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So a while ago I proved that every closed subscheme of a (quasicompact) projective scheme is a projective scheme (i.e. of the form $\operatorname{Proj A}$). I often hear the classical definition that projective varieties over $k$ are closed subschemes of $\mathbb P^n_k$ for some $n$, so I would hope that this also true with the more modern definitions.

To be precise, I take a variety to be a scheme $X$ over a field $k$ which is reduced, separated, and of finite type. In particular, a projective variety is a variety is which of the form $\operatorname{Proj}A$. I suspect that I could potentially relax the reduced and separated conditions, and get a similar result, but I don't care that much about that.

Let $X$ be a projective variety, so that $X=\operatorname{Proj}A$. In particular $X$ is Noetherian, and there is a finite open cover of the form $\operatorname{Spec}(A_{f_i})_0$ for some fixed graded ring $A$, and $f_i$ homogenous of positive degree, $0\leq i\leq m$. Each $(A_{f_i})_0$ should be isomorphic to $k[x_0,\dots, x_{n_i}]/I_i$ by finite type. Without loss of generality, I can take $n_i=n$ for all $i$ because there is some maximum $n_j$ and for any $n_i<n_j$ I can quotient $k[x_0,\dots, x_{n_j}]$ out by $\langle x_{n_i+1},\dots, x_{n_j},I_i\rangle$ to ge the same $k$ algebra.

I suspect I should be able to then embed $X$ into $\mathbb P^{n+1}_k$ by defining local maps, but I can't quite see how to do this, and how the gluing should work. Any help would be greatly appreciated.

Chris
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2 Answers2

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Proj behaves in enough funny ways that this is not true without (potentially many) further assumptions. For instance, let $R$ be a ring, and let $R[t]$ be the graded ring where $R$ has degree $0$ and $t$ has degree $1$ - then $\operatorname{Proj} R[t] \cong \operatorname{Spec} R$ (ref), so for instance if we take $R=k[x]$, then $\operatorname{Proj} k[x,t]$ where $\deg x = 0$ and $\deg t=1$ is exactly $\Bbb A^1_k$, which is not projective over $k$. This is also unfixable via putting a condition on $A_0$, since Proj is insensitive to alterations of $A_0$ (ref).

The more standard definition of a projective variety $X$ over $k$ is one where the morphism $X\to \operatorname{Spec} k$ is projective; there are then a few different choices of what "projective" mean (see Stacks 01W7) which are all equivalent when working over a field (and more generally, all equivalent when working over a base scheme which admits an ample invertible sheaf).

KReiser
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  • I see...Vakil defines a projective scheme over $A$ to be $\operatorname{Proj}S$ where $S$ is a graded ring finitely generated by $A=S_0$; this seems to eliminate the grading issue in your counter example, but you say $\operatorname{Proj}$ is insensitive to alterations of $A_0$. Does forcing the zero degree elements to be a field not somehow fix this problem? – Chris Nov 10 '24 at 18:14
  • Take the same ring but the $0$-graded piece to $k$ will give you the same Proj. If you add the conditions that $S_0=A$ and $S$ is finitely generated over $S_0$, then you can prove that there is a closed immersion $\operatorname{Proj} S\to\Bbb P^n_A$ for some $n$; but those assumptions aren't in your question. (If you would like to add these assumptions, I would recommend asking a new question, assuming it hasn't already been covered here on MSE.) – KReiser Nov 10 '24 at 18:24
  • thank you, I will ask a new question. – Chris Nov 10 '24 at 18:27
  • It seems from Corollary 5.16 in Hartshorne that we need $A$ to be finitely generated over $A_1$ as an $A_0$ algebra. Also I asked a new question here: https://math.stackexchange.com/questions/4996730/every-projective-shceme-over-a-is-a-closed-subscheme-of-mathbb-pn-a – Chris Nov 12 '24 at 05:59
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The fact that $\mathrm{Proj}\,A$ is covered by $\mathrm{Spec}\,A_{f_i}$ plainly means that any homogeneous prime ideal not containing the irrelevant ideal of $A$ must not contain every $f_i$.

So, any prime ideal containing all of the $f_i$ is irrelevant. Hence as topological spaces, one has $$\mathrm{Proj}\,A=\mathrm{Proj}(k[f_0,…,f_m]).$$ The latter is definitely a closed subscheme of $\mathbb{P}^m_k$.

Further, a commonly known fact is that given a Noetherian scheme $S$ and a closed set $Z\subseteq S$, there is only one reduced closed subscheme of $S$ with underlying topological space $Z$. Since both above schemes are reduced, it follows that in fact, $$\mathrm{Proj}\,A=\mathrm{Proj}(k[f_0,…,f_m])$$ as schemes, so that $\mathrm{Proj}\,A$ is a closed subscheme of $\mathbb P^m_k$.

Simon Pitte
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  • For the commonly known fact I use, Noetherianity might not even be required actually. I’ll leave it there to be safe. – Simon Pitte Nov 10 '24 at 10:13
  • This answer is not correct as currently written - for example, when $A=k[x,t]$ with $\deg x = 0$ and $\deg t =1$, then $\operatorname{Proj} A$ is covered by $D_+(t)$, but $\operatorname{Proj} A$ is not in any sense the same as $\operatorname{Proj} k[t]$. Additionally, even if this were the case, the claim that $\operatorname{Proj}(k[f_0,\cdots,f_m])$ is a closed subscheme of $\Bbb P^m_k$ requires more work - what if not all of the $f_i$ have the same degree? (This of course can be made to work with more argument, but that argument isn't present.) – KReiser Nov 10 '24 at 16:37
  • Thank you for the counterexample. You are right, I should have given more thought to my answer. – Simon Pitte Nov 10 '24 at 17:31