I have the following problem:

Let K be any field. An finite dimensional associative non-unital algebra A is a vector space A, togeter with a K-biliniear associative operation such that there is NO identity element for this operation.

I call such an algebra simple if it has no nontrivial proper ideals.

(Edit: I did not define the term ideal: An ideal is supposed to be a vector subspace, closed under the multiplication with elements of the alebgra (from left and from right))

An easy example would be the ground field K as a 1dimensional K-vector space, tohether with the zero-multiplication, sending everything to zero. This is finite dimensional, associative and it certainly has no identity-element. Furthermore, it is simple since it is one-dimensional.

Now, my question is: Are there any other examples?

The canonical examples of simple associative algebras are matrix algebras but since they contain the identity matrix, they are unital.

Another idea would be to take the non-unital algebra A and adjoint a unit to get a unital algebra and then use the known classification there, but this so obtain unital algebra will never be simple since it contains the original non-unital abgera as an ideal. So, another formulation of the problem would be to classify the unital associative algebras with exaclty 3 ideals, where the nontrivial of the 3 has codimension 1 or something like that.

I would be very grateful if someone could help me out here, Tom

  • $\begingroup$ Your example is not simple unless $K$ is a finite field of prime order, since under the zero multiplication, any subgroup of $(K,+)$ is an ideal. $\endgroup$ – Alexander Bors Sep 25 '14 at 13:46
  • $\begingroup$ Okay, I should have defined what I mean by the term "ideal". An ideal should be a K-vector subspace, I will edit that. $\endgroup$ – Tom Sep 25 '14 at 13:53
  • $\begingroup$ This can be found in Curtis and Reiner. I believe it was first proved by Burnside, maybe assuming algebraically closed. $\endgroup$ – Benjamin Steinberg Sep 25 '14 at 15:23

I suspect there's a simpler argument that doesn't involve adjoining a unit, but ...

Adjoin a unit to get a unital algebra $B=K\oplus A$.

Since $B$ is a finite dimensional algebra, the Jacobson radical $J(B)$ of $B$ is the maximal nilpotent ideal of $B$. This must be an ideal of $A$, since all nilpotent elements of $B$ are in $A$.

So if $A$ is simple, either $J(B)=A$ or $J(B)=0$.

In the first case, $A$ is nilpotent. $A^2$ is an ideal, so $A^2=0$, and so any subspace of $A$ is an ideal, leaving only the example you mention.

In the second case, $B$ is semisimple, and therefore a product of matrix algebras over division algebras over $K$, and any ideal is a subproduct of matrix algebras and so has a unit.

But $A$ is an ideal of $B$.


Here is a direct argument avoiding adjoining a unit and explicit use of the Wedderburn theorem. In what follows left ideals and ideals are always subspaces. If $A^2=0$, then any subspace is an ideal. So $A$ is 1-dim and is the example you have. So assume $A^2\neq 0$. Then $A^2=A$ since $A^2$ is an ideal.

If $L$ is a nilpotent left ideal and $a\in A$ then $La=0$ or $La$ is nilpotent. Thus the sum of all nilpotent left ideals of $A$ is two-sided and hence $A$ or $0$. But $A=A^2$. We conclude that $A$ has no nilpotent left ideals.

Let $L$ be a minimal left ideal. Then $L^2\neq 0$ implies $L=La$ for some $a\in L$. Suppose $ea=a$ with $e\in L$. Then $e^2a=ea$ and $e,e^2\in L$. Since $L$ is finite dimensional and $La=L$ we have right multiplication by $a$ is injective on $L$. Thus $e=e^2$ and $L=La=Lea$ implies $e\neq 0$. Thus $L=Ae$ with $e$ idempotent.

Next we claim by induction on dimension that each left ideal $L$ is of the form $Ae$ for some idempotent $e$. If $L$ is minimal, we are done by above. Else let $L'$ be a minimal left ideal contained in $L$ and write $L'=Ae$ with $e$ idempotent as above. Let $L''$ be the set of elements $x$ of $L$ with $xe=0$. Then $L''$ is a proper submodule of $L$ and hence $L''=Af$ with $f$ idempotent by induction. Then we claim $L=A(e+f-ef)$ and that $e'=e+f-ef$ is idempotent.

If $a\in L$, then $a-ae\in L''=Af$ and so $a-ae=(a-ae)f$. Thus $ae'=a(e+f-ef)=ae+(a-ae)f=ae+a-ae=a$ and therefore $L=Ae'$. Also taking $a=e'$ we see $e'e'=e'$. This completes the induction. Thus $A=Ae$ for some idempotent. Therefore $A$ has a right identity. A dual argument shows $A$ has a left identity and so $A$ has an identity.

  • $\begingroup$ Thank you, too. I also like this proof without using deeper results but just down to earth calculation. I tried things in this direction, too, but I did not get anywhere... $\endgroup$ – Tom Sep 26 '14 at 9:01

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