Nullstellensatz
Idea
For a polynomial $f\in\mathbb{C}[x]$, we can split $f$ into linear factors. If you think about what this really means, it is saying that any finite collection of points, counting multiplicities, uniquely determines a polynomial up to scalar multiplication (e.g. write it as a product of linear factors).
This establishes a correspondence between algebraic sets (of $\mathbb{C}$) the ideal generated by a polynomial $f$ without multiple roots which has roots precisely those in the algebraic set. In particular, the ideal generated by such a polynomial includes the ones with multiple roots. We can say $$I(Z(f))=(f).$$
But what if we took $g$ to be a polynomial with roots the same as $f$ but gave $g$ a multiple root? Then the above is not true; $f\not\in (g)$, for example. It is true, however, that $$I(Z(g))=\text{rad}(g).$$
This establishes a bijection between the radical ideals of $\mathbb{C}[x]$ and the algebraic subsets of $\mathbb{C}$. The roots of a polynomial (the algebraic subsets) are often the most important quality of a polynomial we are interested in, so this gives us a correspondence between the geometric class of polynomials and the algebraic class of radical ideals in $\mathbb{C}[x]$.
Hilbert’s Nullstellensatz generalizes this statement to arbitrary algebraically closed fields and polynomials of multiple variables over them.
Statement
Let $k$ be an algebraically closed field.
Theorem. If $I\subset k[x_1,\dots,x_n]$ is an ideal, then $$I(Z(I))=\text{rad}(I).$$ Thus, the correspondence $I\mapsto Z(I)$ and $X\mapsto I(X)$ induce a bijection between the algebraic subsets of $\mathbb{A}^n_k$ and the radical ideals of $k[x_1,\dots,x_n]$.
Projective version
For all $d$, let $$I_d=(x_0,\dots,x_n)^d.$$ Notice that $Z(I)=\emptyset$ since 0 is not a point in projective space.
Theorem. If $I\subset k[x_0,\dots,x_n]$ is a homogeneous ideal, then
- If $I\supset I_d$ then $V(I)=\emptyset$.
- Otherwise, $Z(I)\neq \emptyset$ and $I(Z(I))=\text{rad}(I)$.