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?

  • $\begingroup$ This is exactly the same as Vakil's notes(5.5.H(b)). $\endgroup$ – 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).

  • $\begingroup$ 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 $I()$. $\endgroup$ – Tom Nov 20 '12 at 0:53
  • $\begingroup$ OK. I know Vakil's very nice notes, but I am too lazy to find the check. I will edit my answer. $\endgroup$ – user18119 Nov 20 '12 at 8:26

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