# When is it true that the ring of global regular functions on a projective variety is just the base ring?

Apologies in advance if this is too elementary.

The following is well known when $A$ is an algebraically closed field:

Let $X$ be an integral closed subscheme of $P^n_A$. Then $\Gamma(X, \mathcal{O}_X) = A$.

My question is: For what other rings does the above statement hold?

There are two proofs of this (for $A$ algebraically closed) in Hartshorne. Both seem to use the fact that the integral closure of $A$ in its quotient field is just $A$ itself in a key way.
So I suspect that having $A$ integrally closed will be crucial, but I do not know. In particular, does the proof in Hartshorne still work, and if so, does it apply to nonnormal domains?

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This holds for any proper morphism $f : X \rightarrow S$ which is flat, with reduced and connected geometric fibres. In particular, the property is stable under base change in this situation. The proof uses the statement when the base scheme is the Spec of an algebraically closed field. –  user1594 May 28 '10 at 14:54
When you say "this holds", what do you mean by "this"? That global functions on $X$ are the same as those on $S$? –  A. Pascal May 28 '10 at 15:01
The statement, as you suspected, is definitely not true if A is not integrally closed. If B is a finite A-algebra, then $P^n_B$ can be written as a closed subscheme of $P^N_A$ for some N. [Given any finitely generated graded B-algebra R with R_0 = B, we can replace R_0 with A and get a f.g. graded A-algebra with an isomorphic Proj.] –  Charles Staats May 28 '10 at 15:12
I can imagine how you might try to prove the statement in the above comment. Under the assumptions on the fibres of $f$, the global functions on the fibres over geometric points will just be constants. Thus $f_{*} \mathcal{O}_{X}$ will have one dimensional fibres and so be a line bundle. You would like to conclude that the natural morphism $\mathcal{O}_{S} \rightarrow f_{*} \mathcal{O}_{Y}$ is in fact an isomorphism. The Leray spectral sequence will then imply that global functions on $X$ are the same as those on $Y$. Is this what JT had in mind? –  A. Pascal May 28 '10 at 15:20
@JT: assuming $f$ finitely presented too? Anyway, the need to use higher sheaf cohomology methods (base change results, theorem on formal functions, etc.) as in above comments reminds me of this: for any ring $A$ with connected spec, Picard group of $\mathbf{P}^n_ A$ is $\mathbf{Z} \times {\rm{Pic}}(A)$, with the infinite cyclic part generated by $O(1)$. The proof begins with the classical case when $A$ is a field, but then one has to make essential use of Grothendieck's results on higher direct images in families (e.g., theorem on formal functions). I am not aware of any simpler proof. –  BCnrd May 28 '10 at 15:40
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The statement is true if and only if $A$ is an algebraically closed field.

Assume the statement is true.

Let $\mathfrak{m}$ be a maximal ideal of $A$. Then $A \to A / \mathfrak{m}$ is certainly finite. Thus, as I explain in my comment above, $\mathbb{P}^n_{A / \mathfrak{m}}$ is an integral subscheme of some $\mathbb{P}^N_A$. Thus, $A = A / \mathfrak{m}$; since $\mathfrak{m}$ was maximal, $A$ is a field.

Suppose $A$ is not algebraically closed. Then $A$ has a finite extension $B$; and again, as remarked in the comment, this implies that $B$ is the ring of global regular functions of some integral closed subscheme of $\mathbb{P}^N_A$, a contradiction.

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Thanks! One question: how does the fact that $P^a_{A/m}$ is an integral subscheme of $P_A^N$ tell us that $A = A/m$? –  Akhil Mathew May 28 '10 at 18:01
For any field $k$, $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k}) = k$. Also $A/m$ is a field so for $X = \mathbb{P}^n_{A/m}$ we have $\Gamma(X,\mathcal{O}_X) = A/m$. On the other hand we 'Assume the statement is true,' i.e. for any closed integral $X \subset \mathbb{P}^m_A$ we have $\Gamma(X,\mathcal{O}_X) = A$. Together these say $A = A/m$. –  solbap May 28 '10 at 20:34
As solbap explains. I would also note that $\Gamma(\mathbb{P}^n_R, \mathcal{O}_{\mathbb{P}^n_R})=R$ holds for arbitrary rings $R$. –  Charles Staats May 28 '10 at 20:57
Ah! Thanks for the explanation. –  Akhil Mathew May 28 '10 at 22:48