Hilbert's Nullstellensatz

Theorem
Let $k$ be an algebraically closed field.

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

Let $k \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.

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


 * $\map I {\map Z J} = \map \Rad J$

Proof
Note first that the operations $\map I {\, \cdot \,}$ and $\map Z {\, \cdot \,}$ are inclusion reversing.

That is:
 * $X \subseteq Y \subseteq k^n \implies \map I X \supseteq \map I Y$
 * $I \subseteq J \implies \map Z I \supseteq \map Z J$

Let $m_a$ be the ideal $\ideal {x_1 - a_1, \ldots, x_n - a_n}$ 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:
 * $\pi_a: k \sqbrk {x_1, \ldots, x_n} \to k: f \mapsto \map f {a_1, \ldots, a_n}$

and note that is an epimorphism of $k$-algebras with kernel:
 * $\map I {\set a} = m_a$

Let now $m$ be a maximal ideal of $k \sqbrk {x_1, \ldots, x_n}$.

Now $\dfrac {k \sqbrk {x_1, \ldots, x_n} } m$ is a field extension of $k$, which is finitely generated.

Hence by a corollary of the Noether Normalization Lemma, we find that $\dfrac {k \sqbrk {x_1, \ldots, x_n} } m$ is a finite field extension of $k$.

Since $k$ is algebraically closed, there is an isomorphism of $k$-algebras:
 * $\dfrac {k \sqbrk {x_1, \ldots, x_n} } m \to k$

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 $I \mapsto \map {\pi^{-1} } I$ 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 $J = \ideal 0$.

It is clear that $\map \Rad {\ideal 0}$ is contained in every maximal ideal.

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

Let $f \in A$ such that it is not nilpotent, that is:
 * $f \notin \map \Rad {\ideal 0}$

Hence:
 * $A_f \cong \dfrac {A \sqbrk x} {\paren {f x - 1} }$

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

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, $\map {\phi^{-1} } \MM$ 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 $\map Z J$ $J \subseteq m_a$.

This implies that the maximal ideals containing $J$ are just the maximal ideals $m_a$ with $a \in \map Z J$.

From Step 2:
 * $\ds \bigcap_{a \mathop \in \map Z J} m_a = \map \Rad J$

Note now also that:

which implies the required result.