In a book on algebra I'm currently working with a proof that uses the binomial theorem for $(x+y)^m$ where $x,y$ are elements of some arbitrary field $k$. This looks strange to me, so I did some research on the Internet, but all pages which state (and prove) this theorem never specify what $x$ and $y$ are. A binomial coefficient is defined by faculties and fractions of integers. This is not evaluable in a general field that does not extend the integers like $\mathbb Q$.
We can use the map $$\mathbb Z \to k,\quad n\mapsto \sum_1^n 1$$ to get a representation of the integers in $k$, but this is only a group, and it is not clear (to me), why the inverses of these internal representations in $k$ are internal integers as well (so that we could define the binomial coefficient via internal integers).
A further question would be, whether we can weaken the requirements from general fields to (not necessarily general) rings. Except for the definition of binomial coefficients, we don't need inverses in the statement, so maybe this is possible as well in some cases (of course this works for all rings that we obtain via a forgetful functor from a field, but I mean rings that are not fields).
Best Answer
I think you're looking at $$ (x+y)^n = \sum_{i=0}^n \binom ni x^i y^{n-i} $$ and thinking that the multiplication of $\binom ni$ with the other factors is the multiplication operation of the field. It's not, really; it's the "scalar" multiplication $\mathbb Z\times k\to k$ that does repeated addition. (Groups are $\mathbb Z$-modules, you see.) The binomial coefficient is really an integer.
Indeed, to prove the binomial theorem, you'd multiply out $(x+y)^n$ and then gather like terms. It's that gathering that produces the binomial coefficients — you count how many terms you have of each type. That's plain old counting, so it yields plain old integers, not elements of the field.
The binomial theorem holds, in this form, in any commutative ring. (We need multiplication to be commutative for the "like terms" to actually be like each other.) (Well, as user26857 pointed out, it's enough that $x$ and $y$ commute with each other, and the scenario where $x$ and $y$ commute but the multiplication is not commutative in general does come up pretty often.)