# Relative proj base ring, closed immersions via Vakil

This is an annoying question and I apologize in advance.

Vakil defines a relative proj sheaf of algebras on $$X$$ to be such that there is a cover of $$\operatorname{Spec}(A)$$ s.t above each we have $$S^A$$ a $$\mathbb{Z}_{\geq 0}$$ graded quasicoherent algebra (this just means each graded component is a quasicoherent module).

Importantly, he requires that $$S^A$$ at grade $$0$$ is precisely $$\operatorname{Spec}(A)$$, and not a quotient (i.e one could fathom a definition that $$S^A$$ just means something $$A$$ acts on and doesn't change degree, but this isn't the case).

Given such a sheaf one defines a scheme by taking the proj construction on each of those, with a natural mapping to $$X$$, so that we get from such a sheaf $$F$$ on $$X$$ a map $$\operatorname{Proj}(F) \to X$$

Projective morphisms are then those that arise in this way for finite-type generated in degree $$1$$ $$F$$.

My question is then why if $$Z \to \mathbb{P}^1_X$$ is a closed embedding then it is a projective morphism. Even in the case $$X=\operatorname{Spec}(A)$$, obviously $$Z$$ is given by a Proj construction which should finish, but at degree $$0$$ it will be a quotient of $$A$$, and not precisely $$A$$. An example to keep in mind is $$X = \operatorname{Spec}(A) = \mathbb{A}^1_k$$, and $$Z$$ is the copy of $$\mathbb{P}^1$$ over a point.

We can try to solve this by just setting in the graded algebra describing $$Z$$ the 0th component to $$A$$, but this doesn't work, it really gives a different scheme.

I think the real solution lies in just instead of requiring $$S_0 = A$$, instead work with graded over $$A$$ to mean there is a map $$A \to S_0$$.

If you want to see this contradicting a specific exercise (or at least the obvious 'solution' to it); though I don't recommand it, since the above baby case holds my concern: Exercise 17.3.A says suppose $$\pi :X \to Y$$ is a morphism, then it is projective if there is a finite type quasicoherent sheaf $$F_1$$ on $$Y$$ and a closed embedding $$X \to \mathbb{P}(F_1)$$ over $$Y$$ (\mathbb{P}(F_1) means take the graded symmetric algebra that $$F_1$$ generates and apply the proj construction to it). The obvious solution wants us to say take the graded module of (\mathbb{P}(F_1)) and divide it by the kernel of the map $$X \to \mathbb{P}(F_1)$$ (i.e $$X$$ is this divided by some ideal). The problem is that at grade $$0$$ we're not supposed to be dividing by anything according to Vakil.

• Here is a similar question: math.stackexchange.com/questions/255215/… Jun 30 at 16:10
• @hm2020 thanks for your comment. Note I'm asking about the easy direction though- why is Harthhorne definition a particular case of Vakils (or if you prefer, Vakil has an exercise that says that projective is the same as closed embeddings in Proj of a finite-type quasicoherent sheaf (i.e the algebra it generates)). My problem is the grading at 0
– Andy
Jun 30 at 16:20
• you should write down precisely what type of definition Vakil is using and the exercise, since "most" people do not have a copy of this book. Jun 30 at 16:22
• @hm2020 Done, is it readable?
– Andy
Jun 30 at 17:03
• Quick comment: If you want to write anything like $\operatorname{Proj}, \operatorname{Spec}, \dots$, you should use \operatorname in LaTeX. Jun 30 at 19:04

The fact that Vakil lets you get away with schemes isomorphic to $$\operatorname{Proj} S_\bullet$$ for $$S$$ a finitely generated algebra over $$A$$ resolves this issue and lets you get past this trouble.

The key thing here is that if $$S_\bullet$$ is a graded ring and $$S_0'$$ is any ring with a homomorphism $$\varphi:S_0'\to S_0$$, the graded ring $$S'_\bullet$$ defined by $$S'_i=\begin{cases} S_0' & i=0 \\ S_i & i > 0 \end{cases}$$ with multiplication defined by $$s'_0\cdot s_d = \varphi(s'_0)\cdot s_d$$ for $$s'_0\in S'_0$$ and $$s_d\in S_d$$ has $$\operatorname{Proj} S_\bullet \cong \operatorname{Proj} S'_\bullet$$.

To prove this, we will show that a graded prime ideal $$P$$ of $$S_\bullet$$ not containing the irrelevant ideal $$S_+$$ is determined by its intersection with $$S_+$$. By the condition that $$P$$ is a graded ideal not containing $$S_+$$, there must be some homogeneous element $$x\in S_+$$ not in $$P$$. Letting $$s_0\in S_0$$ be arbitrary, by primality of $$P$$ we have that $$s_0x\in P$$ exactly when $$s_0\in P$$, and so we can recover $$P\cap S_0$$ from $$P\cap S_+$$.

Letting $$S$$ be a graded $$A$$-algebra where $$S_0$$ isn't necessarily equal to $$A$$, we can apply the above construction with $$S'_0=A$$ and $$\varphi:S_0'\to S_0$$ the structure map for $$S_0$$ as an $$A$$-algebra to see that $$\operatorname{Proj} S_\bullet$$ is isomorphic to $$\operatorname{Proj} S'_\bullet$$, which is a graded ring over $$A$$ in the sense of Vakil.

• Cool I think you're correct: a different explanation (that also shows why everything is an isomorphism): on $D_f$ for $f$ of positive degree we get the same ring, since if there was a difference it would be because of the grade $0$, but for any $s \in S_0$, in $D_f$ $s = s*f/f$ so it comes from some positive grade.
– Andy
Jun 30 at 23:07
• Yes, exactly - this is also what shows you that the structure sheaf is the same, which I probably should have included in the answer as well. Glad to have helped. Jul 1 at 0:21

Question: "My question is then why if $$Z→P^1_X$$ is a closed embedding then it is a projective morphism."

Answer: If $$I:=\oplus_i I_i$$ is a sheaf of graded commutative $$\mathcal{O}_Y$$-algebras with $$I_0:=\mathcal{O}_Y$$ and $$I$$ generated by $$I_1$$ as $$I_0$$ algebra, there is a surjective map

$$\rho: Sym_{\mathcal{O}_Y}^*(I_1) \rightarrow I \rightarrow 0$$

of sheaves of graded $$\mathcal{O}_Y$$-algebras. The map $$\rho$$ gives rise to a closed immersion

$$i: \mathbb{P}(I) \rightarrow \mathbb{P}(I_1).$$

Hence in this case, $$\mathbb{P}(I)$$ is always a closed subscheme of $$\mathbb{P}(I_1):=Proj(Sym_{\mathcal{O}_Y}^*(I_1))$$ in Hartshornes definition. You do not need $$I_1$$ to be coherent for this to make sense.

If you define a map $$f:X \rightarrow Y$$ to be projective iff it factors through a closed immersion

$$i: X \rightarrow \mathbb{P}(J):=Proj(Sym_{\mathcal{O}_Y}^*(J))$$

(where $$J$$ is a quasi coherent sheaf) with $$f:=\pi \circ i$$ where $$\pi$$ is the canonical projection morphism, it follows any scheme on the form $$\mathbb{P}(I)$$ is projective over $$Y$$ if it satisfies the above condition. This is a more general definition than the one in Hartshorne.

More generally $$I_0$$ will be a sheaf of commutative $$\mathcal{O}_Y$$-algebras with $$T:=Spec(I_0)$$ and a canonical map $$p:T \rightarrow Y$$. You will get a closed immersion

$$\mathbb{P}(I) \subseteq \mathbb{P}(I_1)$$

and canonical maps $$\pi: \mathbb{P}(I_1) \rightarrow T \rightarrow Y$$.

If $$I_1 \cong \mathcal{O}_Y^{d+1}$$ is a trivial $$\mathcal{O}_Y$$-module of rank $$d+1$$ you get a closed immersion

$$i: \mathbb{P}(I) \subseteq \mathbb{P}^d_Y.$$

You construct relative projective $$d$$-space by taking the relativ proj of the trivial sheaf of rank $$d+1$$. I believe this is the Hartshorne definition.

• My problem with this is that Vakil only allows you to take the Proj construction when the 0th component of the graded ring is the ring $A$- over $Spec(A) \subset Y$. In your map, your $Sym$ graded ring is perfectly great (since its zeroth component is $A$ really$, but then you divide by some kernel :( – Andy Jun 30 at 17:05 • @Andy - the ideal sheaf$\mathcal{I}$is a global version of the ideal$I \subseteq A[x_0,..,x_n]$with$X:=V(I) \subseteq \mathbb{P}^n_A\$. Jun 30 at 17:37