Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across an interesting problem that basically said something along the lines of ``Show that Hilbert's Nullstellensatz is equivalent to the Fundamental Theorem of Algebra.'' My algebraic geometry is a bit weak, but I was always under the impression that the Nullstellensatz was basically a multi-dimensional form of the fundamental theorem of algebra. I am correct, or am I missing something and need to take a closer look?

share|improve this question
Assume the Nullstellensatz. Let $f(x) \in \Bbb{C}[x]$. Then $V(f) \neq \emptyset$. For if $V(f) = \emptyset$ then $IV(f) = I(\emptyset) = \operatorname{rad}(f)$. Since $I(\emptyset) = \Bbb{C}[x]$ it follows that $\operatorname{rad} (f) = \Bbb{C}[x]$ which is a contradiction. –  user38268 Mar 14 '13 at 2:19
Nice done, @BenjaLim...Perhaps it's only needed to add $\,\deg f\ge 1\,$ –  DonAntonio Mar 14 '13 at 2:42
So it sounds like this was just a stupid question to ask. –  Brent J Mar 14 '13 at 2:57
@DonAntonio Yes I should have added that. –  user38268 Mar 14 '13 at 3:08
I don't think so, @BrentJ ... –  DonAntonio Mar 14 '13 at 3:09

2 Answers 2

Probably the following is meant: If $k$ is a field, then the statement of Hilbert's Nullstellensatz holds for $k[x]$ if and only if $k$ is algebraically closed. This is an easy observation. But it is well-known that then it already holds for $k[x_1,\dotsc,x_n]$ for arbitrary $n$.

share|improve this answer

Yeah, although it might be worth noting that you don't really need the "full force" formulation of the Nullstellensatz. It is perhaps more accurate (perhaps a little pedantic, of course) to say it is most closely the weak Nullstellensatz, because a nontrivial polynomial gives rise to an ideal $I$ that sits under a maximal ideal $\mathfrak{m}$. The weak Nullstellensatz tells us exactly what these are (the points on the variety, in our setting), whence $\emptyset \neq V(\mathfrak{m}) \subseteq V(I)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.