Radical of an ideal
In commutative algebra, the radical of an ideal I in a commutative ring R with identity, denoted \sqrt{I} or \mathrm{rad}(I), is defined as the set of all elements x \in R such that x^n \in I for some positive integer n \geq 1.[1][2] This construction yields an ideal that contains I and is the smallest ideal with this property that equals its own radical, meaning an ideal J is radical if J = \sqrt{J}.[1][3] The radical possesses several key algebraic properties that underscore its importance. Notably, \sqrt{I} equals the intersection of all prime ideals of R containing I, providing a prime-filtering characterization.[2][1] It satisfies \sqrt{\sqrt{I}} = \sqrt{I}, ensuring idempotence under the radical operation, and for ideals I and J, \sqrt{IJ} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}, which facilitates computations in ideal operations.[2][1] In a unique factorization domain, the radical of a principal ideal generated by a product of irreducibles simplifies to the ideal generated by their product without exponents.[1] The nilradical, specifically \sqrt{0}, consists of all nilpotent elements in R, forming the prime ideals' intersection over the zero ideal.[2] Beyond pure algebra, the radical bridges to algebraic geometry via Hilbert's Nullstellensatz. For a finitely generated algebra over an algebraically closed field, the radical \sqrt{I} determines the same affine variety V(I) as I, since V(\sqrt{I}) = V(I), and the ideal of polynomials vanishing on a variety is always radical.[3] This correspondence establishes a bijection between radical ideals and algebraic sets, foundational for studying solution sets of polynomial systems. Primary ideals have radicals that are prime, linking decomposition theory to geometric irreducibility.[1] These aspects highlight the radical's role in both theoretical structures and applied contexts like scheme theory.Definition and Fundamentals
Formal Definition
In a commutative ring R with identity and an ideal I \subseteq R, the radical of I, denoted \sqrt{I}, is defined as the set \sqrt{I} = \{ r \in R \mid \exists n \geq 1 \text{ such that } r^n \in I \}. [4] This set consists of all elements of R whose some positive power lies in I. The radical arises naturally as the preimage under the quotient map R \to [R/I](/page/Quotient_ring) of the nilradical of the quotient ring R/I, where the nilradical is the set of nilpotent elements in R/I.[5] Thus, \sqrt{I} captures precisely those elements of R that become nilpotent modulo I. To verify that \sqrt{I} is itself an ideal, first note that I \subseteq \sqrt{I} since if r \in I, then r^1 = r \in I. For closure under multiplication by elements of R, let r \in \sqrt{I} and s \in R. There exists n \geq 1 such that r^n \in I, so (sr)^n = s^n r^n \in I because I absorbs multiplication by elements of R. Hence, sr \in \sqrt{I}. For closure under addition, let r, t \in \sqrt{I}, so there exist n, m \geq 1 with r^n \in I and t^m \in I. Set k = nm. By the binomial theorem, (r + t)^{2k} = \sum_{i=0}^{2k} \binom{2k}{i} r^i t^{2k - i}. For each term, if i \geq n, then r^i = r^{i - n} r^n \in I (since r^{i - n} \in R); similarly, if $2k - i \geq m, then t^{2k - i} \in I. Since k = nm \geq n, m, in all cases at least one factor sends the term into I, so each term is in I. As I is closed under addition, (r + t)^{2k} \in I, whence r + t \in \sqrt{I}. The additive inverse follows similarly, confirming \sqrt{I} is an ideal.[6] Finally, \sqrt{I} is the smallest radical ideal containing I, in the sense that if J is any ideal with I \subseteq J and \sqrt{J} = J, then \sqrt{I} \subseteq J. This follows because \sqrt{I} equals the intersection of all prime ideals containing I, and prime ideals are radical, so their intersection is radical and minimal with respect to containing I.[4]Basic Properties
The radical \sqrt{I} of an ideal I in a commutative ring R always contains the ideal itself, that is, I \subseteq \sqrt{I}. This follows directly from the definition, as every element of I satisfies the condition that its first power lies in I. Equality holds if and only if I is a radical ideal, meaning no element outside I has a power in I.[7] The radical operation preserves inclusions of ideals: if J \subseteq I, then \sqrt{J} \subseteq \sqrt{I}. To see this, suppose x \in \sqrt{J}, so x^n \in J \subseteq I for some positive integer n; thus x \in \sqrt{I}. This monotonicity ensures that the radical enlarges smaller ideals in a controlled manner.[8] For sums of ideals, the radical satisfies \sqrt{I} + \sqrt{J} \subseteq \sqrt{I + J}. If a \in \sqrt{I} and b \in \sqrt{J}, then a^m \in I and b^n \in J for some m, n \geq 1; raising a + b to the power k = m + n - 1 yields (a + b)^k \in I + J by the binomial theorem, as each term involves either at least m factors of a (landing in I) or at least n factors of b (landing in J). More generally, \sqrt{I + J} = \sqrt{\sqrt{I} + \sqrt{J}}, reflecting the stability of the radical under such operations.[7] The radical preserves finite intersections: for a finite family of ideals I_1, \dots, I_r, \sqrt{\bigcap_{i=1}^r I_i} = \bigcap_{i=1}^r \sqrt{I_i}. The inclusion \sqrt{\bigcap I_i} \subseteq \bigcap \sqrt{I_i} holds for arbitrary families, since if x^n \in \bigcap I_i, then x^n \in I_i for each i, so x \in \sqrt{I_i} for each i. For the reverse inclusion in the finite case, suppose x \in \bigcap \sqrt{I_i}, so for each i there exists n_i \geq 1 with x^{n_i} \in I_i; let n be the least common multiple of the n_i, then x^n = (x^{n_i})^{n/n_i} \in I_i for each i (since I_i is an ideal), so x^n \in \bigcap I_i and thus x \in \sqrt{\bigcap I_i}.[7] In the case of a principal ideal (a) generated by a single element a \in [R](/page/R), the radical consists of all b \in [R](/page/R) such that b^k = a c for some c \in [R](/page/R) and integer k \geq 1. This explicitly captures elements whose powers are multiples of a, illustrating how the radical "fills in" nilpotent-like behaviors relative to the generator.[7]Examples
In Principal Ideal Domains
In principal ideal domains (PIDs), such as the ring of integers \mathbb{Z} or the polynomial ring k over a field k, every ideal is principal, and the radical of a principal ideal (f) can be explicitly computed using the prime or irreducible factorization of the generator f. The radical \sqrt{(f)} is the principal ideal generated by the square-free part of f, which is the product of its distinct irreducible factors (each taken to the first power). This follows from the fact that in a PID, the radical is the intersection of the minimal prime ideals containing (f), and these primes correspond exactly to the distinct irreducible factors of f.[9] Consider the case of \mathbb{Z}. For the principal ideal n\mathbb{Z} where n \geq 0 has prime factorization n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r} with distinct primes p_i and exponents a_i \geq 1, the radical is \sqrt{n\mathbb{Z}} = (p_1 p_2 \cdots p_r) \mathbb{Z}. For example, take n = 12 = 2^2 \cdot 3; then \sqrt{12\mathbb{Z}} = 6\mathbb{Z}, since $6^2 = 36 \in 12\mathbb{Z} but $2 \notin \sqrt{12\mathbb{Z}} (no positive power of 2 is a multiple of 3, as 3 does not divide any $2^k).[9] Another illustration is \sqrt{4\mathbb{Z}} = 2\mathbb{Z}, where $4 = 2^2, demonstrating strict containment $4\mathbb{Z} \subsetneq 2\mathbb{Z} (consistent with the general property I \subseteq \sqrt{I} for any ideal I).[9] In the polynomial ring k over a field k, the situation is analogous. If f \in k factors as f = c \cdot g_1^{e_1} g_2^{e_2} \cdots g_s^{e_s} where c \in k^\times is the leading coefficient and the g_i are distinct monic irreducible polynomials with exponents e_i \geq 1, then \sqrt{f k} = (g_1 g_2 \cdots g_s) k, the ideal generated by the square-free part of f. This extracts the product of all distinct roots of f (counting multiplicity at least once), effectively removing higher powers in the factorization. For instance, if f(x) = (x-1)^2 (x-2), then \sqrt{f k} = ((x-1)(x-2)) k.[10] To compute the radical in these PIDs, factorize the generator into irreducibles (primes in \mathbb{Z}, irreducibles in k) and form the product of the distinct factors; this yields the square-free kernel directly. In k, the square-free part can also be obtained via the gcd process: iteratively compute \gcd(f, f') to isolate square-free components, though full factorization provides the explicit generator.[10] These methods highlight the computational simplicity in univariate PIDs compared to higher dimensions.In Polynomial Rings
In multivariate polynomial rings over a field k, the computation of the radical of an ideal generally requires finding its primary decomposition, as the radical is the intersection of the associated prime ideals; this contrasts with univariate cases, where principal ideal domains allow simpler direct determination. For instance, consider the ideal (y^4) in k[x, y]. This is a primary ideal with associated prime (y), so its radical is (y). To see this, note that y^4 \in (y^4) implies y \in \sqrt{(y^4)} by the definition of the radical, and more generally, any f \in \sqrt{(y^4)} satisfies f^n \in (y^4) for some n, forcing f to lie in (y) since the ring is a domain.[11] Another example is the ideal (x^2, xy) in k[x, y]. Its primary decomposition is (x) \cap (x^2, y), where (x) is prime (hence primary) and (x^2, y) is primary with radical (x, y). The minimal associated prime is (x), so \sqrt{(x^2, xy)} = (x). This follows because x^2 \in (x^2, xy) implies x \in \sqrt{(x^2, xy)}, and the decomposition confirms no other minimal primes.[11] Non-principal radicals arise naturally in such rings; for example, in \mathbb{C}[x, y], the ideal (xy) equals (x) \cap (y), the intersection of its minimal primes, and since the generator is square-free, (xy) is already radical: \sqrt{(xy)} = (xy). In contrast, for (x^2 y), the primary decomposition is (x^2) \cap (y), with associated primes (x) (from the (x)-primary component (x^2)) and (y). Thus, \sqrt{(x^2 y)} = (x) \cap (y) = (xy). For example, (xy)^2 = x^2 y^2 = y \cdot (x^2 y) \in (x^2 y), confirming xy \in \sqrt{(x^2 y)}, while powers of x or y alone are not in the ideal. The primary decomposition establishes the associated primes.[11] To compute radicals in polynomial rings, one effective method uses primary decomposition via Gröbner bases, which allows algorithmic determination of associated primes even for non-monomial ideals.Advanced Properties
Relation to Prime and Primary Ideals
The radical of an ideal I in a commutative ring R is equal to the intersection of all prime ideals of R containing I, that is, \sqrt{I} = \bigcap \{ P \mid P \text{ prime ideal of } R, \, I \subseteq P \}. [12] One inclusion follows directly from the definition: if f \in \sqrt{I}, then f^n \in I \subseteq P for every such prime P, so f \in P because P is prime. For the reverse inclusion, suppose f \notin \sqrt{I}. Then no power of f lies in I, so the collection of ideals containing I but avoiding all powers of f is nonempty (containing I itself) and inductive under union. By Zorn's lemma, it admits a maximal element M, which can be shown to be prime. This M contains I but no power of f, hence f \notin M. Thus, f fails to lie in every prime containing I.[12] If Q is a primary ideal of R, then \sqrt{Q} is a prime ideal, known as the associated prime of Q. To see this, suppose ab \in \sqrt{Q}. Then (ab)^n \in Q for some n \geq 1, so a^n b^n \in Q. By primarity of Q, either a^n \in Q (hence a \in \sqrt{Q}) or some power of b^n lies in Q (hence b \in \sqrt{Q}). Thus, \sqrt{Q} satisfies the prime condition.[13] In a Noetherian ring, \sqrt{I} equals the intersection of the minimal prime ideals containing I. This follows from the primary decomposition theorem: I decomposes as a finite intersection of primary ideals Q_i with associated primes P_i = \sqrt{Q_i}, where the minimal P_i over I determine \sqrt{I}, as any embedded primes contain a minimal one.[14] More generally, for any ideal I in a commutative ring R, the radical \sqrt{I} is the intersection of the associated prime ideals of the module R/I, that is, \sqrt{I} = \bigcap_{P \in \mathrm{Ass}(R/I)} P. The associated primes \mathrm{Ass}(R/I) arise as the radicals of the primary components in any primary decomposition of I, and since embedded associated primes contain minimal ones, the intersection over all associated primes coincides with that over the minimal ones containing I.[14]Idempotence and Radical Ideals
The radical operation on ideals in a commutative ring exhibits idempotence, meaning that for any ideal I, \sqrt{\sqrt{I}} = \sqrt{I}.[15] This property follows from the characterization of the radical as the intersection of all prime ideals containing I, as previously discussed; the prime ideals containing \sqrt{I} are precisely those containing I, yielding the same intersection.[15] A direct verification proceeds as follows: if x \in \sqrt{\sqrt{I}}, then x^n \in \sqrt{I} for some n \geq 1, so (x^n)^m \in I for some m \geq 1, implying x^{nm} \in I and thus x \in \sqrt{I}.[15] An ideal J is called a radical ideal (or semiprime ideal) if \sqrt{J} = J.[16] Equivalently, a radical ideal is the intersection of prime ideals.[15] For such an ideal J, the quotient ring R/J is reduced, meaning its nilradical is zero: \text{nil}(R/J) = 0.[15] The radical map I \mapsto \sqrt{I} forms a closure operator on the lattice of ideals in a commutative ring, satisfying three key properties: it is extensive (I \subseteq \sqrt{I}), monotonic (if I \subseteq K, then \sqrt{I} \subseteq \sqrt{K}), and idempotent (\sqrt{\sqrt{I}} = \sqrt{I}).[16] These ensure that \sqrt{I} is the smallest radical ideal containing I.[16]Applications
In Algebraic Geometry
In algebraic geometry over an algebraically closed field k, Hilbert's Nullstellensatz establishes a profound connection between the radical of an ideal and the geometry of algebraic varieties. Specifically, for an ideal I in the polynomial ring k[x_1, \dots, x_n], the radical \sqrt{I} equals the ideal I(V(I)) consisting of all polynomials that vanish on the variety V(I) defined by I.[17] This equivalence implies that the zero set of an ideal depends only on its radical, as V(I) = V(\sqrt{I}).[17] The weak form of the Nullstellensatz complements this by stating that if V(I) = \emptyset, then I = k[x_1, \dots, x_n], guaranteeing that proper ideals yield nonempty varieties.[17] Geometrically, radical ideals encode reduced structures in algebraic varieties, meaning schemes without nilpotent elements in their coordinate rings, which represent points with no infinitesimal thickenings.[18] This captures the essential zero locus of the ideal, extracting its "square-free" part that defines the variety purely in terms of its support, free from multiplicities or embedded components. For instance, in a polynomial ring over k, the radical of the ideal generated by (f)^m for an irreducible polynomial f and integer m > 1 is simply (f), corresponding to the hypersurface variety V(f) without multiplicity.[17] In the broader context of scheme theory, the construction \operatorname{Spec}(R / \sqrt{I}) yields the reduced scheme associated to \operatorname{Spec}(R / I), preserving the topological space while eliminating all nilpotents from the structure sheaf.[19] This reduction process highlights how the radical ideal defines the underlying reduced geometry, aligning affine schemes with classical varieties.[18]In Commutative Algebra
In commutative algebra, the radical of an ideal I in a ring A, denoted \sqrt{I}, is intimately connected to primary decomposition theory. If I admits a primary decomposition I = \bigcap_{i=1}^n Q_i, where each Q_i is primary with associated prime P_i = \sqrt{Q_i}, then \sqrt{I} = \bigcap_{i=1}^n P_i. This intersection consists precisely of the associated prime ideals of I, simplifying the structure by reducing the radical to the primes minimal over I.[20] In Noetherian rings, where every ideal has a finite primary decomposition, this implies that \sqrt{I} is the intersection of finitely many prime ideals. Specifically, the minimal primes over I determine \sqrt{I}, and any embedded primes do not contribute to the radical. For a radical ideal itself, the primary decomposition collapses to an irredundant intersection of distinct minimal prime ideals, ensuring no embedded components.[20] The Artin-Rees lemma further underscores the role of radical ideals in module theory over Noetherian rings. For finitely generated modules M and submodule N \subseteq M, and ideal I \subseteq A, there exists an integer k \geq 0 such that I^n \cap N = I^{n-k} (I^k \cap N) for all n \geq k. This stability property aids in analyzing the support of modules, where \operatorname{Supp}(M) = \{ P \text{ prime} \mid M_P \neq 0 \} = V(\sqrt{\operatorname{Ann}(M)}), showing that radical ideals precisely capture the geometric support via the vanishing set.[21] In computational contexts, testing whether an ideal I is radical can be performed using Gröbner bases. Compute a Gröbner basis G of I; if the initial monomial ideal \langle \operatorname{in}(g) \mid g \in G \rangle is square-free (generated by square-free monomials), then I is radical. Radical ideals are idempotent, aiding their identification in decompositions.[22]Generalizations and Variants
Nilradical as a Special Case
The nilradical of a commutative ring R, denoted \mathfrak{n} or \sqrt{0}, is the radical of the zero ideal and serves as the prime example of a radical ideal. It consists of all nilpotent elements in R, that is, \mathfrak{n} = \{ r \in R \mid r^n = 0 \text{ for some positive [integer](/page/Integer) } n \}. This set forms an ideal, and equivalently, \mathfrak{n} is the intersection of all prime ideals of R.[23][24] A key property of the nilradical is that it vanishes in reduced rings, where R has no nonzero nilpotent elements, so \mathfrak{n} = 0. In Artinian commutative rings, the nilradical coincides with the Jacobson radical (the intersection of all maximal ideals) and is nilpotent, meaning some power of it is zero.[25][26] For computation, consider polynomial rings: if R is reduced, then the polynomial ring R is also reduced, so its nilradical is zero. More generally, the nilradical of R comprises all polynomials whose coefficients lie in \mathfrak{n}, the nilradical of R.[23] The quotient ring R / \mathfrak{n} is reduced, as the nilradical precisely captures and eliminates all nilpotent elements.[25]Jacobson Radical
The Jacobson radical of a commutative ring R with identity, denoted J(R), is defined as the intersection of all maximal ideals of R:J(R) = \bigcap \{ M \mid M \text{ is a maximal ideal of } R \}.
This ideal consists precisely of those elements r \in R such that $1 - r s is a unit in R for every s \in R.[27][24] For a proper ideal I of R, the Jacobson radical of I, denoted \mathrm{jac}(I), is the intersection of all maximal ideals containing I:
\mathrm{jac}(I) = \bigcap \{ M \mid I \subseteq M \text{ and } M \text{ is maximal in } R \}.
By the correspondence theorem for ideals, \mathrm{jac}(I) is the preimage of J(R/I) under the canonical quotient map R \to R/I, so \mathrm{jac}(I)/I \cong J(R/I).[28] In commutative rings, the Jacobson radical J(R) properly contains the nilradical \mathfrak{n}(R) in general, though equality holds in specific cases such as Artinian rings.[29] For local rings—those with a unique maximal ideal m—the Jacobson radical coincides with m, so J(R) = m.[27][30] A fundamental property is that the quotient ring R/J(R) is Jacobson semisimple, meaning J(R/J(R)) = 0. In the commutative case, R/J(R) embeds as a subdirect product of fields, reflecting the structure of semisimple commutative rings.[29][30]