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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.

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Over projective space $\mathbb{P}^n$, you can define basic open affines $D(x_i)$ (or more generally $D(f)$ for $f$ homogeneous) that correspond to homogeneous prime ideals that do not contain $f$. All of projective space is the union of the $D(x_i)$, as any prime ideal that contains all the $x_i$ (i.e. fails to lie in the union of the $D(x_i)$) must be the irrelevant ideal $(x_0, \dots, x_n)$, and that is not included in the Proj.

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Thanks for your response, Akhil. Perhaps, I wasn't sufficiently clear with my question. Suppose, $Proj(S)=\cup D(f_i)$ where the $f_i$ are homogeneous, how do we conclude they generate the homogeneous maximal ideal. The argument I made above only shows they generated an ideal which is not contained in any prime ideal. In the affine case, the only ideal not contained in any prime is the unit ideal. In the projective case, there are several candidates, for example, any power of the homogeneous maximal ideal. So we need something more to say that the $f_i$ generate the homogeneous maximal ideal. –  user9449 Apr 12 '11 at 2:47
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