Recently I have been working with a certain subgroup of $GL\_{10}(\mathbb{F}\_2)$ and for various reasons was fairly sure it contained a normal subgroup isomorphic to $A\_5$. Today I was able to affirmatively show that this presumed copy of $A\_5$ does indeed exist in the bigger group. Now the matrices composing this copy of $A\_5$ are themselves $2\times 2$ matrices over a certain commutative 5-dimensional algebra $\mathcal{S}$ (necessarily with zero divisors) over $\mathbb{F}\_2$; this gives rise to a 2-dimensional representation of $A\_5$ over $\mathcal{S}$ which I am fairly sure is irreducible (as opposed to just indecomposeable).


Now it is a fairly standard exercise early in learning representation theory to show that a simple group cannot have an irreducible representation of dimension 2. As the proof of this simple fact relies only on considerations involving the characters of elements of order 2, does the existence of the above representation rely only on the fact that I am working in characteristic 2, or is it related to the fact that I am working with rings with zero divisors as opposed to an algebraically closed field, or something else more subtle? I am primarily curious as to whether it is solely related to the characteristic, as I am also working with some related groups defined over commutative rings in other characteristics which I suspect also contain subgroups isomorphic to $A\_5$. Hopefully someone with more background in modular representation theory than I can shed some helpful light on this situation.



Failure of a basic fact from Representation Theory
To reinforce Geoff's answer, I'd emphasize that the "fairly standard exercise" mentioned in the question is only standard when you consider characteristic 0 irreducible representations as is usually done in a first course (then it's also usual to start with a big field like $\mathbb{C}$ to avoid extra complications).
When dealing with modular representations, there are lots of further issues.


Qiaochu attempts an outline of the "standard exercise" which shows clearly where problems arise. In his first step, you need characteristic 0 (and a splitting field) to be confident that the degree of an irreducible representation divides the group order. This is definitely false in general. And then of course characteristic 2 is especially dangerous for dealing with the negative of the $2 \times 2$ identity matrix. 


The bottom line is that you have to avoid this "standard exercise" and focus just on the special situation you are studying, in order to avoid confusion.


[ADDED] 1) Concerning the "standard exercise" (say over $\mathbb{C}$), I recall that decades ago it appeared as a *Math Monthly* problem (which I wrote an answer for) and later became an exercise in at least one advanced algebra text. Given an absolutely irreducible representation of degree 2 for a (necessarily nonabelian) finite simple group, classical character theory implies that the group order is even, so you can apply Cauchy etc. Of course, Feit-Thompson gets you there as well, but that's overkill. However, in odd characteristic that does help to get through the other steps of the proof, but in characteristic 2 you'd still face $I = -I$. 


2) While the degrees of irreducible representations of simple groups of Lie type are far from being known explicitly in general, over a splitting field of the defining characteristic you can see easy examples where degrees fail to divide the group order. For instance, take $\mathrm{PSL}\_2(\mathbb{F}\_7)$ in characteristic 7. Here you have an irreducible of degree 5 whereas 5 doesn't divide the group order 168.


