Note first that the operations $I \left({\cdot}\right)$ and $Z \left({\cdot}\right)$ are inclusion reversing. That is: :$X \subseteq Y \subseteq k^n \implies I \left({X}\right) \supseteq I \left({Y}\right)$ :$I \subseteq J \implies Z \left({I}\right) \supseteq Z \left({J}\right)$ {{explain|Link to a page where the above is demonstrated.}} 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: :$I \left({\left\{{a}\right\}}\right) = m_a$ {{explain|Link to a page where the above is demonstrated.}} 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. {{qed|lemma}} === 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 $J$. {{explain|Link to a page explaining the above.}} 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. {{handwaving|"it is clear that" -- needs a link to a page demonstrating it}} 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 $f$. {{qed|lemma}} === Step 3 === Note now that a point $a \in k^n$ belongs to $Z \left({J}\right)$ {{iff}} $J \subseteq m_a$. 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: {{begin-eqn}} {{eqn | l = I \left({Z \left({J}\right)}\right)       | r = \bigcap_{a \mathop \in Z \left({J}\right)} I \left({\left\{{a}\right\}}\right)       | c =  }} {{eqn | r = \bigcap_{a \mathop \in Z \left({J}\right)} m_a       | c =  }} {{end-eqn}} {{explain|justification for the above equation block}} which implies the required result. {{qed}} {{Namedfor|David Hilbert|cat = Hilbert}} '''Nullstellensatz''' is German for '''zero locus theorem'''. Category:Ideal Theory Category:Algebraic Geometry plzaiadqnelufoeyfzukib94qv6k4gi