Radicals, An Introduction

Introduction

Throughout this topic, rings are assumed commutative. Yes, noncommutative rings have radical ideals too, but most of the theorems presented here will not apply in that world. You probably want to stick with commutative rings for now, where the algebra is simpler, then explore radicals in noncommutative rings later.

Radical Ideal

Let H be an ideal of the ring R. The radical of H, written rad(H), is the intersection of all prime ideals containing H. If there are no prime ideals containing H then rad(H) = R. However, as long as H is a proper ideal, disjoint from the multiplicative set 1, some maximal prime ideal contains H, and rad(H) ⊂R.

If H is already the intersection of prime ideals than rad(H) = H. It follows that rad(rad(H)) = rad(H).

Here is an equivalent definition of rad(H). The radical of H is the set of elements x such that xn lies in H for some positive n.

If H = R then every x is in H, and rad(H) = R, as it should. So assume H is a proper ideal.

Let xn lie in H, hence in every prime ideal containing H. This means x is in every prime ideal containing H, and x is in rad(H). Conversely, assume xn is never in H, and let S be the multiplicative set consisting of the powers of x. Since H and S are disjoint, some maximal prime ideal contains H and misses S, whence x is not in rad(H).

This characterization is probably the origin of the term "radical ideal", for radical means nth root, and rad(H) is the set of radicals of elements of H. This characterization only applies when R is commutative.

Nil Radical, Reduced

Set H = 0 to get the nil radical of R. This is written nil(R). If R is noncommutative there is an upper nil radical and a lower nil radical, but these are equal for commutative rings, and that's the world we're in right now. To learn about radicals in noncommutative rings, click here.

Using our alternate definition, the nil radical is the ideal consisting of all nilpotent elements.

A ring is reduced if its nil radical is 0. An integral domain has no zero divisors, no nilpotent elements, and is therefore reduced.

Since prime ideals correspond, R/rad(H) has 0 as the intersection of prime ideals, and is reduced.

Self Radical

The ideal H is "self radical" if H = rad(H). In other words, H is the intersection of prime ideals.

One Prime Ideal

Assume R has one prime ideal P. Every maximal ideal is prime, hence P is maximal and unique, and R is local. Furthermore, P = nil(R). Divide by the nil radical and find a field.

Conversely, if R/nil(R) is a field, then nil(R) is a maximal ideal, which is the only prime ideal.

Intersection

Let H be the intersection of finitely many ideals in R. We will show that the radical of the intersection is the intersection of the radicals.

If xn is in H it is in each of the ideals containing H. Conversely, if some power of x is in each of the ideals, the largest power of x is in all of them, and xn is in H.

Product

The radical of the product of finitely many ideals is the radical of their intersection.

Since the product ideal is contained in the intersection, the radical of the product is contained in the radical of the intersection. Conversely, let x lie in the radical of the intersection, hence xn is in each of the ideals. If there are k ideals, xnk is included in the product, thus the radical of the intersection is contained in the radical of the product.

As a corollary, rad(Hn) = rad(H).

When xn = x

If every x in R satisfies xn = x, for some n > 1, then prime ideals are maximal.

Divide out by any prime ideal, and everything in the quotient ring satisfies xn = x. Remember that this is an integral domain. For x nonzero, xn-1 = 1, and xn-2 is the inverse of x. Everything is invertible, the quotient ring is a field, and our prime ideal is maximal.