Radicals, An Introduction

Introduction

Radical ideals and nil radicals can be defined for commutative or noncommutative rings. The definitions and theorems are more complicated in the noncommutative world, so you may want to look at radicals in commutative rings first, then return to this topic.

Nil/Nilpotent Ideals

You've probably seen these definitions before, but let's run through them again.

An element x in a ring R is nilpotent if some power of x = 0.

A left ideal H is nil if every x in H is nilpotent. The entire left ideal H is nilpotent if Hⁿ = 0. Here Hⁿ is shorthand for H*H*H… n times. Remember that the product of two left ideals is another left ideal. After finitely many multiplications of H, the result is 0.

If Hⁿ = 0 then xⁿ = 0 for every x in H. Thus a nilpotent ideal is a nil ideal. The converse does not hold, even for commutative rings. Consider Z adjoin infinitely many indeterminants, with x_iⁱ set to 0. Let all the indeterminants generate H. Given a polynomial in H, raise it to the n^th power using the multinomial theorem. When n is sufficiently large, all the terms drop to 0. Thus H is a nil ideal. However, Hⁿ is not 0, and H is not nilpotent.

Assume G and H are nilpotent left ideals with a common exponent n, and consider G+H. Raise a+b in G+H to the 2n. Remember that a and b may not commute, so the binomial theorem does not apply. Expand the product, and each term contains at least n instances of a or b, with the other variable interspersed. This is at least n elements from G or from H, multiplied together. The result is 0. Therefore G+H is nilpotent, with exponent no larger than 2n.

By induction, the finite sum of nilpotent left ideals is another nilpotent left ideal. However, H could be generated by infinitely many nilpotent ideals, without being nilpotent. Let the i^th ideal be generated by the i^th indeterminant x_i, in the earlier example.

If an ideal is generated by nil ideals, it is nil. Let's start with a nil ideal H and a nil left ideal J. Take a from H and b from J, and look at a+b mod H. This is the same as b mod H, hence (a+b)ⁿ is 0 in R/H. Pulling back, (a+b)ⁿ lies in H and is nilpotent, hence a+b is nilpotent.

Induction makes a finite sum of nil ideals nil, and any x in an infinite sum of nil ideals is a finite sum, contained in the sum of finitely many nil ideals, and is nilpotent. Thus the sum of nil ideals is nil.

Kothe's conjecture says the sum of two left nil ideals is a left nil ideal. This remains an open question.

Upper Nil Radical

The sum of all the nil ideals in R is itself a nil ideal - the largest nil ideal in R. This is called the upper nil radical of R, and is sometimes written upnil(R).

Note that x is in upnil(R) iff the principal ideal generated by x is nil.

Nil and Finitely Generated

Let H be a nil ideal that is finitely generated as a left ideal, and assume the generators lie in the center of R. Write x as a linear combination of generators, and consider xⁿ. Expand this into a sum of products. In each term, move the n generators to the right. If n is sufficiently large, at least one of the generators will become 0, since all the generators are nilpotent in H. The terms drop to 0, xⁿ = 0, Hⁿ = 0, and H is nilpotent.

There are infinitely generated nilpotent ideals. Adjoin infinitely many indeterminants to Z, and set all products of indeterminants to 0. Let the indeterminants generate H, and H² = 0. An infinite ascending chain of ideals shows H is not finitely generated.