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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.

Tereza Tizkova
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$I(V(J))$ 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.

Hank Scorpio
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  • Why is $I(V(J))$ the intersection though? I thought it is defined as ${ f \in K[x_1, ... x_n] \mid f(P) = 0 \forall P \in V(J) }$? And why is it the same as the intersection of all MAXIMAL ideals containing $J$? – Tereza Tizkova Sep 02 '22 at 19:59
  • @TerezaTizkova - by the weak Nullstellensatz, when $K$ is algebraically closed, the maximal ideals correspond exactly to the sets of all functions vanishing at a given point $P$. Then the fact that what you wrote is the intersection over all $P \in V(J)$ of the ideals of functions vanishing at $P$ can be shown just by inclusion in both directions – stillconfused Sep 02 '22 at 20:24
  • @Hank Scorpio - is the converse not a little more complicated than running this argument backwards? The OP's statement is that $K[x_1, \dots, x_n]$ is a Hilbert ring for any field $K$. I think there should be some argument about extending from any field to an algebraic closure in order to use the Nullstellensatz over only an algebraically closed field to get the given statement over any field – stillconfused Sep 02 '22 at 20:26
  • @stillconfused But weak Nullstellensatz states that $V(I)$ cannot be empty for a proper ideal $I$, again, how does this say what you wrote in your comment? I dont know why I havent come accross this version of the theorems before. – Tereza Tizkova Sep 02 '22 at 20:32
  • @stillconfused yeah, you have to upgrade your viewpoint and/or argument a little bit if you're not working over an algebraically closed field. I kind of assumed the OP was working in that context, though - I'm happy to adjust my post if they can clarify their setup. – Hank Scorpio Sep 02 '22 at 21:19
  • @HankScorpio I do work over an algebraically closed field. Have updated that in the question to be clear. – Tereza Tizkova Sep 04 '22 at 13:55
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    @TerezaTizkova - from that version of the weak Nullstellensatz, let $I$ be any ideal. Then $V(I)$ contains some point $P = (p_1, \dots, p_n)$. Since ${P} \subset V(I)$, we have that the ideal of $P$ contains the ideal of $V(I)$, but the ideal of ${P}$ is $M_P = (x_1 - p_1, \dots, x_n - p_n)$, so $I$ is contained in the ideal of $V(I)$ is contained in $M_P$. Thus every ideal is contained in the ideal of some point, so all maximal ideals must be ideals of a point. And these ideals are maximal because the quotient is a field. Thus these are all the maximal ideals. – stillconfused Sep 06 '22 at 01:43
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    I've been careful in the above comment to say "the ideal of $V(I)$" and not $\sqrt{I}$, just because the weak Nullstellensatz is often used to prove the strong one, so we don't want to be circular – stillconfused Sep 06 '22 at 01:44