Radical of an ideal


In ring theory, a branch of mathematics, the radical of an ideal of a commutative ring is another ideal defined by the property that an element is in the radical if and only if some power of is in . Taking the radical of an ideal is called radicalization. A radical ideal (or semiprime ideal) is an ideal that is equal to its 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 preimage of the ideal of nilpotent elements (the nilradical) of the quotient ring   (via the natural map  ). The latter proves that   is 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   (the evens).
    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 homomorphism   must have   in the kernel in order to have a well-defined homomorphism (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 homomorphism   must factor through  , so we only have to 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   that 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[4]  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 that 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 that   is an ideal. 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 so  . 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 1994, Proposition 7.14
  2. ^ Aluffi, Paolo (2009). Algebra: Chapter 0. AMS. p. 142. ISBN 978-0-8218-4781-7.
  3. ^ Atiyah & Macdonald 1994, Proposition 4.2
  4. ^ Lang 2002, Ch X, Proposition 2.10