Corollary 9.3.8 (Analytic Nullstellensatz)BE77FEFA9B1C

Proof. Every element of $\sqrt I$ vanishes on $V(I)$. For the converse we first show that $\sqrt I$ is the intersection of the prime ideals containing $I$. A prime ideal containing $I$ contains $\sqrt I$. Let $f\notin \sqrt I$, so that the ideals containing $I$ and disjoint from $S=\{ 1,f,f^2,\ldots \} $ form a nonempty family; by proposition 9.2.4 it has a maximal element $\mathfrak p$, since otherwise there would be a strictly increasing infinite sequence of ideals. The ideal $\mathfrak p$ is prime: if $a,b\notin \mathfrak p$, then $\mathfrak p+(a)$ and $\mathfrak p+(b)$ meet $S$, say $f^ k=p_1+xa$ and $f^ l=p_2+yb$ with $p_1,p_2\in \mathfrak p$; multiplying, we see that $ab\in \mathfrak p$ would give $f^{k+l}\in \mathfrak p$, which is impossible. Thus $f$ lies outside some prime ideal containing $I$. Now let $g$ vanish on $V(I)$. For every prime ideal $J\supset I$ we have $V(J)\subset V(I)$, so $g$ vanishes on $V(J)$ and therefore $g\in J$, by the equality $I_{V(J),0}=J$ proved in theorem 9.3.4. Hence $g$ lies in the intersection of all these prime ideals, which is $\sqrt I$. $\square$