Radical of an ideal
From Academic Kids

In ring theory, a branch of mathematics, the radical of a ring isolates certain bad properties of the ring. There are several different kinds of radicals, such as the nilradical and the Jacobson radical, as well as a theory of general radical properties.
Nilradicals
Let R be a commutative ring. First we will show that the nilpotent elements of R form an ideal N. Let a and b be nilpotent elements of R with a^{n} = 0 and b^{m} = 0. We will show that a+b is nilpotent. We can use the binomial theorem to expand (a+b)^{n+m1}:
 <math>(a+b)^{n+m1}=\sum_{i=0}^{n+m1}{n+m1\choose i}a^ib^{n+m1i}<math>
For each i, exactly one of the following conditions will hold:
 i≥n
 n+m1i≥m
This says that in each expression a^{i}b^{n+m1i}, either the exponent of a will be large enough to make the expression vanish, or the exponent of b will be large enough to make the expression vanish. Thus a+b is nilpotent, and hence in N.
To finish checking that N is an ideal, we take an arbitrary element r∈R. (ra)^{n} = r^{n}a^{n} = 0, so ra is nilpotent, and hence in N. Thus N is an ideal.
N is defined to be the nilradical of R. It is written Rad(R) or √R.
It is useful to define the nilradical of an ideal I in R. To do so we let Rad(I) be the preimage of the Rad(R/I) under the projection map R→R/I. Rad(I), also written √I, is the nilradical of I. Under this definition, the nilradical of R is just the nilradical of the zero ideal (0).
It is equivalent to define the nilradical of I as
 <math>\hbox{Rad}(I)=\{r\in Rr^n\in I\ \hbox{for some}\ n\}<math>
To see this, note first that if r is in Rad(I), then for some n, r^{n} is zero in R/I, and hence r^{n} is in I. Second, if r^{n} is in I for some n, then the image of r^{n} in R/I is zero, and hence r^{n} is in Rad(I).
The nilradical is the most common radical in commutative algebra, and there it is usually called just the radical. An ideal that is equal to its radical is called a radical ideal and is said to be radical.
If P is a prime ideal, then R/P is an integral domain, so it cannot have zero divisors, and in particular it cannot have nilpotents. Hence all prime ideals are radical.
By using localization, we can see that Rad(I) is the intersection of all the prime ideals of R that contain I: Every prime ideal is radical, so the intersection J of the prime ideals containing I contains Rad(I). If r is an element of R which is not in Rad(I), then we let S be the set {r^{n}n is a nonnegative integer}. S is multiplicatively closed, so we can form the localization S^{1}R. Form the quotient S^{1}R/S^{1}I. By Zorn's lemma we can choose a maximal ideal P in this ring. The preimage of P under the maps R→S^{1}R→S^{1}R/S^{1}I is a prime ideal which contains I and does not meet S; in particular, it does not meet r, so r is not in J.
Jacobson radicals
Let R be any ring, not necessarily commutative. The Jacobson radical of R is the intersection of the annihilators of all simple right Rmodules.
There are several equivalent characterizations of the Jacobson radical, such as:
 J(R) is the intersection of the regular maximal right (or left) ideals of R.
 J(R) is the intersection of all the right (or left) primitive ideals of R.
 J(R) is the maximal right (or left) quasiregular right (resp. left) ideal of R.
As with the nilradical, we can extend this definition to arbitrary twosided ideals I by defining J(I) to be the preimage of J(R/I) under the projection map R→R/I.
If R is commutative, the Jacobson radical always contains the nilradical. If the ring R is a finitely generated Zalgebra, then the nilradical is equal to the Jacobson radical, and more generally: the radical of any ideal I will always be equal to the intersection of all the maximal ideals of R that contain I. This says that R is a Jacobson ring.