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If I have a set of matrices, call this set U, how can I make this a UFD (unique factorization domain)? In other words, given any matrix $X \in U$, I would be able to factorize X as $X_1 X_2 ... X_n$ where $X_i \in U$ and this factorization is unique?

We may assume the matrix entries are real or complex, but I'd prefer not to add additional restrictions on the numbers.

I guess there are many ways to do this, trivially I can take the set $\{pI\}$ where $I$ is the identity matrix and $p$ is a prime number. But I want to have matrices that are more "general". Thanks!

Edit: There would be a distinguished subset $P \subset U$ which are "primes". To rephrase the question: what I want is, if I take a bunch of elements $X_i, Y_i \in P$, I can guarantee that $\prod X_i \neq \prod Y_i$, if at least there exists one $X_j \neq Y_j$. Can it be done?

Some of the matrices must be non-commutative, ie $A,B \in P$ and $AB \neq BA$.

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Unique factorisation in a UFD is usually defined "up to invertible elements", e.g. in the integers, the difference between positive and negative factorisations is ignored. You should probably make more clear how this affects your question. – Ben Millwood Dec 6 '12 at 17:11
Unique factorization pretty much requires commutativity. It's very hard to define uniqueness if $pqp\neq ppq\neq qpp$. In particular, the definition of "Unique Factorization Domain" requires that the ring be commutative. – Thomas Andrews Dec 6 '12 at 17:11
It's not clear what you want. Let U' be the completion of U by matrix multiplication. If U' is commutative and has unique factorisation, then you already have what you want. If U' is not commutative or doesn't have unique factorisation, there is no way to get a UFD with your starting U. So you cannot make U have the feature you want. It's completion is already determined by U. You may want to ask how to choose U such that the completion is a UFD, or something else entirely. – ex0du5 Dec 6 '12 at 17:16
If they are not commutative, then it is not a Unique Factorization Domain. See the definition: – Thomas Andrews Dec 6 '12 at 17:22
I'm just gonna throw this out there…… – Alexander Gruber Dec 6 '12 at 17:26

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