1
$\begingroup$

If $R$ is a commutative ring with identity, then $Spec(R)=\cup D(f_i)$ $i\in A$ (where $D(f)$ is the principal open set in $Spec(R)$ consisting of prime ideals of $R$ not containing $f$.) is equivalent to $({f_i:i\in A})=(1)$. What is the projective analog of this result. The proof for this results proceeds by showing that if $p$ is a prime in $R$, then $p$ does not contain one of the $f_i$ and hence no prime contains the ideal $({f_i:i\in A})$. In the projective case, there are several ideals in $S$ not contained in $Proj(S)$ which is the set of homogeneous prime ideals not equal to the maximal ideal, $S=k[x_0,x_1,...,x_n]$. For example, any power of the homogeneous maximal ideal. So I am unable to carry out an analogous argument.

1 Answers 1