, \ldots, x_n]$. (there is another definition in the book too, just glueing $n+1$ affine space). Then in serveral exercises, the author uses the symbol $[x_0, x_1, \ldots, x_n]$ without giving the meaning of it. For example(added a screenshot too for voiding misquote):
7.3.F Make sense of the following sentence: $\Bbb{A}_k^{n+1} \backslash \{\vec{0}\} \to\Bbb{P}^n_k$ given by $(x_0, x_1, \ldots, x_n) \mapsto [x_0, x_1, \ldots, x_n]$ is a morphism of schemes.
I did get a morphism $\Bbb{A}_k^{n+1} \backslash \{\vec{0}\} \to\Bbb{P}^n_k$ by glueing the morphism $$D(x_i) = \operatorname{Spec} A[x_0, x_1, \cdots, x_n]_{x_i} \to \operatorname{Spec} (A[x_0, x_1, \ldots, x_n]_{x_i})_0 = D_+(x_i)$$. But I cannot see how is it given by $(x_0, x_1, \ldots, x_n) \mapsto [x_0, x_1, \ldots, x_n]$.
Related paragraphs are in FOAG subsection 4.5.7 (page 150) and exercise 7.3.F (page 202) (and exercise 7.3.O uses the notation $[f_0, f_1, \ldots, f_n]$)
There are more examples using notations of this style, e.g., 7.4.1 Example, the morphism $\Bbb{C}\Bbb{P}^1 \to \Bbb{C}\Bbb{P}^2$ given by $[s, t] \to [s^{20}, s^{9}t^{11}, t^{20}]$. I interpret it as the morphism given by $\Bbb{C}[x, y, z] \to \Bbb{C}[s, t]$ given by $x=s^{20}, y=s^9t^{11}, z=t^{20}$. Nevertheless, the usage of notation $[x_0, x_1, \ldots, x_n]$ in exercise 7.3.F seems unable to be interpret in this way.
Thank you very much.
Update:
After Robert and hm2020's excellent answers, I want to update the question and make my question more specific to the sentence "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$". I think I know how to construct the morphism, here is a proof(skip it if too long, my concrete question is after the proof):
Denote $D(x_i)$ be the distinguished open subset of $\mathbb{A}_{k}^{n+1}$ that $x_i$ does not vanished. Then one has that $\cup D(x_i) = \mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\}$: if some point $p \notin \cup D(x_i)$, then for all $x_i$ one has $p \notin D(x_i)$, i.e., $x_i \in p$. Then $(x_0, x_1, \ldots, x_n) \subset p$. Since $(x_0, x_1, \ldots, x_n)$ is a maximal ideal, one have $(x_0, x_1, \ldots, x_n) = p$. Then $p = \overrightarrow{0}$ by the notation and $p \notin \mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\}.$ If $p \in \cup D(x_i)$, then there is some $x_i$ such that $p \in D(x_i)$. Hence $x_i \notin p$. Hence $p \neq \overrightarrow{0}$ and $p \in \mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\}$.
On the other hand, denote $D_+(x_i)$ be the homogeneous distinguished open subset of $\Bbb{P}^n_k := \operatorname{Proj} k[x_0, x_1, \ldots, x_n]$ that $x_i$ does not vanished. One has $\Bbb{P}^n_k = \cup D_+(x_i)$.
By $D(x_i) \cong \operatorname{Spec} k[x_0, x_1, \ldots, x_n]_{x_i}$
and $D_+(x_i) \cong \operatorname{Spec}(( k[x_0, x_1, \ldots, x_n])_{x_i})_0$ and since any map on rings induce a map on the schemes of their spectrums on the opposite direction, from the inclusion map $(( k[x_0, x_1, \ldots, x_n])_{x_i})_0 \hookrightarrow k[x_0, x_1, \ldots, x_n]_{x_i}$ one has$$D(x_i) \cong \operatorname{Spec} k[x_0, x_1, \ldots, x_n]_{x_i}
\to D_+(x_i) \cong \operatorname{Spec}(( k[x_0, x_1, \ldots, x_n])_{x_i})_0$$. Composing it with $D_+(x_i) \hookrightarrow \Bbb{P}^n_k$. One has morphisms $\phi_i: D(x_i) \to \Bbb{P}^n_k$.
These morphisms in fact can be glued into a morphism $\mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\} \rightarrow \mathbb{P}_{k}^{n}$, i.e, they agree on the overlaps: Since $D(x_i) \cap D(x_j) = D(x_i x_j) \cong \operatorname{Spec} k[x_0, x_1, \ldots, x_n]_{x_i x_j}$ and $D_+(x_i) \cap D_+(x_j) = D_+(x_i x_j) \cong \operatorname{Spec} (( k[x_0, x_1, \ldots, x_n])_{x_i x_j})_0$, one has $\phi_i |_{D(x_i) \cap D(x_j)} = \phi_j |_{D(x_i) \cap D(x_j)}: \operatorname{Spec} k[x_0, x_1, \ldots, x_n]_{x_i x_j} \to \operatorname{Spec} (( k[x_0, x_1, \ldots, x_n])_{x_i x_j})_0 \hookrightarrow \Bbb{P}^n_k$.
$\square$
But from the proof, one see that the morphism $\mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\} \rightarrow \mathbb{P}_{k}^{n}$ is just "given"(one construct it from glueing of pieces, and the morphism on pieces are just given like "from one's mind", It's like "aha. as we already know this and this, we construct it from that and that"). It's not as Vakil said in the text, the morphism is "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$". I can make sense construction a morphism $\mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\} \rightarrow \mathbb{P}_{k}^{n}$, but I cannot make sense how it "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$".
So my question is about how "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$" really make sense scheme-theoreticly? I know that:
- In classical algebraic geometry, it just like Robert's answer.
- For affine scheme, since the opposite category of rings is equivalent to the category of affine schemes, I understand how does morphism given by maps on affine coordinates work too.
- For morphism on projective schemes, I understand it too I think, it's via Vakil's section 7.4(here I have to attach two screenshots to make the context complete):
But back to the case "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$", the map $\mathbb{A}_{k}^{n+1}\backslash\{\overrightarrow{0}\} \rightarrow \mathbb{P}_{k}^{n}$ is neither affine scheme to affine scheme, nor project scheme to project scheme. How can it make sense saying "given by $(x_0, x_1, \cdots, x_n) \mapsto [x_0, x_1, \cdots, x_n]$"?
I hope this update state my question more clear. I mainly study algebraic geometry by myself using Vakil's FOAG. It may be a little weird question and maybe I hit some wrong path. Thank you very much again!
Best Answer
In classical, i.e. non-scheme-theoretic algebraic geometry, the projective space of dimension $n$ over some field $k$ is usually defined as $\mathbb{P}^n(k) = (k^{n+1} \setminus \lbrace 0 \rbrace) / k^\times = (\mathbb{A}^{n+1}(k) \setminus \lbrace 0 \rbrace) / k^\times$. This means that an element of $\mathbb{P}^n(k)$ is an equivalence class of points of $\mathbb{A}^{n+1}(k) \setminus \lbrace 0 \rbrace $, where we identify two points $(x_0,\dotsc,x_n)$ and $(y_0,\dotsc,y_n)$, if there exists $\lambda \in k^\times$ with $y_i = \lambda x_i$ for all $0 \leq i \leq n$. These equivalence classe are most commonly denoted by $[x_0,\dotsc,x_n]$ or by $[x_0 : \dotsc : x_n]$ and are called homogeneous coordinates. The quotient map $\mathbb{A}^{n+1}(k) \setminus \lbrace 0 \rbrace \rightarrow \mathbb{P}^n(k)$ is thus given by mapping $(x_0,\dotsc,x_n)$ to $[x_0,\dotsc,x_n]$. Also here one works with affine charts to study its properties.
In scheme-theoretic language one can also make sense of maps defined in terms of coordinates, see e.g. this post here. Vakil wants the readers to construct the above quotient map scheme-theoretically, which for example boils down to gluing it together from the standard affine charts as you did. In particular, he assumes familiarity with the classical situation. I would now suggest you to compare the local pieces of the two maps and how they are glued to see that they encode the same kind of map. One can also interpret the task to check that the right thing happens on maximal ideals, i.e. closed points, of course.