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.

enter image description here

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.


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 $\phi_i |_{D(x_i) \cap D(x_j)} = \phi_j |_{D(x_i) \cap D(x_j)}: \operatorname{Spec} k[x_0, x_1, \ldots, x_n]_{x_i x_j} \to \operatorname{Spec} (( k[x_0, x_1, \ldots, x_n])_{x_i x_j})_0 \hookrightarrow \Bbb{P}^n_k$.


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:

  1. In classical algebraic geometry, it just like Robert's answer.
  2. 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.
  3. 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): enter image description here enter image description here

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!


2 Answers 2


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.

  • $\begingroup$ Thank you very much. I do skip lots of content in the classical algebraic geometry, and the verification of how the local pieces of the affine maps glued. Thinking of them I kind of get some feeling about the notation of $[x_0, x_1, \cdots, x_n] $: it means the affine coordinates and how they glued as an "equivalence class" right? $\endgroup$
    – onRiv
    Commented Dec 22, 2022 at 9:59
  • $\begingroup$ More for the last comment: It's $[1, x_1/x_0, \cdots, x_n/x_0]$ or $[x_0/x_1, 1, \cdots, x_n/x_0]$ or ... or $[x_0/x_n, x_1/x_n, \cdots, 1]$. Then everything are affine now and I can understand it via affine coordinates. $\endgroup$
    – onRiv
    Commented Dec 22, 2022 at 10:06
  • $\begingroup$ does checking the map on closed points rely on $k$ being algebraic closed? I realize that maybe math.stackexchange.com/questions/4255702/… is related to the confusion I get too. $\endgroup$
    – onRiv
    Commented Dec 22, 2022 at 10:07

Question: "Thank you very much."

Answer: If $V:=\{e_0,..,e_n\}$ is a vector space over a field $k$ and if $V^*:=\{x_0,..,x_n\}$ is the dual vector space it follows the symbols $x_i:=e_i^*$ are the coordinate functions on $V$ in the sense that for a vector $v:=\sum_i a_ie_i$ it follows $x_i(v):=e_i$. The ring of polynomial functions on $V$, denoted $Sym_k^*(V^*)$ is isomorphic to the polynomial ring $A:=k[x_0,..,x_n]$ and if $\mathbb{A}^{n+1}_k:=Spec(A)$ it follows for any $n+1$ tuple of numbers $p:=(a_0,..,a_n):=\sum_i a_ie_i \in V$ we get a maximal ideal $\mathfrak{m}_p:=(x_0-a_0,..,x_n-a_n) \subseteq A$. A map of affine schemes $\phi: Spec(R) \rightarrow Spec(T)$ is unquely determined by a map of commutative rings $f: T \rightarrow R$. You should not confuse the point $p \in V$ and the maximal ideal $\mathfrak{m}_p \subseteq A$.

Example: When writing down 2-uple embedding $f: \mathbb{P}^1_k \rightarrow \mathbb{P}^2_k$ some authors write

$$(1) f([x_0:x_1]):=[x_0^2:x_0x_1:x_1^2]$$

where $x_i$ are "homogeneous coordinates" on the projective line. Some authors write

$$(2) f([a_0: a_1]):=[a_0^2: a_0a_1:a_1^2]$$

where $a_0,a_1 \in k$ are elements in the field and where not both $a_i=0$.

You may also write down $\pi: \mathbb{A}^{n+1}_k-\{(0)\} \rightarrow \mathbb{P}^n_k$ by

$$\pi(a_0,..,a_n):=[a_0: \cdots : a_n].$$

Since not all $a_i$ are zero it follows you get a well defined map. Moreover $\pi(\lambda a_0,..,\lambda a_n)=\pi(a_0,..,a_n)$.

Some authors do not distingush between the point $p \in V$ and the coordinate functions $x_0,..,x_n$ which are elements in the dual $V^*$.

The exercise asks: You should convince yourself that you understand the distinction between the "set theoretic map" defined in $(2)$ and the map of schemes $f: \mathbb{P}^1_k \rightarrow \mathbb{P}^2_k$. When you have written down the map as in $(2)$ this does not mean you have written down a "map of schemes". Hence: You must figure out yourself what a morphism of schemes is and why some authors write a morphism as in $(1))$ and $(2)$.

If you want to construct this map as a "map of schemes" you must do this in terms of coordinate rings and maps of rings. Since the projective line and plane are not affine schemes, what you must do is to define the map on an open affine cover and then prove your construction glue to give a global map. This is explained in Ch II in Hartshorne.

In this case you construct projective space as a quotient of $X:=\mathbb{A}^{n+1}-{(0)}$ by a group action, and $X$ is not an affine scheme: You want to construct $\mathbb{P}^n_k$ as a quotient $X/G$ where $G$ is the "multiplicative group" $k^*$. In the "scheme language" you define $G:=Spec(k[t,1/t])$. There is an action $\sigma: G \times_k X \rightarrow X$ and to define this action you must write it down "dually" in terms "coactions". The open subscheme $X \subseteq \mathbb{A}^{n+1}_k$ is not affine hence a morphism $\sigma$ must be defined locally, on an open affine cover of $X$. You find this discussed on this site. Here is an example how to do this for the projective line:

Why is the projection map $\mathbb{A}^{n+1}_k\setminus \{0\} \to \mathbb{P}^n_k$ a morphism of schemes?

Here you construct the "line bundle" $\mathbb{V}(\mathcal{O}(m)^*)$ using "quotients":

A description of line bundles on projective spaces, $\mathcal{O}_{\mathbb{P}^n}(m)$ defined using a character of $\mathbb{C}^*$.

What I do here is to construct an action of group schemes

$$\sigma: G \times \mathbb{A}^2_k \rightarrow \mathbb{A}^2_k$$

for any commutative ring $k$ and then we restrict this to the open subscheme $U:=\mathbb{A}^2_k-\{(0)\}$ to get an action

$$\sigma: G \times U \rightarrow U.$$

If you want to define such group actions for more general group schemes, such as $G:=Spec(\mathbb{Z}[t,1/t ])$ and $\mathbb{P}^n_{\mathbb{Z}}$ etc., you must understand this construction.

Comment: "Now I only understand Pn via Proj (and via glueing n+1 affine schemes, and the classical set-theoretic definition). Sad I'm still lacking some backgrounds for (actions and) quotients of schemes. Is it related to how to scheme-theoreticly interpret (x0,x1,⋯,xn)↦[x0,x1,⋯,xn] rigorously? If so, I return the question later and maybe vaguely use (x0,x1,⋯,xn)↦[x0,x1,⋯,xn] just as a hint rather than a definition temporarily (since I'am mainly following Vakil's FOAG)."

Response: Over more general base schemes there is no "quick and easy" way to define morphisms of schemes. You must define maps wrto an open affine cover and prove that maps glue to a global map.

If $X:=\mathbb{A}^2_{\mathbb{Z}}$ with $U:=X-\{(0)\}$ there is a well defined action of $G:=Spec(\mathbb{Z}[t,1/t])$ on $U$ and a well defined $G$-invariant map

$$ \pi: U \rightarrow \mathbb{P}^1_{\mathbb{Z}}.$$

In Mumford's "Geometric invariant theory" he defines the notion "geometric quotient" for arbitrary actions of group schemes, and I suspect the map $\pi$ is such a geometric quotient. You must verify that Mumford's axioms are fulfilled for $\pi$.

Note: If you take a look in Hartshorne's book, CHI, you fill find the following result (Prop.I.3.5): If $Y$ is any affine variety and $X$ any quasi projective variety (over an algebraically closed field $k$), there is a 1-1 correspondence of sets

$$Hom_{var/k}(X,Y) \cong Hom_{k-alg}(A(Y), \Gamma(X, \mathcal{O}_X))$$

But in the above case the varieties/ (or schemes) $U, \mathbb{P}^1_k$ are not affine varieties (or schemes). Hence a morphism of such varieties must be defined locally: You choose open affine subvarieties of $U$ and $\mathbb{P}^1_k$, define maps locally and then prove that your maps agree on intersections and glue to a globally defined morphism. Once you have understood the definition of a morphism, you can continue and study the relation between a morphism to projective space and global sections of invertible sheaves. In fact: If you take a look at Prop.II.7.1 in HH.Ch.II you will find they use this principle to prove the relation between global sections of invertible sheaves and maps to projective space.

  • 1
    $\begingroup$ @onriv - in order to fully understand the "map of schemes" you should try to construct the projective line over $\mathbb{Z}$ as a quotient: There "should be" an isomorphism $(\mathbb{A}^2_{\mathbb{Z}}-\{(0)\})/G \cong \mathbb{P}^1_{\mathbb{Z}}$. $\endgroup$
    – hm2020
    Commented Dec 22, 2022 at 12:02
  • 1
    $\begingroup$ There is a well defined action $\sigma:G \times \mathbb{A}^2_{\mathbb{Z}}\rightarrow \mathbb{A}^2_{\mathbb{Z}}$ and the "quotient" $U/G$ should be the projective line over the integers. What differs from the "field"-case is that $G(\mathbb{Z})=\mathbb{Z}^*=\{-1,1\}$. $\endgroup$
    – hm2020
    Commented Dec 22, 2022 at 12:04
  • 1
    $\begingroup$ ...one should get an isomorphism of schemes $U/G \cong Proj(\mathbb{Z}[x_0,x_1])$. $\endgroup$
    – hm2020
    Commented Dec 22, 2022 at 12:08
  • 2
    $\begingroup$ -1: This is needlessly confusing and not really correct. It's perfectly fine to write down a map between projective spaces using the homogeneous coordinates, since an $S$-map from an $S$-scheme $X$ to $\Bbb P^n_S$ is given by a choice of line bundle $\mathcal{L}$ on $X$ and $n+1$ global sections of $\mathcal{L}$ which generate the sheaf. $\endgroup$ Commented Dec 22, 2022 at 14:46
  • 1
    $\begingroup$ @onriv - Post another question on this if you want more information on the relation with GIT - one may prove directly that the quotient morphism $U \rightarrow \mathbb{P}^n_k$ is a geometric quotient. $\endgroup$
    – hm2020
    Commented Dec 23, 2022 at 11:19

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