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.
(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 . 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 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
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
^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.