Let \(k\) be an algebraically closed field of characteristic zero and let \(X\subseteq\AA^n_k\). Show that \(I(X)\) is radical in \(\mco(\AA^n_k)=k[x_1,\ldots,x_n]\), and maximal if and only if \(X\) is a point.
Solution
Put \[ A=k[x_1,\ldots,x_n] \] and \[ I(X) = \{f\in A:f(a)=0\text{ for every }a\in X\}. \]
1. The ideal \(I(X)\) is radical.
Let \[ f^m\in I(X) \] for some integer \(m\geq1\). Then for every \[ a\in X \] one has \[ 0=f^m(a)=f(a)^m. \] Since \(k\) is a field, it has no nonzero nilpotents, so \[ f(a)=0 \] for every \(a\in X\). Hence \[ f\in I(X). \] Thus \[ \sqrt{I(X)}=I(X), \] so \(I(X)\) is radical.
2. If\[ X=\{a\}, \qquad a=(a_1,\ldots,a_n), \]then\[ I(X) = (x_1-a_1,\ldots,x_n-a_n) \]is maximal.
Evaluation at \(a\) defines a surjective homomorphism \[ \operatorname{ev}_a:A\longrightarrow k, \qquad f\longmapsto f(a). \] Its kernel is \(I(\{a\})\). Hence the first isomorphism theorem gives \[ A/I(\{a\})\cong k. \] Because the quotient is a field, \(I(\{a\})\) is maximal.
The kernel is explicitly \[ (x_1-a_1,\ldots,x_n-a_n), \] because quotienting by these generators sends every polynomial to its constant value at \(a\).
3. If \(I(X)\) is maximal, then \(X\) is a singleton.
Since a maximal ideal is proper, \[ I(X)\ne A. \] But \[ I(\varnothing)=A, \] so \(X\) is nonempty. Choose \[ a=(a_1,\ldots,a_n)\in X. \] Every polynomial vanishing on all of \(X\) vanishes at \(a\), hence \[ I(X)\subseteq I(\{a\}). \] By step 2, \(I(\{a\})\) is a proper ideal. Since \(I(X)\) is maximal, the inclusion forces \[ I(X)=I(\{a\}). \]
For each \(i\), \[ x_i-a_i\in I(\{a\})=I(X). \] Thus every \[ b=(b_1,\ldots,b_n)\in X \] satisfies \[ b_i-a_i=0 \] for every \(i\). Therefore \(b=a\), and hence \[ X=\{a\}. \]
4. Therefore\[ \boxed{ I(X)\text{ is maximal} \quad\Longleftrightarrow\quad X\text{ is a point}. } \]
Step 2 proves that a singleton has maximal vanishing ideal, and step 3 proves that a maximal vanishing ideal comes from a singleton.
5. Q.E.D.
Step 1 proves radicality, and step 4 proves the maximality characterization.