# (Possibly) alternative statement of Hilbert's Nullstellensatz

In my notes, there is a statement entitled "Nulstellensatz version 2":

If $k = \bar{k}$, and $\mathfrak{m} \subseteq k[x_1,\ldots,x_n]$ is a maximal ideal, then $k[x_1,\ldots,x_n]/\mathfrak{m} \cong k$.

I assume "version 2" implies it is equivalent to the usual version, which states:

If $k = \bar{k}$, $\mathfrak{a} \subseteq k[x_1,\ldots,x_n]$ is an ideal and $f \in k[x_1,\ldots,x_n]$ is a polynomial which vanishes on all points in $Z(\mathfrak{a})$, then $f^r \in \mathfrak{a}$ for some positive integer $r$.

I can see that "version 2" follows from the "usual version" as follows:

Usual version $\implies$ $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$ for ideals $\mathfrak{a}$ of $k[x_1,\ldots,x_n]$. We also have the following easily verified facts:

i) $Y_1 \subseteq Y_2 \subseteq k^n \implies k[x_1,\ldots,x_n] \supseteq I(Y_1) \supseteq I(Y_2)$;

ii)$T_1 \subseteq T_2 \subseteq k[x_1,\ldots,x_n] \implies k^n \supseteq Z(T_1) \supseteq Z(T_2)$ and

iii) $Y \subseteq k^n$ is irreducible if and only if $I(Y) \subseteq k[x_1\ldots x_n]$ is prime.

Combining this information gives that $Z$ and $I$ give an inclusion-reversing correspondence between (irreducible) affine subvarieties of $k^n$ and prime ideals in $k[x_1,\ldots,x_n]$. So if $\mathfrak{m}$ is a maximal ideal, then it corresponds to a minimal irreducible closed subset of $k^n$ (since $\mathfrak{m}$ is prime and maximal), which must be a point $\{p = (p_1,\ldots,p_n)\}$. So $\mathfrak{m} = I(p)$, which one can see is just $\langle (x_1-p_1)\cdots(x_n-p_n) \rangle$. So $k[x_1,\ldots,x_n]/\mathfrak{m} \cong k$ via the "evaluate at $p$" map.

My question is this: Does the "usual version" follow from "version 2" and, if so, how?

Thanks.

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I'm pretty sure the notation I'm using is standard, but in case it isn't: if $T \subseteq k[x_1,...,x_n]$, then $Z(T) = \{ p \in k^n \ | \ f(p) = 0 \ \forall \ f \in T \}$. If $Y \subseteq k^n$, then $I(Y) = \{f \in k[x_1,...,x_n] \ | \ f(p) = 0 \ \forall p \in Y \}$. –  Matt Feb 7 '12 at 21:42

Then I grew up and, in particular, wrote some notes on commutative algebra. The Nullstellensatz is treated in some detail in $\S 11$. Hilbert's Nullstellensatz is Theorem 267, and you can see that part a) of that result is precisely the weak version you are asking about. Part c) of this result is the statement that for any ideal $J$ of $\overline{k}[t_1,\ldots,t_n]$, $I(V(J)) = \operatorname{rad} J$.
Denote the statement I call the "usual version" in my original post by "$u$", and parts a) and c) in your notes by $a$ and $c$. I mentioned $u \implies c$ (and can prove it), and I've shown $u \implies a$. Your notes show that $a \implies c$. I'd be happy if I could show $c \implies u$... Also, the wikipedia page I linked to in my original post has another statement which it calls the "weak" Nullstellensatz. Where does that fit into this? –  Matt Feb 7 '12 at 22:33
@Matt: what you call u is a restatement of the relation $I(V(J)) = \operatorname{rad} J$. –  Pete L. Clark Feb 7 '12 at 22:46
@Pete: Regarding the Nullstellensatz: if $S\!\subseteq\!K^n$, does $V(I(S))\!=\!\overline{S}$ hold always, or just when $K$ is algebraically closed? Could you please provide an example of $S$, such that $V(I(S))\!\supsetneq\!\overline{S}$? Possibly when $K=\mathbb{R}$. I can't think of an example. Do we need $n\!\geq\!2$? –  Leon Feb 28 '12 at 17:41