We say that a ring is a Hilbert ring if any radical in it is the intersection of maximal ideals.
I quote my university lecture notes:
It is actually possible to show that if $K$ is any field, then the ring $K[x_1, . . . , x_n]$ is a Hilbert ring. The (strong) Hilbert Nullstellensatz is equivalent to this fact.
My question is, how is Hilbert Nullstellensatz equivalent to this?
My thoughts: Hilbert´s Nullstellensatz says that for any $K$ algebraically closed and any proper ideal in $K[x_1, . . . , x_n]$, $I(V(J)) = \sqrt{J}$. That is, any ideal generated by an affine algebraic set generated by $J$ is in a form of racical of $J$.
In other words, if a polynomial $f \in V(J)$ has $f(P) = 0 \forall P \in V(J)$, then $\exists n \in \mathbb{N}$ such that $f^n \in J$. (Or it says "$f^n \in I$", not $J$ on Wikipedia which I also dont understand why.)
The $J$ (I think) "lives" in $K[x_1, . . . , x_n]$ too and maybe we could use that "maximal ideal implies prime ideal, which impplies radical ideal" to achieve the mentioned equivalence?
I am not really sure though, maybe it is just a mistake in the notes or my wrong interpretation of something.
Thank you for your help.
EDIT: To clarify, I am working over an algebraically closed field.
$I(V(J))$ is the intersection of all the functions which vanish at each point of $V(J)$. But that's the same as the intersection of all the maximal ideals containing $J$. By the definition of a Hilbert ring, that's the same thing as $\sqrt{J}$. For the converse, run this backwards.