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


 * $$Z \left({J}\right)$$ is the zero-locus of the ideal $$J$$, defined by:
 * $$Z \left({J}\right) = \left\{{\left({z_1,\ldots, z_n}\right) \in k^n : \forall g \in I: g \left({z_1,\ldots, z_n}\right)}\right\}$$


 * $$\sqrt{J}$$ 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 $$a \in k^n$$.

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 a surjective morphism 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 $$\frac{k \left[{x_1,\ldots, x_n}\right]} {\mathfrak{m}}$$ 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 $$\frac{k \left[{x_1,\ldots, x_n}\right]}{\mathfrak{m}}$$ is a finite field extension of $$k$$.

Since $$k$$ is algebraically closed, there is an isomorphism of $$k$$-algebras:
 * $$\frac {k \left[{x_1,\ldots, x_n}\right]} {\mathfrak{m}} \to k$$

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.

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
 * $$\pi : A \to \frac{A}{J}$$

induces a bijection $$I \mapsto \pi^{-1}(I)$$ from the sets of radical, prime and maximal ideals of $$\frac{A}{J}$$ 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 :
 * $$A_f \cong \frac{A[x]}{(fx-1)}$$

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 $$f$$.

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 $$\bigcap_{a \in Z \left({J}\right)} \mathfrak{m}_a = \sqrt{J}$$.

Note now also that :
 * $$I \left({Z \left({J}\right)}\right) = \bigcap_{a \in Z \left({J}\right)} I \left({\left\{{a}\right\}}\right) = \bigcap_{a \in Z \left({J}\right)} \mathfrak{m}_a$$,

which implies the required result.

Nullstellensatz is German for zero locus theorem.