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

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3 Answers 3

up vote 10 down vote accepted

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).

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By far the simplest and the only one I could understand without having to attempt to learn new maths –  Casebash Nov 8 '09 at 5:24

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.

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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, .. xn]/I is a finitely-generated C-algebra with corresponding variety V, then the ring of functions on V is C[x1, ... xn]/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.

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this answer actually extends to arbitrary commutative rings in a sense. the jacobson radical is the intersection of all of the maximal ideals of your ring R. maximal ideals are closed, or geometric, points of spec R. so the jacobson radical consists of functions on spec R that vanish at every geometric point. –  Ian Shipman Oct 31 '09 at 0:40
And the exact statement extends to Jacobson rings for which the notions of nilradical and Jacobson radical coincide so a function is nilpotent iff it vanishes at the closed points. –  Greg Stevenson Oct 31 '09 at 3:26
What do you mean by variety? –  Casebash Nov 8 '09 at 2:59

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