# Projective Nullstellensatz

I'm confused about the proof of the Nullstellensatz for projective varieties.

If $J \subset k[x_0, \ldots , x_n]$ is a homogeneous ideal, we may regard $V(J)$ as a closed subset $V(J) = V \subset \mathbb P_k^n$ and also as a closed subset $V(J) = V^a \subset \mathbb A_k^n$. If $\pi : \mathbb A_k^{n+1} \backslash \{ 0\} \to \mathbb P_k^n$ is the residue class map used to define projective space, then we have that $V^a = \pi^{-1}(V) \cup \{ 0\}$ ($0$ is always a root of a homogeneous polynomial). The statement I have of the Projective Nullstellensatz is:

Let $k$ be an algebraically closed field and $J \subset k[x_0, \ldots , x_n]$ be a homogenous ideal. Then we have

i) $V(J) = \emptyset \iff \sqrt{J} \supset \langle x_0, \ldots , x_n \rangle$

ii) If $V(J) \neq \emptyset$, then $I(V(J)) = \sqrt{J}$

The proof of i) goes: $V(J) = \emptyset \iff V^a \subset \{0 \} \iff \sqrt{J} \supset \langle x_0 , \ldots , x_n \rangle$. But mustn't $V^a$ always contain $0$, so we always have equality $\sqrt{J} = \langle x_0 , \ldots , x_n \rangle$?. But then that isn't true, since we could let $J = k[x_0, \ldots , x_n]$ What's going on?

Thanks

• Can you tell me where you got the statement for this? My book doesn't state the result explicitly, so I would like to use your reference which does have an explicit statement. – Chill2Macht Jun 21 '17 at 14:20
• It is in Fulton's book in algebraic curves, page 46 – yamete kudasai Oct 26 '18 at 12:22

But mustn't $V^a$ always contain $0$, so we always have equality $\sqrt{J}=\langle x_0,\ldots,x_n\rangle$?
Not necessarily. There is a single exception, which you mentioned, that is $J=(1)$. Then $V^a=\emptyset$ so $V\subset \{0\}$ and $\sqrt{J}\supset \langle x_0,\ldots,x_n\rangle$ where both inclusions are proper. In all other cases the degree of $J$ is at least $1$ so $0\in V^{a}$ giving us equality.