### Introduction

The jacobson radical of a ring R, written jac(R), is the intersection of all the maximal left ideals of R.

Given an element y in R, several conditions are equivalent to y lying in jac(R). The first is the definition, and the second assures the left invertability of 1-xy.

1. y is in every maximal left ideal.

2. For every element x, 1-xy is left invertible.

Let y lie in jac(R) and assume 1-xy has no left inverse. Thus 1-xy generates a proper left ideal. Let H be a maximal left ideal containing R*(1-xy). Now H contains 1-xy and y, hence H contains 1. This contradiction shows 1-xy is left invertible, for any x.

Suppose M is a simple left R module, and y*M is nonzero. In other words, y does not kill M. Let v be an element of M with yv nonzero. Since M is simple, v generates M, and yv also generates M. There is some x satisfying xyv = v. Thus 1-xy kills v, and 1-xy has no left inverse. This contradicts condition (2) above. Therefore y kills every simple left R module.

Now to complete the circle from (3) back to (1). Assume y kills every simple left R module. Let H be a maximal left ideal. Let M be the left R module consisting of the cosets of H in R, as R acts on these cosets. If M contains a proper nonzero submodule, this becomes a proper left ideal containing H, which is impossible. Therefore M is a simple left R module.

Remember that y kills every simple left R module. This means y*1 lies in H. Thus y lies in H. This holds for every maximal left ideal H, thus y is in jac(R).

To summarize, y is in jac(R) iff y kills every simple left R module.

Remember, a module is simple iff it is isomorphic to the cosets of H, for some maximal left ideal H, viewed as a left R module. Let's summarize again. The element y is in jac(R) iff y kills every left module R/H, where R/H represents the cosets of a maximal left ideal H.

Bringing all this together, y is in jac(R) iff y is in the intersection of the annihilators of all the simple left modules R/H.

Let's look at annihilators in general. If M is a left module, let K be the annihilator of M, i.e. the elements of R that take M to 0. Clearly K is a left ideal, but it is also a right ideal. Thus the annihilator of a left module, or a left ideal, is a two sided ideal.

Since jac(R) is the intersection of annihilators, it is a two sided ideal. This is counterintuitive. Take the intersection of all the maximal left ideals, and find a two sided ideal.

Next, we have a fourth condition for y ∈ jac(R).

4. 1-xyz is a unit for every x and z.

Remember that jac(R) is an ideal, so xy, and yz, and xyz, are already in jac(R). It is enough to show 1-y is a unit, for y in jac(R).

By condition (2), 1-y is left invertible. Let u be its left inverse. Write u-uy = 1, or u = 1+uy. Since u is right invertible, 1+uy is right invertible. By condition (2), 1+uy is also left invertible. therefore 1+uy, or u, is a unit, and so is its inverse 1-y.

Conversely, if 1-xyz is a unit for every x and z, then 1-xy is left invertible for every x, and y belongs to jac(R).

Using condition (4), jac(R) is characterized by 1-xyz being a unit. The same criteria falls out if we define jac(R) as the intersection of maximal right ideals. this yields another counterintuitive result, the left jacobson radical is the same as the right jacobson radical. That is why it is simply called the jacobson radical. It is the intersection of the maximal left ideals, or the intersection of the maximal right ideals, or the intersection of the maximal left and right ideals.

Finally we have the fifth condition for y lying in jac(R). Notice that this too is symmetric, showing once again that the left jacobson radical equals the right jacobson radical.

5. jac(R) is the largest ideal H, such that the elements 1-y, for all y in H, are units. The same holds if H is the largest left ideal or right ideal having this property.

Let H be any left ideal having this property. This means 1-y is left invertible, and since xy is in H for any x, 1-xy is left invertible. By condition (2), y is in jac(R), and our left ideal H lies in jac(R).

If H is a right ideal, then 1-yx is right invertible for every x, y is in the intersection of maximal right ideals, y is in jac(R), and H is in jac(R).

Conversely, jac(R) is an ideal that we will call H. You can think of it as a left ideal or right ideal if you like. By condition (4), 1-xyz is a unit for every y in H. This means 1-y is a unit, satisfying condition (5). Jac(R) is the largest (left/right/two-sided) ideal satisfying condition (5).