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More explicitly, if $M_{2 \times 2}(\mathbb{R})[x]$ denotes the ring of polynomials over the ring of 2x2 matrices with real coefficients (with indeterminate x a 2 by 2 matrix with real coefficients), how do i properly define multiplication? e.g. suppose $A_0,A_1,B_0,B_1 \in M_{2 \times 2}(\mathbb{R})$, and let $f[x] = A_0 + A_1x$ and $g[x] = B_0 + B_1x$ be elements of $M_{2 \times 2}(\mathbb{R})[x]$. Then I would assume the product $fg[x]$ would be

$ fg[x] = (A_0 + A_1x)(B_0 + B_1x) = A_0B_0 + A_0B_1x + A_1xB_0 + A_1xB_1x $

But then complications arise due to $M_{2 \times 2}(\mathbb{R})$ being non-commutative. As far as I know (I've only taken one course so far on abstract algebra), this ring is well-defined (in that a polynomial ring can have coefficients in any ring, not just commutative ones). I've checked google, wikipedia, etc and haven't found anything relevant to this topic. Is there any standard literature on this topic?

My plan was to eventually be able to investigate cases in which unique factorizations may hold (if any), or maybe polynomials having unique left or right inverses, etc.

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I won't vote to close, since no one else has yet, but I think this question could use a lot of sharpening. If you're just asking how the multiplication is defined, then I don't think this is an MO level question (one point you may be missing is that x commutes with matrices of scalars). You seem to just be confusing yourself by thinking about polynomials in matrices rather than matrices with polynomial entries, which are the same thing. – Ben Webster Dec 30 '10 at 7:59
If you want to ask for references for this stuff, that's a separate question; I'm sure a lot has been written about matrix rings over commutative rings, though I don't personally know of any good references. – Ben Webster Dec 30 '10 at 8:01
How does x commute with matrices of scalars if i define x as a 2x2 matrix as above? – Brian Hepler Dec 30 '10 at 8:08
Ah, now I see that I missed that part. Now I'm just confused by your definitions. The standard terminology is that for a ring A (not necessarily commutative), A[x] (called "polynomials in A") is the ring gotten by adding an indeterminate x which commutes with elements of A. It sounds like you want something else. – Ben Webster Dec 30 '10 at 8:20
The use of "polynomial" here is confusing (see my non-answer below). Perhaps you are really asking a more general question: Given a ring $R$, does there exist a ring $S$ containing $R$ as a subring and an indeterminate $x\in S$ (which does not necessarily commute with elements of $R$)? – Mark Grant Dec 30 '10 at 11:11
up vote 8 down vote accepted

I have two answers for you, depending on what you have in mind.

You want to add an $x$ to the ring of 2x2 matrices, such that while $x$ commutes with multiples of the identity, it doesn't commute with anything else. You can adjoin a noncommuting indeterminate by using what's called the free product. You take the two rings $M_2(\mathbb{R})$ and $\mathbb{R}[x]$, and then you form the free product over $\mathbb{R}$. Wikipedia has an entry on the free product of groups. The construction for rings is fairly similar.

That construction has one weakness: $x$ will not satisfy any relations. There relations that all 2x2 matrices will satisfy, but $x$ in the free product won't. Rings all of whose elements satisfy identities are known as polynomial identity rings. For example, any four 2x2 matrices satisfy an identity of degree four (described in this blog post). So any three elements of $M_2(\mathbb{R})$ and $x$ should satisfy that relation. (I don't know if all possible relations are generated from specializing this one relation. There could be other relations that rely on specific properties of specific matrix elements.)

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Also, a good general reference for noncommutative ring theory is Louis Rowen's 2 volume Ring Theory. – arsmath Dec 30 '10 at 12:45
+1: This seems the right answer to the right question. – Mark Grant Dec 30 '10 at 14:16
Is it not possible to quotient out by a suitable ideal to force x to satisfy certain relations though? For instance, if we have the ideal J generated by $\{ (\lambda I ) x − x (\lambda I) : \lambda \in R \}$, where R is the underlying ring of the matrix ring, then take the quotient by J, would we necessarily get something with unwanted relations? – Zhen Lin Dec 30 '10 at 14:28
You definitely could. You just have to figure out what the relations are. The standard four-variable identity from the theory of polynomial identity rings gives a you a family of relations, but there may be others. – arsmath Dec 30 '10 at 15:17

Ben has already answered the question in the comments by noting that the indeterminate $x$ commutes with elements of $M_2(\mathbb{R})\subseteq M_2(\mathbb{R})[x]$. But I wanted to add some comments because the construction of polynomial rings is a subtle business.

The polynomial ring $R[x]$ should be thought of a ring containing $x$ and a subring isomorphic to $R$, such that $R$ commutes with $x$, and an element $a_0 + a_1x + \cdots +a_nx^n\in R[x]$ is zero if and only if each $a_i\in R$ is zero. Note that this implies that the indeterminate $x$ is transcendental over $R$, that is, $x$ is not a root of any polynomial with nonzero coefficients in $R$. In your example, this cannot be true if $x$ is an element of $M_2(\mathbb{R})$.

A good reference treating the construction of $R[x]$ for an arbitrary rings $R$ (and lots more besides) is Hungerford's Algebra book.

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The OP didn't actually want to talk about the usual polynomial ring, contrary to what the first phrase says, because he wants $x$ to be an indeterminate matrix. – Alex B. Dec 30 '10 at 9:39

There are already issues with $f[X] =X$ and $g[X] =B_0$. (I switched the variable to upper case, that seems less confusing in this context) Then $fg$ can not be written as $A_0 + A_1X$. So you have to allow "coefficents" on both sides and even between. Addition $AXB+CXD$ would also be a problem.

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