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I am going through William Fulton's "Algebraic Curves" and when he started talking about projective spaces, I got a little confused as the idea I think is right is not mentioned anywhere in the book. I started to think it might be just wrong and decided to ask this question here. If similar question already appeared somewhere I would be glad to be redirected there.

So in affine case we have Nullstellensatz which says $$I(V(I)) = \sqrt{I}.$$ It is very easy to check that an ideal of any algebraic set is radical. Also we have that $\sqrt{I}$ is the smallest radical ideal, containing I (if $J$ is any radical ideal containing I, then $I \subseteq \sqrt{I} \subseteq J$). So in a sense, $\sqrt{I}$ is the smallest possible answer to the problem of figuring out what $I(V(I))$ is.

So in projective case my reasoning was as follows. We know that ideal of a projective algebraic set is homogeneous. Also, for an ideal I, let's denote by $H(I)$ the intersection of all homogeneous ideals containing $I$ (which is, plainly, homogeneous too). So the analogy suggests that we must have $$I(V(I)) = H(I).$$ If it is even true, what assumptions have to be made in order for this to work? Clearly, the example of $x^2 + 1$ over $\mathbb{R}$ suggests that field has to be algebraically closed. But is there something else?

Also why I initially found it peculiar is that there is no particular notation for what I've just called "$H(I)$". Or is there any standard choice which I am not aware of?

Thank you for your answers in advance.

Aleksei Kubanov
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    $H(I)$ is just the usual radical as explained in the linked duplicate (Keenan Kidwell's answer in particular). Taking your field to be algebraically closed will indeed suffice, and this is typically known as the projective nullstellensatz. – KReiser Mar 01 '22 at 18:43
  • @KReiser this is exactly what I was looking for. Thank you a lot! – Aleksei Kubanov Mar 01 '22 at 20:56

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