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What is Hilbert's Nullstellensatz in the sense of the generalization of "fundamental theorem of algebra"?

I've seen that in some texts it was referred to as the generalization of the fundamental theorem of algebra in several variables.

How exactly does Hilbert's Nullstellensatz relate to the fundamental theorem of algebra? Also could you please provide some examples to show that it is related?

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I find it strange to say that the Nullstellensatz is a generalization of the fundamental theorem. To prove the Nullstellensatz you need the field to be algebraically closed, so you have to first prove the fundamental theorem, before you can apply the Nullstellensatz to $\mathbb C$. – Fredrik Meyer Apr 1 '14 at 11:47
up vote 15 down vote accepted

Let's prove that the Nullstellensatz implies the fundamental theorem of algebra in the 1D case.

Let $p \in \Bbb C[z]$. The Nullstellensatz says that if we have another polynomial $f \in \Bbb C[z]$, such that $f$ has the same zeroes as some $g \in \langle p \rangle$, then $f^r \in \langle p \rangle$ for some $r \in \Bbb N$.

Now assume there exists a polynomial $p \in \Bbb C[z]$ that has no zeroes. Clearly the polynomial $1$ has the same zero set (the empty set!); the Nullstellensatz says that $1 = 1^r \in \langle p \rangle$ for some $r \in \Bbb N$. Since $1 \in \langle p \rangle$, $p$ is constant.

Thus by contraposition every nonconstant polynomial in $\Bbb C[z]$ has a root.

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What about the application to the several variable cases? – Victor Apr 1 '14 at 2:55
@Victor The fundamental theorem of algebra doesn't apply in that case. That's why this is a "generalization": the Nullstellensatz is equivalent to the FTA in the 1D case (prove the other direction!) so one can think of the Nullstellensatz in several variables as a "several variable" FTA. – Mike Miller Apr 1 '14 at 2:58
It is usually the "weak Nullstellensatz" that is referred to as a multivariable analogue of FTA. For $k$ an algebraically closed field the "weak Nullstellensatz" can be stated as: "For any proper ideal $I$ in $k\left[X_1, X_2, \ldots, X_n\right]$, $V(I)$ is non-empty." Since ideals in $k[X]$ are principle, the $1D$ analogue contains precisely the same content as FTA. A proper ideal $I$ in $k[X]$ is given by a non-constant polynomial with zeros $V(I)$. – Marcel T. May 9 '14 at 6:18
Doesn't using the Nullstellensatz here already rely on $\mathbb{C}$ being algebraically closed, hence making this a circular argument? Essentially what you've proved is that non-constant polynomials in algebraically closed fields have roots. – jaakhaamer Jun 2 '15 at 15:56
@jaakhaamer: The point was to show that the Nullstellensatz implies that nonconstant polynomials in fields where this form of the Nullstellensatz is true (algebraically closed fields) have roots. I'm not assuming that we can actually prove the Nullstellensatz, only showing that it being true implies FTA over our field. – Mike Miller Jun 2 '15 at 16:08

The fundamental theorem of algebra says that a non-constant polynomial over an algebraically closed field has a root.

Note that $f(x) \in \mathbb C[x]$ is a non-zero constant if and only if the ideal $\bigl(f(x)\bigr)$ is the unit ideal. Since any ideal in $\mathbb C[x]$ is principal, another way to phrase the FTA is as follows: if $I \subset \mathbb C[x]$ is a non-unit ideal, then there exists $z \in \mathbb C$ such that $f(z) = 0$ for all $f(x) \in I$. (The converse also holds, more or less obviously.)

The Nullstellensatz says the same thing for an ideal in $\mathbb C[x_1,\ldots,x_n]$, namely: an ideal $I$ in this ring is non-unit if and only if there exists $(z_1,\ldots,z_n)$ such that $f(z_1,\ldots,z_n) = 0$ for all $f \in I$. (Again, the if direction is obvious, but the only if direction is non-trivial --- although actually, in the case of when the field is $\mathbb C$, it is not so difficult to prove, because $\mathbb C$ is so much bigger than its prime subfield $\mathbb Q$.)

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