2 questions about the morphism $\mathbb{A}_k^{n+1}\backslash\{0\} \rightarrow \mathbb{P}_k^n$

Question

An exercise in Ravi Vakil's algebraic geometry notes says "Make sense of the following sentence: $$\pi: \mathbb{A}_k^{n+1}\backslash\{0\} \rightarrow \mathbb{P}_k^n$$ given by $$(x_0,x_1...,x_n)\mapsto [x_0,x_1...,x_n]$$ is a morphism of schemes.'"

I assume he means the morphism with $$Spec~ k[x_0,x_1,..,x_n]_{x_i}\rightarrow Spec~(k[x_0,x_1,..,x_n]_{x_i})_0 $$ induced by the inclusion $$(k[x_0,x_1,..,x_n]_{x_i})_0 \subset k[x_0,x_1,..,x_n]_{x_i}$$ (where here $(k[x_0,x_1,..,x_n]_{x_i})_0$ is the degree zero elements of the ring $k[x_0,x_1,..,x_n]_{x_i}$).

Is this correct?

And I have another question: I would have thought that the image $\pi(\mathfrak{p})$ of a point $\mathfrak{p}\in\mathbb{A}_k^{n+1}\backslash\{0\}$ could be described as follows: If $x_i\notin \mathfrak{p}$, and $$\mathfrak{p} = (f_1,...,f_m)$$ then we use $x_i$ to "homogenize" each $f_j$ -- that is we multiply each term of $f_j$ by some power of $x_i$ so that the resulting $f_j'$ is homogeneous. Then $$\pi(\mathfrak{p}) = (f_1',...,f_m') $$ (Here we are thinking of $\mathbb{P}_k^n$ as the space of all homogeneous primes of $k[x_0,..,x_n]$ not containing the irrelevant ideal)

Is this correct?

This is my intuition, but I can't seem to prove it using the map in the first question.

Answer 1

$\DeclareMathOperator{\Spec}{Spec}$Let's call $R=k[x_0,\dots x_n]$. We have $\mathbb{A}^{n+1}-\{0\} = \Spec(R)-(x_0,\dots x_n)$ and $\mathbb{P}^n$ corresponds to homogenous prime in $R$ different from $(x_0 \dots x_n)$.

Now, given a prime ideal $\mathfrak{p} \in \Spec(R)$ we can take its homogenization $\mathfrak{p}^h$ which is a prime homogenous ideal.

Now, the map $$ \pi: \mathbb{A}^{n+1}-\{0\} \to \mathbb{P}^n $$

$$\mathfrak{p} \to \mathfrak{p}^h$$ is at least well defined for topological spaces.

You can check that this map does exactly what you correctly pointed out at the end of your question. Now, to define the map between sheaves ,you can notice that $\pi(D(x_i))=D_+(x_0)$ where $D(x_i)=\{\mathfrak{p} \in \Spec(R) \mid x_i \not\in \mathfrak{p} \}$.

So, to let's say we want to define our morphism of sheaves $$\pi: \mathcal{O}_{\mathbb{P}^n} \to \pi_*\mathcal{O}_{\mathbb{A}^{n+1}-\{0\}} .$$

We start to define it restricted to the open sets $\pi(D_+(x_i))$ so that because of the observation above and the fact that both $U_i=D_+(x_i)$ and $V_i=D(x_i)$ are affine , it is sufficient to define a morphism of rings $ \mathcal{O}_{\mathbb{P}^n}(U_i) \to \pi_*\mathcal{O}_{\mathbb{A}^{n+1}-\{0\}}(V_i)$.

Now, with the standard identification, this is the standard morphism of immersion you pointed out.

It's a standard verification that these morphisms patch locally so that we get the global $\pi$ between sheaves we wanted.

Let's look at what happens to primes here: let $i=0$ for the sake of simplicity. So we have $\mathcal{O}_{\mathbb{P}^n}|_{U_0} \cong \Spec\left(k\left[\dfrac{x_i}{x_0}\right]\right)$. This isomorphism goes like this: you take a homogenous prime $\mathfrak{p}$, you localize it in $x_0$ and then you take the degree zero part.

To get what Vakil is saying, try to look a what happens at maximal ideals.