Hilbert's Nullstellensatz

Theorem

Let $k$ be an algebraically closed field.

Then we have for every ideal $J \subseteq k \left[{x_1,\ldots, x_n}\right]$ that:

$I \left({Z \left({J}\right)}\right) = \sqrt{J}$

where:

• $I$ is the ideal associated with a subset of $k^n$ defined by:

$I \left({Z}\right) = \left\{{f \in k \left[{x_1, \ldots, x_n}\right]}:\forall a \in Z: f(a)=0\right\}$

• $Z \left({J}\right)$ is the zero-locus of the ideal $J$, defined by:

$Z \left({J}\right) = \left\{{\left({z_1,\ldots, z_n}\right) \in k^n : \forall g \in I: g \left({z_1,\ldots, z_n}\right)}=0\right\}$

• $\sqrt{J}$ denotes the radical of the ideal $J$.

Proof

Note first that the operations $I \left({\cdot}\right)$ and $Z \left({\cdot}\right)$ are order reversing.

That is, if $X \subseteq Y \subseteq k^n$, we find that $I \left({X}\right) \supseteq I \left({Y}\right)$ and if $I \subseteq J$ we have that $Z \left({I}\right) \supseteq Z \left({J}\right)$.

Let $\mathfrak{m}_a$ be the ideal $\left({x_1 - a_1, \ldots, x_n - a_n}\right)$ with $a \in k^n$.

Step one

Claim

$\mathfrak{m}_a$ are the only maximal ideals.

Proof

Let $a \in k^n$.

Define now:

$\pi_a : k \left[{x_1,\ldots, x_n}\right] \to k : f \mapsto f \left({a_1, \ldots, a_n}\right)$

and note that is an epimorphism of $k$-algebras with kernel:

$I \left({\left\{{a}\right\}}\right) = \mathfrak{m}_a$

Let now $\mathfrak{m}$ be a maximal ideal of $k \left[{x_1,\ldots, x_n}\right]$.

Now $\displaystyle \frac{k \left[{x_1,\ldots, x_n}\right]} {\mathfrak{m}}$ is a field extension of $k$, which is finitely generated as a $k$-algebra.

Hence by a corollary of the Noether Normalization Lemma, we find that $\displaystyle \frac{k \left[{x_1,\ldots, x_n}\right]}{\mathfrak{m}}$ is a finite field extension of $k$.

Since $k$ is algebraically closed, there is an isomorphism of $k$-algebras:

$\displaystyle \frac {k \left[{x_1,\ldots, x_n}\right]} {\mathfrak{m}} \to k$

Let $a_i$ denote the image $x_i$. Hence we find that $\mathfrak{m}_a \subseteq \mathfrak{m}$, which implies an equality since the first one is a maximal ideal.

$\Box$

Step two

Claim

The radical of an ideal $J$ in a finitely generated $k$-algebra $A$ is equal to the intersection of the maximal ideals that contain $J$.

Proof

Note that the projection morphism

$\pi : A \to \dfrac{A}{J}$

induces a bijection $I \mapsto \pi^{-1}(I)$ from the sets of radical, prime and maximal ideals of $\dfrac{A}{J}$ to the sets radical, prime and maximal ideals of $ A$ that contain $J$.

Hence we need to prove this only if $J = (0)$.

It is clear that $\sqrt{(0)}$ is contained in every maximal ideal.

Hence we need to prove that every element that is not in $\sqrt{(0)}$ is not contained in some maximal ideal.

Let $f $ be in $A$ such that it is not nilpotent, i.e. $f \not \in \sqrt{(0)}$.

Hence we find that :

$A_f \cong \dfrac{A[x]}{(fx-1)}$

is a non-trivial $k$-algebra, which thus has a maximal ideal $\mathfrak{M}$.

Consider now the morphism

$\phi : A \to A_f$

which is a morphism of finitely generated $k$-algebras.

Hence by a corollary of the Noether Normalization Lemma we find that $\phi^{-1}(\mathfrak{M})$ must also be maximal.

This is a maximal ideal $A$ that does not contain $f$.

$\Box$

Step three

Note now that a point $a \in k^n$ belongs to $Z \left({J}\right)$ if and only if $J\subseteq \mathfrak{m}_a$. This implies that the maximal ideals containing $J$ are just the maximal ideals $\mathfrak{m}_a$ with $a \in Z(J) $.

Step two then tells us that $\displaystyle \bigcap_{a \in Z \left({J}\right)} \mathfrak{m}_a = \sqrt{J}$.

Note now also that :

$\displaystyle I \left({Z \left({J}\right)}\right) = \bigcap_{a \in Z \left({J}\right)} I \left({\left\{{a}\right\}}\right) = \bigcap_{a \in Z \left({J}\right)} \mathfrak{m}_a$,

which implies the required result.

$\blacksquare$

Source of Name

This entry was named for David Hilbert.

Nullstellensatz is German for zero locus theorem.