I doubt that there is a simple answer to this question. If you proceed as in your calculus course and simply write out the equations for the denominators betai, then the maps sigma and delta defining your Ore algebra will appear in them. Thus everything depends on how complicated these maps are. In any case, for a non-trivial Ore algebra, these equations will no longer form a simple linear system.

For the fun of it, I looked at the special case of linear differential operators with K=Q(y) and x=d/dy. For n=2 you obtain then a combination of a linear algebraic equation for beta1 and beta2 and a linear first order differential equation for beta2. It is easy to see that a solution exists, but that it is not unique, as you are free to choose an integration constant. For higher n, things should be similar, but you will have even more choice, as now more and higher order differential equations will appear.

Thanks a lot for your answer!

Yes, we can asumme the special case of linear differential operators. And n = 2 for the moment.

Well, what I think is that we can always find beta1 and beta2 from a set of linear algebraic equations. We can do a trick and work with the adjoint polynomial ring K*[x]. In plain words, instead of expanding x(beta1+beta2) we can rewrite original polynomial b(x) = b1x + b0 as b(x) = x b1 - db1/dy + b0 and compare these coefficients. Then we get beta1+beta2 = b1 and instead of having derivatives of beta's in the second equation we only have derivative of b1 which is not problem at all.

Thus, we obtain finally a linear algebraic equation for beta1 and beta2, and a linear algebraic equation for beta2.

However, for me, the problem lies somewhere else. I do not know how to find alpha1 and alpha2 at the first place. These are not a1 and a2 respectively, in general. If we know alpha1 and alpha2 then a(x) = (x-a2)(x-a1) is LCLM of x-alpha1 and x-alpha2. But we need the converse here. Sure, alpha1 and alpha2 might not be unique. I just like to find some, at the beginning.

You are right: for n=2 (and linear differential operators) you get around solving differential equations. However, there are two problems with your solutions. I do not think that the same trick will help for larger n (although one might do some kind of iteration) and one should still keep in mind that the trick just produces in a simple manner one of many solutions. Thus one still has a difference to the classical decomposition which is unique.

Concerning your final question: I am sure that I have seen some work on factorisation in Ore algebra. There has definitely been quite some work on the factorisation of linear differential operators. I do not have any reference handy, but if you do an internet search on factorisation you should come up with quite a number of hits. If you have trouble finding something, just tell me and I will ask some colleagues who are closer to this subject than I.

Just our of curiosity: for what do you need the decomposition and in what kind of non-commutative polynomial rings are you interested? Ore algebras are just a special kind of polynomial rings of solvable types and most things you can do with Ore algebras you can also do in this larger class of non-commutative rings.

Working with the adjoint skew polynomial ring, I can get around for higher n as well. Computing beta's this way is OK for me.

Concerning the factorization of a(x) I suppose the factorization is already known, just like to do this partial fraction decomposition somehow. But I asked personally some other colleagues, and one of them, prof. Petkovsek, recommended to look at

reu.dimacs.rutgers.edu/~darlayne/SecondREUPresentationNEWEST.ppt

I believe that is exactly what I needed. But having this discussion here was very helpful for me anyway. Thanks a lot.

To explain in what kind of Ore rings I am interested in needs a lot of technicalities. I can only give you the following rather intuitive explanation, or otherwise refer you to some papers.

Assume we have a nonlinear control system descirbed by a nonlinear differential equation

y(n) = f(y,...,y(n-1),u,...,u(m))

where f is meromorphic, and the derivative operator is d/dt. Then, after differentiating it, we can write

dy(n) + an-1 dy(n-1) + ... + a0 dy = bm du(m) + ... + b0 du

for some functions ai, bj. The two associated polynomials

a(s) = sn + an-1 sn-1 + ... + a0

b(s) = bm sm + bm-1 sm-1 + ... + b0

are from a (non-commutative) skew polynomial ring. I now use s as an indeterminate in polynomials. Elements of the associated skew field of left fractions

F(s) = a-1(s) * b(s)

are the so-called transfer functions of nonlinear systems, as we have

dy = F(s) du

This represents an analogy of the formalism introduced for linear systems (i.e. f being linear) through Laplace transforms. The transfer function F(s) of such a system is defined as

Y(s) = F(s) U(s)

and is useful for plenty of things related to control theory and engineering. Finding partial fraction decomposition of F(s) is also helpful in many ways.

The transfer functions of nonlinear systems as introduced above have many analogous properties. So in my case K does not equal directly to e.g. Q(t) but the structure is similar.