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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?

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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

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 $n$.

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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 $\emptyset \neq V(\mathfrak{m}) \subseteq V(I)$.

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