# Understanding projective scheme as representable functor

I cannot understand the notes that my teachers use for class.

Theorem: let $S$ be a scheme and $E$ a quasicoherent $O_S$-module. The functor $$Sch_S^o\rightarrow Set$$ $$(\phi :X\rightarrow S)\mapsto \{[\phi ^*E\rightarrow L]\}$$ is representable by an $S$-scheme $\mathbb{P}(E)\rightarrow S$. To elaborate on the meaning on the RHS of $\mapsto$ an $S$-scheme $X$ is mapped to the set of maps from $\phi ^*E$ to $L$, where $L$ is invertible, modulo isomorphisms i.e. two maps $a$ and $b$ are same iff $\exists s:L_1\rightarrow L_2$ such that $b=sa$.

For a ring $A$, we denote $\mathbb{P}^n _A :=\mathbb{P} ((A^{(n+1)})^*)$. Firstly how is this seen as a quasicoherent sheaf on $\mathrm{Spec}(A)$ As an example they take $\mathbb{P} ^n _\mathbb{C}(\mathbb{C}) _{\mathbb{C}}$ which I don't understand. Firstly, can anyone guess the meaning of the last $\mathbb{C}$? I have given my guess below. The one in the brackets is talking about $\mathbb{C}$ points of the scheme. Apparently, the following sequence of equalities hold $\mathbb{P}^n _\mathbb{C}(\mathbb{C})_\mathbb{C}=\{\mathrm{Spec}(\mathbb{C})\mathrm{-morphisms.of. }\mathrm{Spec}(\mathbb{C})\mathrm{to.}\mathbb{P}^n _\mathbb{C}\}$ $=\{\mathrm{invertible.quotients.of.rank.1.of.}(\mathbb{C}^{n+1})^*\}=\{\mathrm{subspaces.of.rank.1.of.}\mathbb{C}^{n+1}\}$ which is indeed the projective space. Can anyoune explain meaning of the first and second $=$? I think the first is just definition. What would be obtained if we would take $\mathbb{P} ^n _\mathbb{C} (\mathbb{C})$ ?

How actually to work with this (in my view) cumbersome definition? How is $\mathrm{Proj}$ related to this?

• Do you have a link to the notes in question? – Daniel McLaury Dec 25 '16 at 17:02
• I have, but it is in French. However the part on $\mathbb{P}^n _{\mathbb{C}}$ is not visible to the general public. – ralleee Dec 25 '16 at 17:16

If we define the functor $\mathbb{P}$ by its representable functor property, then $\mathbb{P}(\mathcal{E})$ is by definition the $S$-scheme such that, for any $S$-scheme $\pi : X \to S$, we have $$\operatorname{Hom}_S(X, \mathbb{P}(\mathcal{E})) = \{ \text{invertible quotients of } \pi^\ast\mathcal{E} \}$$ for all quasicoherent $\mathcal{E}$ (and such that the right thing happens for morphisms). You've written this as a theorem above, so you're working with a different definition, but at any rate this is a true statement we can use.
Now $(A^{n+1})^\ast$ is an $A$-module, so the sheaf on $\operatorname{Spec} A$ associated to it is quasicoherent by definition. So we can apply the above result to it with no trouble.
For the example you give, we want to consider $\mathbb{P}^n_{\mathbb{C}} = \mathbb{P}((\mathbb{C}^{n+1})^\ast)$. By definition its $\mathbb{C}$-points are given by $$\operatorname{Hom}_\mathbb{C}(\operatorname{Spec} \mathbb{C}, \mathbb{P}^n_\mathbb{C})$$ so let's take $S = \operatorname{Spec} \mathbb{C}$ and $X = \operatorname{Spec} \mathbb{C}$ and $\mathcal{E} = (\mathbb{C}^{n+1})^\ast$. Here $\mathbb{C}$ denotes the trivial line bundle on $S$, which is of course just a copy of the vector space $\mathbb{C}$ since $S$ consists of a single point. From the defining property of $\mathbb{P}(\mathcal{E})$ above we have $$\operatorname{Hom}_{\operatorname{Spec} \mathbb{C}}(\operatorname{Spec} \mathbb{C}, \mathbb{P}^n_\mathbb{C}) = \{\text{invertible quotients of } \pi^\ast (\mathbb{C}^{n+1})^\ast\}$$ Since $X$ consists of a single point, $\pi^\ast (\mathbb{C}^{n+1})^\ast$ is just a single copy of $(\mathbb{C}^{n+1})^\ast$. And the rest is now linear algebra.
I'm not a hundred percent sure what they were going for with the notation. There's at least one error -- a $\mathbb{P}^n$ where a $\mathbb{P}$ is meant -- so it could be a mistake. It could also mean that $\mathbb{P}^n_\mathbb{C}$ is taken as a scheme per se, whereas the other thing denotes the same space regarded as a $\mathbb{C}$-scheme.
• Awesome. Thanks! Can you just add me one more thing, if it is not too difficult, what is relation with the $\mathrm{Proj}$ functor? – ralleee Dec 25 '16 at 21:15