# Semisimple and Jacobson Semisimple

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

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!

# Highlights in Linear Algebraic Groups 12: Radical, Reductive

To analyse the structure of a group G
you will need the radical and a torus T.
The group of Weyl may also may also suit
to prevent the scattering of many a root.
Functors are nice including the one of Lie
Parabolics bring in the ge-o-metry!
The theory of weights may seem oh so eerie
Until you start representation theory!

The structure of reductive and semisimple groups over an algebraically closed field will be pinnacle of this post series. After we have finished with this, this series will end and we will start to learn about algebraic groups from the perspective of group schemes, and we shall use some of the results we have seen so far by using that we really have just been studying the $\overline{k}$-points of group schemes (classical algebraic geometry).

The topic for today is the radical and unipotent radical, that will allow us to define the concept of semisimple and reductive group. We will then use the roots, which are certain characters of a maximal torus. These will give us a root system, so we will take a break to study these, and classification of root systems will enable us to classify algebraic groups.
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