The radical of an ideal in a commutative ring, denoted by or, is defined as . Intuitively, is obtained by taking all roots of elements of within the ring. Equivalently, is the pre-image of the ideal of nilpotent elements in the quotient ring . The latter shows is itself an ideal. 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.
The radical of the ideal of integer multiples of is .
The radical of is.
The radical of is.
In general, the radical of ' is ', where is the product of all distinct prime factors of, the largest square-free factor of . In fact, this generalizes to an arbitrary ideal.
Consider the ideal It is trivial to show , but we give some alternative methods. 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. 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
Properties
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 . 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 is equal to the intersection of all prime ideals of This property is seen to be equivalent to the former via the natural map which yields a bijection defined by
An ideal in a ring is radical if and only if the quotient ring is reduced.
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 where and 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.
