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?



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


