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.

  • $\begingroup$ Do you assume that your algebras are finite-dimensional? And what do you mean by a soluble algebra? $\endgroup$ – Tom De Medts Sep 4 '14 at 9:44
  • 1
    $\begingroup$ Oh, and it's cross-posted: math.stackexchange.com/questions/908449/… $\endgroup$ – Tom De Medts Sep 4 '14 at 9:44
  • $\begingroup$ One more thing: if you refer to some article, you should give a complete reference, not just the title... $\endgroup$ – Tom De Medts Sep 4 '14 at 9:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Benjamin Steinberg Sep 4 '14 at 14:02
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

There might be a clever argument to show it is a direct product of a quaternion division algebra and a soluble algebra using Mobius inversion on the subgroup lattice. I will have to think about it.

  • $\begingroup$ Thanks for this example. Since we have zero characteristics there is also an complemen by the Wedderburn-Malcev-Theorem. Do you know such a complement? Are this the algebras you mentioned in your previous answer? $\endgroup$ – Sven Wirsing Sep 4 '14 at 14:41
  • $\begingroup$ I know the complement. If I have a chance later I will write something. $\endgroup$ – Benjamin Steinberg Sep 4 '14 at 16:31
  • $\begingroup$ The semisimple quotient is the subalgebra spanned by the cosets of subgroups. $\endgroup$ – Benjamin Steinberg Sep 4 '14 at 16:47
  • $\begingroup$ Thanks Benjamin, I guess there is already some theory in place for such algebras? The background of my example is that I am looking for the Cartan-Subalgebras of the associated Lie-algebra. In this case one has to determine the maximal commutative subalgebras of the complement and then calculating the centralizer of these in the whole algebra. $\endgroup$ – Sven Wirsing Sep 4 '14 at 16:56
  • 2
    $\begingroup$ ams.org/journals/tran/2009-361-03/S0002-9947-08-04712-0 will tell you which monooids have reduced algebras. You can also see from there and group theory which ones are not solvable. I am not sure how easy it is to see whether it is a direct product of division algebras and solvable algebras. $\endgroup$ – Benjamin Steinberg Sep 4 '14 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.