How does Hilbert's Nullstellensatz generalize the "fundamental theorem of algebra"? 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?
 A: 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.
A: 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 $\mathbb Q$.)
