Algebra - Decomposition of a matrix polynomial

Dear All,

This is related with a problem that I'm trying to solve on my PhD dissertation in econometrics, and I thought that some mathmatician can know the answer.

What is known about a possible extension, $E$ , of the ring, $A$ , of all n-by-n matrices with entries in $\mathbb{C}$ such that any non-constant polynomial of $A[x]$ splits in a product of linear factors in $E[x]$?

$ax = xa$ iff $a$ is in the commutator of $E$. Moreover, $A$ is unitary.

Thanks a lot,

Federico Carlini

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What is $C?$ The complex numbers? – Igor Rivin Dec 12 2011 at 17:47
What do you mean by $A[x]$? Since $A$ is non-commutative, this does not have a standard meaning. Do you want $x$ to commute with elements of $A$, or are $ax,xa$ distinct? And, by the way, what do you mean by linear factors? – Felipe Voloch Dec 12 2011 at 18:42
C is complex number. – Federico Carlini Dec 12 2011 at 19:07
Dear Felipe, $A$ is non-commutative with respect to $x$. A linear factor is a polynomial $A0+A_{1,s}xA_{1,d}$, with $A_{1,s}$ and $A_1,d$ non zero factors. Sorry for the notation. I try to do better the next time. It's my first time I'm posting here – Federico Carlini Dec 15 2011 at 18:00
Wait a minute. If a linear factor is this, what is a polynomial? I mean $AxBxC+DxExF+\dots$ is not convertible into one quadratic term in general... – fedja Dec 16 2011 at 21:03
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This is true because of following: any matrix over principal ideal domain can be reduced by elementary Gauss transformations to the Smith normal form (see this: http://en.wikipedia.org/wiki/Smith_normal_form). The ring of polynomials over any field is such a domain. Smith normal form can be splitted to linear factors. Elementary Gauss transformations may be of the form $(z, w)\mapsto (z+p(x)w, w)$, where $p(x)$ is not linear. I think that these transformations cannot be splitted in linear factors (I don't know how to prove it - please let me know if you do know this, I'm curious about).

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 I forgot to write that this can be done with elementary transformations if your principal ideal domain is euclidian - and the univariate polynomial ring is euclidian. – zroslav Dec 16 2011 at 21:05

Dear Alfonso,

I'm not understanding why $A$ has to be a domain.

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 It is not. Ignore my reply, as it was completely wrong. – Alfonso Gracia-Saz Dec 13 2011 at 5:06

If you only need to show that $E$ exists, that is easy. $A$ is a domain, so it has a field of quotients $K$. Let $E$ be the algebraic closure of $K$. Then $E$ satisfies the condition you want. This is non-constructive, however.

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 $A$ is a non-commutative ring and is has zero divisors. – Felipe Voloch Dec 12 2011 at 17:50 Oops. Silly me. – Alfonso Gracia-Saz Dec 13 2011 at 5:06