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It's just all the rings (with 1) I know can be written as a matrix, i.e., find some matrix representation of it (not necessary commutative). Complex numbers for example is written as obvious matrix. E.g. like this

enter image description here http://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_complex_numbers

However, is it? possible allowing the matrix to be infinite or just have finite amount of columns.

Like complex numbers are uncountable and it gets around the problem as you can write it as a 2x2 matrix. Even the most exotic example of a ring I know can be written as column finite matrix.

Also, I suppose can all groups be represented as $n \times n$ matrix?

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What exactly do you mean by saying a ring can be 'written as a matrix'? –  Daniel Freedman Dec 30 '11 at 12:18
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@DanielFreedman I mean isomorphism. –  simplicity Dec 30 '11 at 12:29
    
@DanielFreedman sorry should explained more. I should have said has every ring with 1 has a matrix representation? –  simplicity Dec 30 '11 at 12:32
    
So are you asking if it is always possible to find a non-zero homomorphism from a ring into a matrix group? –  Daniel Freedman Dec 30 '11 at 12:33
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@simplicity: You should be more precise about your question. Perhaps you should specify that $R$ is possibly non-commutative, and that the matrix ring should be over a commutative ring, or even a field. –  Zhen Lin Dec 30 '11 at 12:45

3 Answers 3

up vote 5 down vote accepted

If $R$ is a $k$-algebra for a field $k$ and you have an injective $k$-algebra homomorphism $R\to M_n(k)$, then $R$ is necessarily finite dimensional. This is a very strong restriction on $R$.

However, if $R$ is finite dimensional, when we have a canonical representation $\varphi:R\to \operatorname{End}_k(R)$ sending $r\in R$ to the linear operator given by left multiplication by $r$, that is, $\varphi(r)(s)=rs$. This is a ring homomorphism because $\varphi(r+r')(s)=(r+r')(s)=rs+r's=\varphi(r)(s)+\varphi(r')(s)$ and $\varphi(rr')(s)=(rr's)=r(r's)=\varphi(r)\circ \varphi(r')(s)$. It is easy to check that it is $k$-linear. Additionally, if $R$ is unital, this map is injective: $\varphi(r)(1)=r$, so if $r\neq 0$, then $\varphi(r)\neq 0$.

Note that we don't even need $R$ to be commutative. We can also generalize to when $k$ is a commutative ring and $R$ is finitely generated and free over $k$ (and probably to finitely generated projective).

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The representation of $R$ via linear maps (endomorphisms) on $R$ in the second paragraph of Aaron's answer is known as the (left) regular representation of $R$. It is a ring-theoretic analogue of Cayley's theorem, that a group $G$ may be represented as a subgroup of the group of bijections (permutations) on $G$. In both cases the representation arises by sending each element $\rm\:r\:$ into its associated (left) multiplication map $\: x\:\to\:r\:x\:.\:$ See this MO question for further discussion.

More concretely (for readers unfamiliar with endomorphism rings), the representation arises simply by viewing each element of $R$ as a $1$-by-$1$ matrix $\:[\:r\:]\:,\:$ with the usual matrix algebra. This allows one to view the ring as a subring of the ring of all linear maps on its underlying additive abelian group - in analogy to viewing a group as a subgroup of the group of all bijections on its underlying set.

For a more general view of such representation theorems from the perspective of category theory see the following interesting book, and its review by John Isbell.

Aleš Pultr and Věra Trnková. Combinatorial, algebraic and topological representations of groups, semigroups and categories. North Holland. 1980.

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@simplicity See also this answer on the regular matrix representation $\rm\,\Bbb C\cong \Bbb R[i].\ \ $ –  Math Gems Feb 5 '13 at 16:01

As Bill Dubuque and Aaron say, any ring can be mapped to matrices if you allow "infinity by infinity" matrices, by considering the regular representation. And, as Aaron says, only a finite dimensional ring can have an injective homomorphism to $n \times n$ matrices, for reasons of dimension.

It is worth mentioning that there are rings with NO homorphisms to finite dimensional matrices. The basic example is the $\mathbb{Q}$ algebra generated by $X$ and $Y$ modulo the relationship $XY=YX+1$. If we had any homomorphism $\rho$ to $n \times n$ matrices, then we would have $\rho(X) \rho(Y) = \rho(Y) \rho(X) + \mathrm{Id}_n$. Taking the trace of both sides, we get $\mathrm{Tr} \ \rho(X) \rho(Y) = \mathrm{Tr} \rho(Y) \rho(X) + n$ or $0=n$, a contradiction.

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