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Radical of an ideal

From Wikipedia, the free encyclopedia

In commutative ring theory, a branch of mathematics, the radical of an ideal is an ideal such that an element is in the radical if and only if some power of is in (taking the radical is called radicalization). A radical ideal (or semiprime ideal) is an ideal that is equal to its own radical. The radical of a primary ideal is a prime ideal.

This concept is generalized to non-commutative rings in the Semiprime ring article.


The radical of an ideal in a commutative ring , denoted by or , is defined as

(note that ). Intuitively, is obtained by taking all roots of elements of within the ring . Equivalently, is the pre-image of the ideal of nilpotent elements (the nilradical) in the quotient ring (via the natural map ). The latter shows is itself an ideal.[Note 1]

If the radical of is finitely generated, then some power of is contained in .[1] In particular, if and are ideals of a noetherian ring, then and have the same radical if and only if contains some power of and contains some power of .

If an ideal coincides with its own radical, then is called a radical ideal or semiprime ideal.


  • Consider the ring of integers.
  1. The radical of the ideal of integer multiples of is .
  2. The radical of is .
  3. The radical of is .
  4. In general, the radical of is , where is the product of all distinct prime factors of , the largest square-free factor of (see radical of an integer). In fact, this generalizes to an arbitrary ideal (see the Properties section).
  • Consider the ideal It is trivial to show (using the basic property ), but we give some alternative methods.[clarification needed] The radical corresponds to the nilradical of the quotient ring which is the intersection of all prime ideals of the quotient ring. This is contained in the Jacobson radical, which is the intersection of all maximal ideals, which are the kernels of homomorphisms to fields. Any ring morphism must have in the kernel in order to have a well-defined morphism (if we said, for example, that the kernel should be the composition of would be which is the same as trying to force ). Since is algebraically closed, every morphism must factor through so we only have the compute the intersection of to compute the radical of We then find that


This section will continue the convention that I is an ideal of a commutative ring :

  • It is always true that , i.e. radicalization is an idempotent operation. Moreover, is the smallest radical ideal containing .
  • is the intersection of all the prime ideals of that contain

    and thus the radical of a prime ideal is equal to itself. Proof: On one hand, every prime ideal is radical, and so this intersection contains . Suppose is an element of which is not in , and let be the set By the definition of , must be disjoint from . is also multiplicatively closed. Thus, by a variant of Krull's theorem, there exists a prime ideal that contains and is still disjoint from (see prime ideal). Since contains , but not , this shows that is not in the intersection of prime ideals containing . This finishes the proof. The statement may be strengthened a bit: the radical of is the intersection of all prime ideals of that are minimal among those containing .
  • Specializing the last point, the nilradical (the set of all nilpotent elements) is equal to the intersection of all prime ideals of [Note 2]

    This property is seen to be equivalent to the former via the natural map which yields a bijection

    defined by [2][Note 3]
  • An ideal in a ring is radical if and only if the quotient ring is reduced.
  • The radical of a homogeneous ideal is homogeneous.
  • The radical of an intersection of ideals is equal to the intersection of their radicals: .
  • The radical of a primary ideal is prime. If the radical of an ideal is maximal, then is primary.[3]
  • If is an ideal, . Since prime ideals are radical ideals, for any prime ideal .
  • Let be ideals of a ring . If are comaximal, then are comaximal.[Note 4]
  • Let be a finitely generated module over a noetherian ring . Then

where is the support of and is the set of associated primes of .


The primary motivation in studying radicals is Hilbert's Nullstellensatz in commutative algebra. One version of this celebrated theorem states that for any ideal in the polynomial ring over an algebraically closed field , one has



Geometrically, this says that if a variety is cut out by the polynomial equations , then the only other polynomials which vanish on are those in the radical of the ideal .

Another way of putting it: the composition is a closure operator on the set of ideals of a ring.

See also


  1. ^ Here is a direct proof. Start with with some powers . To show that , we use the binomial theorem (which holds for any commutative ring):
    For each , we have either or . Thus, in each term , one of the exponents will be large enough to make that factor lie in . Since any element of times an element of lies in (as is an ideal), this term lies in . Hence , and . To finish checking that the radical is an ideal, take with , and any . Then , so . Thus the radical is an ideal.
  2. ^ For a proof, see the characterisation of the nilradical of a ring.
  3. ^ This fact is also known as fourth isomorphism theorem.
  4. ^ Proof: implies .


  1. ^ Atiyah–MacDonald 1969, Proposition 7.14
  2. ^ Aluffi, Paolo (2009). Algebra: Chapter 0. AMS. p. 142. ISBN 978-0-8218-4781-7.
  3. ^ Atiyah–MacDonald 1969, Proposition 4.2
  4. ^ Lang 2002, Ch X, Proposition 2.10


This page was last edited on 17 December 2020, at 13:51
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