# What are your favorite finite non-commutative rings?

When you are checking a conjecture or working through a proof, it is nice to have a collection of examples on hand.

There are many convenient examples of commutative rings, both finite and infinite, and there are many convenient examples of infinite non-commutative rings. But I don't have a good collection of finite non-commutative rings to think about. I usually just think of a matrix ring over a finite field.

Do YOU have other examples that you particularly like/find easy to use/find to be a good source of counterexamples?

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Community wiki? – Steve Huntsman Jun 7 '10 at 15:06
Good call. Fixed! – Oliver Jun 7 '10 at 15:28

2 families of examples that are sometimes useful to have in mind:

(1) The group ring of a non-abelian finite group over a finite commutative ring.

and

(2) the incidence algebra of a finite poset over a finite commutative ring (the ring of upper triangular matrices is a basic example of this).

Of course, both of these are special cases of the same more general categorical (or quiver) definition. Before I wrote that I'd never dealt with the more general concept, but that was a lie...

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From one point of view, the classification of finite non-commutative rings or even finite commutative rings is wild and tons of things can happen. From another point of view, finite rings are highly restricted and not all that much can happen.

By the Chinese remainder theorem, every finite ring is unique direct sum of $p$-primary finite rings, i.e., rings whose cardinality is a power of some prime $p$. Such a ring is an algebra over $\mathbb{Z}/p^k$ for some $k$, and the finiteness condition is then equivalent to the statement that this algebra is Artinian and finitely generated. We can look at the case $k=1$ for a start, because then you get a finite-dimensional algebra over a field. Rings over $\mathbb{Z}/p^k$ for higher $k$ will look similar to these algebras.

A finite-dimensional algebra has Jacobson radical and a maximal semisimple quotient if you quotient by that radical. By the theorems of Wedderburn and Artin-Wedderburn, the semisimple quotient is a direct sum of matrix algebras over a finite field. So that wraps up the semisimple examples.

Let's suppose at the other end that the semisimple quotient is a finite field $\mathbb{F}_q$. Any such algebra is a finite-dimensional quotient of free polynomial algebra $\mathbb{F}_q\langle x, y, \ldots \rangle$. In fact it will be a quotient of the truncation defined by killing all monomials of degree $k$, for some $k$. Such a truncation is already an interesting example, e.g. the algebra spanned by $1,x,y,x^2,y^2,xy,yx$, with all larger monomials set to 0. The wild part of the classification is that a general quotient of this type can be very complicated, because it is defined by an ideal or a truncation in a very complicated position. (It is like cutting a riser pipe: The cut can be smooth or very jagged.)

The general finite-dimensional algebra will be a combination of these various ideas. That is kind-of glib because you can get things like truncated path algebras, but it is true.

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I don't care how dumb this makes me look: Could you (or someone else) explain the comparison with riser pipe? Also, what is riser pipe? I looked it up, but it just said it was vertical pipe in a building. – Harry Gindi Jun 7 '10 at 18:38
A riser pipe is any pipe sticking up vertically. I was making a reference to the oil spill. Let's say that a riser pipe is a cylinder. You can cut the cylinder either with a clean curve (a round circle) or a messy curve (a fractal, topologists's sine curve, etc.). The analogy is that the clean surves are a cofinal family, just as clean co-Artinian ideals (monomial ideals) in a polynomial ring are cofinal with all co-Artinian ideals. – Greg Kuperberg Jun 7 '10 at 19:06
Yes, this is an interesting answer. Therefore the most upvotes. But it does not answer the question. The question is about giving nice examples which can be used in practice, and not a classification of all examples. – Martin Brandenburg Jun 7 '10 at 22:52
Martin, I somewhat disagree: it's true that there should be some basic examples, but many examples are formed by a small number of constructions from the ones already known, and constructions are important to keep in mind. – Victor Protsak Jun 7 '10 at 23:59
I'm going to agree with Martin. This is a lovely answer, and certainly the most enlightening one, but it doesn't really address the question I thought I was asking. Stephen's answer is more directly applicable, which is what I was looking for. – Oliver Jun 8 '10 at 23:15

Kent Fuller gives a nice way of constructing non-commutative artinian rings via diagrams in his "Algebras from Diagrams" paper (Journal of Pure and Applied Algebra, Volume 48, Issues 1-2, September 1987, Pages 23-37). Using this approach, you can create a ring with a preconceived structure, which is visualized via a diagram. There are some restrictions on exactly which diagrams could be used, but if you are looking for easy examples then this shouldn't be a concern. Then using any finite field, you can create (essentially) a (semi)group ring over that field where the associated algebra semigroup is given by the diagram. This might be nice/easy/good source of counter-examples for you to use.

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Here's a link: dx.doi.org/10.1016/0022-4049(87)90105-8 – Chris Phan Jun 7 '10 at 17:41

The path algebra of an acyclic quiver (no loops or closed circuits) over a finite field. These rings are hereditary, i.e. they have projective dimension $\leq 1$ and occupy an intermediate position between Greg's semisimple and wild examples. If a quiver is "arboreal" (underlying undirected multigraph is in fact a tree) then its path algebra may be identified with the incidence algebra of the poset of vertices in which $i\leq j$ $\iff$ there is a directed path from $i$ to $j$ (by the assumption, there is at most one).

For any ring $R$ you can form matrix algebra $M_n(R)$ and they will be Morita equivalent. In particular, matrix rings of path algebras are also hereditary, but their simple modules are not all one-dimensional.

By the way, I think that in the noncommutative setting, when Frobenius morphism is not available, this is a wrong question: all finite-dimensional algebras over a field should be considered on a similar footing, regardless of whether the field is finite or infinite.

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It's worth pointing out that this is the same as an incidence algebra (take the Hasse diagram of the poset as your quiver). Of course, it's good to have multiple ways of visualizing the same algebra! – GS Jun 7 '10 at 17:21
Yes and no. For example, the Kronecker quiver has 2 vertices and a double edge, so it's not an incidence algebra. ADE quivers are trees, so it's true for them. I've added an explicit connection. – Victor Protsak Jun 7 '10 at 17:56
You're right! Sorry about that. – GS Jun 7 '10 at 18:01
Actually, that there be no multiple edges is not enough for a path algebra on an acyclic quiver to be an incidence algebra: you have also to prohibit paths parallel to arrows. – Mariano Suárez-Alvarez Jun 7 '10 at 22:21
Thank you Mariano, I've fixed it. The quiver <---> incidence algebra correspondence isn't very natural from the quiver point of view. – Victor Protsak Jun 7 '10 at 23:55