Wrapping up the Jacobson Radical « A Mind for Madness

Posted by: hilbertthm90 | February 28, 2009

Wrapping up the Jacobson Radical

We now have the following equivalent definitions of the Jacobson radical. Remember right now we assume R is commutative with 1.

1) Intersection of all maximal ideals
2) Intersection of the annihilators of all simple left R-modules
3) The set of non-generators of R
4) The set of elements, x, such that $1-rx$ has a left inverse for all r.

I think I already pointed out that from at least two of these definitions we automatically get that $J(R)$ is a two-sided ideal. Two basic examples are now that if R is any field, then $J(R)=\{0\}$ . And if K is a field, and $R=K[[x_1, \ldots x_n]]$ , then $J(R)=\{f\in R : f \ has \ 0 \ constant \ term\}$ . An important generalization is that in any local ring the Jacobson radical is the set of non-units.

An important result called Nakayama’s Lemma is that if $M$ is finitely generated, then $M=\Phi(M)+N$ implies that $M=N$ . Special case: If $M= J(R)M+N$ , then $M=N$ . Corollary to that special case: If $M=J(R)M$ , then $M=\{0\}$ (this last form is what is sometimes called Nakayama’s Lemma).

Proof: Suppose $M=\langle x_1+n_1, x_2+n_2, \ldots, x_m+n_m\rangle$ . Where $x_j\in \Phi (M)$ , and $n_j\in N$ for all j. Define $S=\{n_1, \ldots, n_m\}$ .

Then with this setup, we exploit the non-generator definition. Note that
$M=\langle x_1, n_1, x_2, n_2, \ldots, x_m, n_m\rangle$
$= \langle S, x_1, \ldots, x_m\rangle$
$= \langle S, x_1, \ldots, x_{m-1}\rangle$
… etc
$= \langle S\rangle \subset N$ .

And we are done! It may have seemed a little roundabout to go through the “Frattini submodule” in developing the Jacobson radical, but it certainly pays off to have lots of definitions as we see here.

The last little bit I wanted to say was that we can define the Jacobson radical for a ring without identity. I don’t want to go through the details, but a standard trick is to define a new ring (with identity) $S=\mathbb{Z}\times R$ with the standard addition, and then $(a,b)(c,d)=(ac, ad+cb+bd)$ . It is pretty basic to check that $J(S)=\{0\}\times I$ where $I$ is some ideal in R (by the fact that $J(\mathbb{Z})=\{0\}$ ). It is also just algebraic manipulation to check that $I$ is the largest ideal in R such that for every $x\in I$ there is a $y\in I$ such that $x+y-yx=0$ . This then is our definition. $J(R)=\cap_{\mathfrak{I}} I$ where $\mathfrak{I}$ is the collection of ideals satisfying that property.