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The radical of an ideal $I$ isWe first establish the properties needed below. If $a^m,b^n\in I$, every term of the binomial theorem expansion of $(a+b{)}^{m+n-1}$ contains either $a^m$ or $b^n$, so $a+b\in \sqrt{I}$. Also $(-a{)}^m\in I$ and $(ra{)}^m=r^m{a}^m\in I$. Thus $\sqrt{I}$ is an ideal. The definition gives $I\subseteq \sqrt{I}$ and shows that $I\subseteq J$ implies $\sqrt{I}\subseteq \sqrt{J}$. Finally, if $a^n\in \sqrt{I}$, then $(a^n{)}^m\in I$ for some $m$, so $a\in \sqrt{I}$. Hence $\sqrt{\sqrt{I}}=\sqrt{I}$; in other words, $\sqrt{I}$ is a radical ideal.

We prove the stronger fact that the radical of an ideal is the intersection of the prime ideals containing it:If $a\in \sqrt{I}$ and $P\supseteq I$ is a prime ideal, then $a^n\in P$ for some $n$, and repeated application of the definition of a prime ideal gives $a\in P$. This proves one inclusion.

Conversely, fix $a\notin \sqrt{I}$. The multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. Consider the ideals containing $I$ and disjoint from $T$. This collection is nonempty, and the union of any chain in it is another such ideal. By Zorn's lemma, there is a maximal member $P$. It is proper because $1\in T$.

To prove that $P$ is a prime ideal, suppose $bc\in P$ but $b,c\notin P$. Maximality forces both $P+(b)$ and $P+(c)$ to meet $T$. Thus, for some $u,v\ge 0$,Multiplication gives $a^{u+v}\in P$, since every term lies in $P$, including $rsbc$. This contradicts disjointness from $T$. Therefore $P$ is a prime ideal containing $I$ and avoiding $a$. This proves the other inclusion, including the required prime ideal avoiding a multiplicative subset argument without using any unproved facts about the nilradical.

If $I=\sqrt{I}$, the displayed intersection represents $I$ as required. Conversely, an intersection of ideals that are all prime ideals is a radical ideal: if $a^n$ belongs to every one of them, then so does $a$. ConsequentlyThe unit ideal $I=R$ is included by interpreting the empty intersection as $R$.

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.

WriteThe equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the cuspidal cubicIt is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.

If $y=0$, the other equations becomeTheir solutions are $(0,0,0)$, which already lies on $C$, and the two pointsThus the irreducible components areEquivalently, the radical ideal of the cusp with two isolated-point components is

For an ideal $I$ in a commutative ring $R$, put $I^0=R$ and let $I^n$ for $n\ge 1$ be the product of ideals with $n$ factors equal to $I$. It consists of finite sums of products of $n$ elements of $I$. These ideals form a descending chain $R\supseteq I\supseteq I^2\supseteq \cdots$.

Every positive power has the same radical of an ideal as $I$:One inclusion follows from $I^n\subseteq I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, giving the other inclusion. Different ideals can therefore define the same radical ideal, even when their successive powers are all distinct.

For a finitely generated algebra over an algebraically closed field $k$, radical ideals correspond, with inclusion reversed, to closed subsets of the maximal spectrum. If $f\notin \sqrt{I}$, the nonzero algebra $(A/I)[1/f]$ is still finitely generated over $k$. Its maximal residue field is $k$ by the Zariski lemma, so contracting a maximal ideal gives a maximal ideal of $A$ containing $I$ and avoiding $f$. Hence the intersection of all maximal ideals containing $I$ is $\sqrt{I}$. This proves the inverse correspondence, including the empty closed set and unit ideal.

For any ideal $I$ in a commutative ring, every prime ideal containing $I$ contains $\sqrt{I}$, because $a^n\in P$ implies $a\in P$. Conversely, if $a\notin \sqrt{I}$, the multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. By Zorn's lemma, there is an ideal $P\supseteq I$ maximal among those disjoint from $T$.

This $P$ is a prime ideal: if $bc\in P$ but $b,c\notin P$, maximality makes both $P+(b)$ and $P+(c)$ meet $T$. Multiplication of representatives from these two intersections puts an element of $T$ in $P$, a contradiction. Thus there is a prime ideal containing $I$ but avoiding $a$, establishing the identity. For $I=R$, the intersection is empty and means $R$. It follows that an ideal is a radical ideal exactly when it is an intersection of ideals that are all prime ideals.