# Definition of $I(\cdot)$ for projective schemes

How should we define the function $I(\cdot)$ for ${\rm Proj} S_\bullet$(the homogeneous prime ideals not containing $S_+$) for a $\mathbb{Z}^{\ge 0}$-graded ring $S_\bullet=S_0\oplus S_+$?

I know the functions $V(\cdot)$ and $I(\cdot)$ for affine schemes. I want the projective version of them. I know the projective version of $V(\cdot)$, i.e. $V(T):=\{p| p\supset T\}\subset {\rm Proj} S_\bullet$ for $T\subset S_+.$

If we follow the affine case, we might define $$I(Z):=\cap_{p\in Z}p$$ for $Z\subset {\rm Proj} S_\bullet$.

However this definition does not satisfy $$I(Z)\subset S_+$$ because if $S_0:=\mathbb{Z}, S_\bullet:=\mathbb{Z}[x]$ and $Z=\{ (2)\}$ then $I((2))=(2)=(2\mathbb{Z})[x]\supset 2\mathbb{Z} \not\subset S_+.$

$\bullet {\bf EDIT}$ (added just after my first comment to the first answer):

How about the following? $$I(Z):=\langle (\cap_{p\in Z}p)\cap \cup_{i>0}S_i\rangle$$ Here $\cup_{i>0}S_i$ means the homogeneous elements of positive degree, and the bracket means the ideal generated by the ingredients.

How do people define $I(\cdot)$ usually?

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This is exactly the same as Vakil's notes(5.5.H(b)). –  Tom Nov 19 '12 at 14:09

Your definition of $I(Z)$ is correct. One doesn't ask $I(Z)$ to be contained in $S_+$.
Over a field, $I(Z)$ not contained in $S_+$ means $Z=\emptyset$. In general, if $S$ is an $S_0$-algebra. Suppose $S_0$ is reduced. If $I(Z)$ is not contained in $S_+$, then there exists $f\in I(Z)_0\subseteq S_0$ non-zero. Thus the image of $Z\to \mathrm{Spec}(S_0)$ is contained in the proper closed subset $V(f)$. In your example, $f=2$.
Edit. Whatever is the definition of $I(Z)$, what you want I guess is $V_+(I(Z))=Z$ (I prefer to put underscore $V_+$ to stress on the homogeneous ideals), as sets when $Z$ is a closed subset. This is the case with your definition. Indeed, if $Z=V_+(I)$, then $$V_+(I(Z)) = Z$$ set-theoretically. By Krull's intersection theorem $\sqrt{I}$ is the intersectio of all (unhomogeneous) primes $q\supseteq I$. So $\sqrt{I}\subseteq I(Z)$ and $V_+(I(Z))\subseteq V_+(\sqrt{I})=Z$. Conversely, if $p\in Z$, then $I\subseteq p$, so $I(Z)\subseteq p$ and $p\in V_+(I(Z))$.
Now to reconcile with other possible definitions, we have $$V_+(I)=V_+(I\cap S_+)$$ for any homogeneous ideal $I$ of $S$ (the RHS is the ideal in the edited part of your OP). This can be found in "Algebraic geometry and arithmetic curves", Lemma 2.3.35 (b).
Thank you so much. It means that Vakil's notes (5.5.H(b)) is wrong without adding an assumption, doesn't it? Hartshorne doesn't seem to contain definition of $I()$. –  Tom Nov 20 '12 at 0:53