On the description of $\mathbb P^n_k$

Question

Vakil's Exercise 4.4.F is asking to show that if $k$ is algebraically closed, then the closed points of $\mathbb P^n_k$ may be interpreted in the traditional way, i.e., as points of the form $(a_,\dots, a_n)$ where not all $a_i$ are zero and $(a_0,\dots, a_n)$ is identified with $(\lambda a_0,\dots, \lambda a_n)$ for $\lambda \in k^\times$.

I'm not sure what exactly I need to do to prove this. The scheme $\mathbb P^n_k$ is glued from $U_i=\operatorname {Spec} k[x_0/x_i,\dots,\widehat{x_i/x_i}, \dots, x_n/x_i] $, but I don't have a clear understanding of how they glue together (one just checks that certain conditions hold and hence they glue, but I don't understand the explicit construction) and what exactly the closed points are. Closed points should correspond to prime ideals of some ring, and some of them should be identified. Is the ring the polynomial ring in $n$ variables? Or one in $n+1$ variables?

I also found the following solution, which I don't really understand:

Are they mapping the point $[a_0:\dots :a_n]$ where $a_j\ne 0$ to the prime ideal $(x_0/x_j - a_0/a_j,\dots , \widehat{x_j/x_j-a_j/a_j}, \dots, x_n/x_j-a_n/a_j)\in U_j$? If so, why is this a closed point (or even any kind of point) of $\mathbb P^n_k$? From what I understand, $U_i s $ don't exactly have to be subsets of $\mathbb P^n_k$, they are only used to construct $\mathbb P^n_k$. Or does the part "which is compatible under isomorphisms" (I don't understand what it means) ensure that this is indeed a closed point of $\mathbb P^n_k$? Any further details would be helpful.

Answer 1

Question: "Closed points should correspond to prime ideals of some ring, and some of them should be identified. Is the ring the polynomial ring in n variables? Or one in n+1 variables?"

Answer: You define projective space using the "Proj"-construction: $\mathbb{P}^n_k:=Proj(k[x_0,..,x_n])$ and you may find it proved (in Hartshorne or any other book) that $D(x_i) \cong Spec(k[\frac{x_0}{x_i},.. \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i},..,\frac{x_n}{x_i}]) \cong Spec(k[y_1,..,y_n])$, hence $D(x_i) \cong \mathbb{A}^n_k$ is affine space. A closed point $x\in \mathbb{P}^n_k$ must "live" in $D(x_i)$ for some $i$, hence you get a closed point

$$x \in D(x_i)\cong \mathbb{A}^n_k$$

and since $k$ is algebraically closed it follows $x$ corresponds to a maximal ideal $\mathfrak{m}=(y_1-a_1,..,y_n-a_n)$ for some $a_i \in k$. The ring $k[y_1,..,y_n]$ is the polynomial ring on $n$ variables.

Example: If $n=1$ and $C:=\mathbb{P}^1_k$, you get two open sets $D(x_0), D(x_1)$ and $D(x_0)\cap D(x_1)=Spec(k[t,1/t])$.

If $x\in D(x_0)\cap D(x_1)$ is a closed point you get a point $x:=(t-\frac{a_1}{a_0})$ with $a_1/a_0\in k$ and since $1/t$ is a unit you get an equality of ideals

$$(t-\frac{a_1}{a_0})=(\frac{1}{t}-\frac{a_0}{a_1}) \subseteq k[t,\frac{1}{t}]$$

in $k[t,1/t]$. This equality of maximal ideals correspond to the "equality"

$$(a_0:a_1) \cong (1:\frac{a_1}{a_0})=(\frac{a_0}{a_1}:1)\text{ in }D(x_0)\cap D(x_1).$$

Question: "Are they mapping the point $p:=[a_0:⋯:a_n]$ where $a_j≠0$ to the prime ideal $\mathfrak{m}_p:=(x_0/x_j−a_0/a_j,……,x_n/x_j−a_n/a_j)∈U_j$? If so, why is this a closed point (or even any kind of point) of $\mathbb{P}^n_k$?

Answer: Yes. The maximal ideal lives in $k[\frac{x_0}{x_j},..,\frac{x_n}{x_j}]$ and is by definition maximal.

Example: If you define (as in HH.CH.I) projective space $\mathbb{P}^n(k)$ as the quotient $k^{n+1}-\{(0)\}/k^*$ where $\alpha \in k^*$ acts as

$$\alpha(a_0,..,a_n):=(\alpha a_0,..,\alpha a_n),$$

it follows by HH.CH.I that the set of $a\in \mathbb{P}^n(k)$ with $a_i \neq 0$ is homeomorphic to affine space $\mathbb{A}^n(k)$. Hence a closed point $a\in \mathbb{P}^n(k)$ corresponds to a point $a\in \mathbb{A}^n(k)$ and to the point $a$ you get a unique maximal ideal $\mathfrak{m}_a \subseteq k[y_1,..,y_n]$ on the form you describe.

Example: Let for simplicity $n=1$ and consider $C:=\mathbb{P}^1_k$. It seems to me Vakil uses the following construction: $U_0:=D(x_0):=Spec(k[\frac{x_1}{x_0}]), U_1:=D(x_1):=Spec(k[\frac{x_0}{x_1}])$ and $U_0\cap U_1:=Spec(k[\frac{x_0}{x_1}, \frac{x_1}{x_0}])$ and there are canonical inclusion

$$ k[\frac{x_1}{x_0}] \subseteq k[\frac{x_0}{x_1}, \frac{x_1}{x_0}] $$

and

$$ k[\frac{x_0}{x_1}] \subseteq k[\frac{x_0}{x_1}, \frac{x_1}{x_0}].$$ By definition $U_i \cong \mathbb{A}^1_k$ and a closed point $x\in C$ must correspond to a maximal ideal $\mathfrak{m}_x:=(\frac{x_1}{x_0}-a) \subseteq k[\frac{x_1}{x_0}]$ or $\mathfrak{m}_x:=(\frac{x_0}{x_1}-b) \subseteq k[\frac{x_0}{x_1}]$.

Note: Using the quotient construction $\pi: \mathbb{A}^{n+1}_k-\{(0)\} \rightarrow \mathbb{P}^n_k$ you must prove that the set $\pi(U_i)$ is open, where $U_i \subseteq \mathbb{A}^{n+1}_k-\{(0)\}$ is the "set" where $x_i \neq 0$. Since $U_i=\tilde{D}(x_i)\cap (\mathbb{A}^{n+1}_k-\{(0)\})$ is an intersection of two open sets it follows $U_i$ is open. Since $\pi$ is a quotient map it follows $\pi(U_i)$ is open. The map $\pi$ has the property that $\pi^{-1}(D(x_i)) \cong D(x_i) \times Spec(k[t,1/t]) \cong \mathbb{A}^n_k \times Spec(k[t, 1/t])$. The induced map

$$\pi_i: \mathbb{A}^n_k \times Spec(k[t,1/t]) \rightarrow \mathbb{A}^n_k$$

is the projection map. Here you find a "scheme theoretic" construction of projective space as a quotient:

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

Comment: "Yes. The maximal ideal lives in $k[x_0/x_j,..,x_n/x_j]$ and is by definition maximal." -- Just as I said in my question, this point lies in $U_j$. But under the definition of projective space in 4.4.9, $U_j$ is not even a subset of $P^n_k$, from what I can tell. So it's not clear why that maximal ideal even lies in $P^n_k$ under that definition.

Answer: Whenever you glue a set of schemes $X_i$ along open subschemes $U_{ij} \subseteq X_i$ (see Hartshorne, Ex.II.2.6), by construction $X_i$ give an open cover of the glued scheme $X$. Hence in your case the scheme $U_j$ is an open subscheme of projective space, giving an open cover.