Suppose we have a maximal ideal $M\subseteq K[X_1, \dots, X_n]$, where $K$ is an abitrary field. We denote the zero locus of $M$ (over the algebraic closure!) by $\mathbf V(M)\subseteq \overline K^n$. Let $J\subseteq K[X_1, \dots, X_n]$ be an ideal such that $\mathbf V(J)\subsetneqq \mathbf V(M)$.
Is it true that $\mathbf V(J)=\emptyset$?
In the case where $K$ is algebraically closed, this would obviously be true. (By applying the Nullstellensatz, we would get $M=\sqrt M\subsetneqq \sqrt J$, thus $\sqrt J=K[X_1, \dots, X_n]$ as $M$ is maximal and therefore $\mathbf V(J)=\mathbf V(\sqrt J) = \emptyset$.)
But what about the general case where $K$ is not algebraically closed?
Best Answer
Let $p \in V (M)$. Then $M \subseteq \ker \operatorname{ev}_p$, where $\operatorname{ev}_p : K [X_1, \ldots, X_n] \to \bar{K}$ is evaluation at $p$. Since $M$ is maximal and $\operatorname{ev}_p$ is not constant $0$, we must in fact have $M = \ker \operatorname{ev}_p$.
On the other hand, since $V (J) \subsetneqq V (M)$, there is some $p$ such that $p \in V (M)$ and $p \notin V (J)$. For such $p$, we can find $f \in J$ such that $\operatorname{ev}_p (f) \ne 0$, which implies $f \notin M$. But $J \subseteq I (V (J))$, so $f \in I (V (J))$, and $I (V (J)) \supseteq I (V (M)) = M$, so $I (V (J))$ must be the unit ideal. Hence, $V (J) = V (I (V (J))) = \emptyset$.