Posted by Jason Polak on 27. August 2017 · Write a comment · Categories: math, modules

Let $R$ be an associative ring with identity. The Jacobson radical ${\rm Jac}(R)$ of $R$ is the intersection of all the left maximal ideals of $R$. So, ${\rm Jac}(R)$ is a left ideal of $R$. It turns out that the Jacobson radical of $R$ is also the intersection of all the right maximal ideals of $R$, and so ${\rm Jac}(R)$ is also an ideal!

The idea behind the Jacobson radical is that one might be able to explore the properties of a ring $R$ by first looking at the less complicated ring $R/{\rm Jac}(R)$. Since the ideals of $R$ containing ${\rm Jac}(R)$ correspond to the ideals of $R/{\rm Jac}(R)$, the ring $R/{\rm Jac}(R)$ has zero Jacobson radical. Often the rings $R$ for which ${\rm Jac}(R) = 0$ are called Jacobson semisimple.

This terminology might be a tad bit confusing because typically, a ring $R$ is called semisimple if every left $R$-module is projective, or equivalently, if every left $R$-module is injective. How does the notion of semisimple differ from Jacobson semisimple? The Wedderburn-Artin theorem gives a classic characterisation of semisimple rings: they are exactly the rings that are finite direct products of full matrix rings over division rings. Since a full matrix ring over a division ring has no nontrivial ideals, the product of such rings must have trivial Jacobson radical. Thus:

A semisimple ring is Jacobson semisimple.

The converse is false: there exists a ring that is Jacobson semisimple but not semisimple. For example, let $R$ be an infinite product of fields. Then ${\rm Jac}(R) = 0$. However, $R$ is not semisimple. Why not? If it were, by Wedderburn-Artin it could also be written as a finite product of full matrix rings over division rings, which must be a finite product of fields because $R$ is commutative. But a finite product of fields only has finitely many pairwise orthogonal idempotents, whereas $R$ has infinitely many.

Incidentally, because $R$ is not semisimple, there must exist $R$-modules that are not projective. However, $R$ does have the property that every $R$-module is flat!

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