Codex @codex  0

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.