# Examples of reduced associative algebras

An associative $K$-algebra A is called reduced (or often basic) if $A/rad(A)$ has no nilpotent elements. It can be shown that this is equivalent to that $A/rad(A)$ is a isomorphic to a direct sum of division algebras. Here $rad(A)$ is the Jacobson radical of $A$.

In the article "Reduced group algebras" and a related one (https://math.stackexchange.com/questions/819466/the-division-algebras-arising-in-the-wedderburn-decomposition-of-a-finite-group) in $char(K)=p>0$ it is shown that for group algebras reduced=soluble holds. (An associatve algebra is called soluble if the factor $A/rad(A)$ is commutative.)

If we take a soluble associative algebra and the direct sum with a division algebra (e.g. lower diagonal matices togehther with real quaternion) we obtain an example of that kind.

My question is whether there are some natural examples of such reduced associative algebras which are not soluble and arising from direct products of division algebras, soluble (or abelian) algebras.

Is it possible to obtain a reduced algebra from every associative algebra?

One source is local algebras and their direct sum. But not all reduced algebras are a direct sum of local algebras.

• Do you assume that your algebras are finite-dimensional? And what do you mean by a soluble algebra? – Tom De Medts Sep 4 '14 at 9:44
• Oh, and it's cross-posted: math.stackexchange.com/questions/908449/… – Tom De Medts Sep 4 '14 at 9:44
• One more thing: if you refer to some article, you should give a complete reference, not just the title... – Tom De Medts Sep 4 '14 at 9:46
• I guess you can get an example by starting with a group which has a Quaternion division algebra over the rationals as the endomorphism algebra of a simple and tweak it into a monoid to get a nontrivial radical. – Benjamin Steinberg Sep 4 '14 at 14:02
• Anyway, you can get a reduced algebra from any finite dimensional associative algebra $A$ by taking $eAe$ where $e$ is the sum of primitive orthogonal idempotents one per isomorphism class of projective indecomposables. – Benjamin Steinberg Sep 4 '14 at 14:25

Let $Q$ be the quaternion group of order $8$ and let $P(Q)$ be the power set of $Q$ viewed as a monoid by $AB=\{ab\mid a\in A,b\in B\}$. Let $\mathbb QP(Q)$ be the contracted monoid algebra of $P(Q)$ over $\mathbb Q$ (so identify the zero of the monoid with the zero of the algebra). I claim that $\mathbb QP(Q)$ is reduced but not solvable.
The algebra $\mathbb QP(Q)$ is not semisimple because $P(Q)$ contains elements which are not von Neumann regular. Its semisimple quotient is the direct product $\prod_{H\leq Q} \mathbb Q[Q/H]$ of the group algebras of the quotients of $Q$. Since it is well known that $\mathbb QQ$ has a quaternion Wedderburn component, and the other direct factors are commutative this is an example of a reduced algebra which is not semisimple.