Intuitive Example of a Jacobson Radical

Question

Can anyone explain what a Jacobson radical is using an intuitive example? I can't quite understand Wikipedia's explanation.

Answer 1

I think my favourite characterization for rings with identity is that y is in the Jacobson radical of R if and only if 1-yx is right invertible for any x in R - so y is sufficiently "zero-like" that moving the unit by its multiples doesn't stop it being invertible.

In fact one can strengthen this to if and only if 1-zyx is actually a unit for any z,x (and deduce from this that the left and right radicals agree).

Answer 2

Lucky timing, I just worked out a couple of examples for the class I'm teaching this semester. Let A be a commutative local ring, m the maximal ideal, and k=A/m the residue field. Let's consider two rings: M = M₂(A) (2-by-2 matrices with entries in A), and I = the Iwahori order consisting of 2-by-2 matrices with entries in A whose lower-left entry is in m, i.e. matrices of the form
[ A A ]
[ m A ]

We'll compute rad(M) and rad(I) using the fact that rad = intersection of annihilators of simple left modules = { x : 1 - xy is a unit for all y }.

First let's compute rad(M). M acts naturally on k^2 by left multiplication (via M₂(A) ->> M₂(k)), and this is a simple M-module. The annihilator of k^2 is M₂(m) = 2-by-2 matrices with entries in m, so rad(M) is contained in M₂(m). On the other hand every element x in M₂(m) has the property that 1-xy is a unit in M₂(A) for all y in M₂(A) (since xy is in M₂(m), the determinant of 1-xy is a unit); therefore rad(M) contains M₂(m), and we've shown rad(M) = M₂(m).

Next let's compute rad(I). Now k is a simple left I-module in two ways: first by multiplication by the upper-left entry (mod m), and second by multiplication by the lower-right entry (mod m). (Check that if x,y are in I then the upper-left entry of xy is congruent mod m to the product of the upper-left entries of x and y, and similar for the lower right.) The annihilator of the first I-module is therefore matrices of the form
[ m A ]
[ m A ]
and the second is matrices of the form
[ A A ]
[ m m ]
and the intersection of these annihilators is the ring of matrices of the form
[ m A ]
[ m m ]
which therefore contains the Jacobson radical. On the other hand every matrix x of that form has the property that det(1-xy) is a unit in A for y in I, and therefore the above collection of matrices is equal to the Jacobson radical.

Answer 3

Well, if R is a finitely generated commutative ring then J(R) is just the nilradical, so for example Z[x]/(x^2) has Jacobson radical (x).

The intuition I have about the nilradical (and by extension, the Jacobson radical) is that it measures how far R is from behaving like the ring of functions on a space. More precisely, if R = C[x₁, .. x_n]/I is a finitely-generated C-algebra with corresponding variety V, then the ring of functions on V is C[x₁, ... x_n]/rad(I) = R/Nil(R) by the Nullstellensatz. This is because elements of the nilradical of R, which is rad(I), are nilpotent on V, and an honest-to-goodness function on a space can't be nilpotent unless it's identically zero.