Hilbert's Nullstellensatz

Theorem
Let $k$ be an algebraically closed field.

Let $n \geq 0$ be an natural number.

Let $k \left[{x_1,\ldots, x_n}\right]$ be the polynomial ring in $n$ variables over $k$.

Then for every ideal $J \subseteq k \left[{x_1,\ldots, x_n}\right]$, the associated ideal of its zero-locus equals its radical:


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

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

That is:
 * $X \subseteq Y \subseteq k^n \implies I \left({X}\right) \supseteq I \left({Y}\right)$
 * $I \subseteq J \implies Z \left({I}\right) \supseteq Z \left({J}\right)$

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

Step 1: Maximal Ideals
It is demonstrated that $m_a$ are the only maximal ideals.

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:
 * $I \left({\left\{{a}\right\}}\right) = m_a$

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

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

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

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

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

Step 2: Radical is Intersection of Maximum Ideals
It is to be demonstrated that the radical of an ideal $J$ in a finitely generated $k$-algebra $A$ is equal to the intersection of the maximal ideals that contain $J$.

Note that the projection morphism:
 * $\pi: A \to \dfrac A J$

induces a bijection $I \mapsto \pi^{-1} \left({I}\right)$ 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 $J = \left({0}\right)$.

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

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

Let $f \in A$ such that it is not nilpotent, that is:
 * $f \notin \sqrt{\left({0}\right)}$

Hence:
 * $A_f \cong \dfrac {A \left[{x}\right]} { \left({f x - 1}\right)}$

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

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, $\phi^{-1} \left({\mathcal M}\right)$ must also be maximal.

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

Step 3
Note now that a point $a \in k^n$ belongs to $Z \left({J}\right)$ $J \subseteq m_a$.

This implies that the maximal ideals containing $J$ are just the maximal ideals $m_a$ with $a \in Z \left({J}\right)$.

From Step 2:
 * $\displaystyle \bigcap_{a \mathop \in Z \left({J}\right)} m_a = \sqrt J$

Note now also that:

which implies the required result.

Nullstellensatz is German for zero locus theorem.