The Hilbert Nullstellensatz

The Nullstellensatz ("zero-locus theorem") is the fundamental bridge between algebra (ideals in polynomial rings) and geometry (algebraic varieties), establishing that maximal ideals correspond to points.

Statement

Theorem5.4Hilbert's Nullstellensatz (weak form)

Let $k$ be an algebraically closed field. Every maximal ideal $\mathfrak{m}$ of $k[x_1, \ldots, x_n]$ has the form $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ for some $(a_1, \ldots, a_n) \in k^n$ .

Theorem5.5Hilbert's Nullstellensatz (strong form)

Let $k$ be algebraically closed and $I \subseteq k[x_1, \ldots, x_n]$ an ideal. Then:

$I(V(I)) = \sqrt{I},$

where $V(I) = \{(a_1, \ldots, a_n) \in k^n : f(a_1, \ldots, a_n) = 0 \text{ for all } f \in I\}$ is the zero set, $I(S) = \{f : f|_S = 0\}$ is the vanishing ideal, and $\sqrt{I} = \{f : f^m \in I \text{ for some } m \geq 1\}$ is the radical.

Proof of the Weak Form

Proof

Let $\mathfrak{m}$ be a maximal ideal of $R = k[x_1, \ldots, x_n]$ . Then $K = R/\mathfrak{m}$ is a field extension of $k$ , and $K$ is a finitely generated $k$ -algebra (generated by the images $\bar{x}_i$ ).

By Zariski's lemma (a consequence of the Noether normalization lemma): a field that is a finitely generated algebra over an algebraically closed field $k$ is equal to $k$ itself. So $K = k$ .

Therefore each $\bar{x}_i = a_i \in k$ , meaning $x_i - a_i \in \mathfrak{m}$ . The ideal $(x_1 - a_1, \ldots, x_n - a_n)$ is maximal (its quotient is $k$ ) and contained in $\mathfrak{m}$ , so $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ . $\blacksquare$

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Consequences

ExampleApplications

$V$ and $I$ are inverse operations on radical ideals and varieties: the maps $I \mapsto V(I)$ and $S \mapsto I(S)$ give a bijection between radical ideals of $k[x_1,\ldots,x_n]$ and algebraic subsets of $k^n$ .
$1 \in I$ iff $V(I) = \emptyset$ : If a system of polynomial equations has no solution (over an algebraically closed field), then $1$ is in the ideal generated by the polynomials.
Irreducible varieties correspond to prime ideals: $V(\mathfrak{p})$ is irreducible iff $\mathfrak{p}$ is prime.

RemarkThe algebra-geometry dictionary

The Nullstellensatz establishes the fundamental dictionary: | Algebra ( $k[x_1,\ldots,x_n]$ ) | Geometry ( $k^n$ ) | |---|---| | Radical ideal $\sqrt{I}$ | Algebraic set $V(I)$ | | Prime ideal | Irreducible variety | | Maximal ideal $(x_i - a_i)$ | Point $(a_1,\ldots,a_n)$ | | $R/I$ | Coordinate ring of $V(I)$ |