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.

  • $\begingroup$ Here is a similar question: math.stackexchange.com/questions/255215/… $\endgroup$
    – hm2020
    Jun 30 at 16:10
  • $\begingroup$ @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 $\endgroup$
    – Andy
    Jun 30 at 16:20
  • $\begingroup$ 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. $\endgroup$
    – hm2020
    Jun 30 at 16:22
  • $\begingroup$ @hm2020 Done, is it readable? $\endgroup$
    – Andy
    Jun 30 at 17:03
  • 2
    $\begingroup$ Quick comment: If you want to write anything like $\operatorname{Proj}, \operatorname{Spec}, \dots$, you should use \operatorname in LaTeX. $\endgroup$
    – Qi Zhu
    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.

  • $\begingroup$ 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. $\endgroup$
    – Andy
    Jun 30 at 23:07
  • 1
    $\begingroup$ 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. $\endgroup$
    – KReiser
    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.

  • $\begingroup$ 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 :( $\endgroup$
    – Andy
    Jun 30 at 17:05
  • $\begingroup$ @Andy - the ideal sheaf $\mathcal{I}$ is a global version of the ideal $I \subseteq A[x_0,..,x_n]$ with $X:=V(I) \subseteq \mathbb{P}^n_A$. $\endgroup$
    – hm2020
    Jun 30 at 17:37

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