Defining binomial coefficients over rings in general I'm looking over my notes, and we proved that the binomial theorem $$(a + b)^n = \sum_{k = 0}^{n} {n \choose k}a^k b^{n-k}$$ holds on all commutative rings. 
However, I am confused on how we are defining  ${n \choose k}$, since the definition $ {n \choose k} = \frac{n!}{(n-k)!k!} $ is no longer well defined over rings in general, since $k!$ may equal $0$. For example, $p! = 0$ in $F_p$, the field of p elements. Any idea how to interpret the binomial coefficients in these cases?
 A: If the division bothers you, it is entirely possible to use a definition for the binomial coefficients that does not involve dividing anything at all. In particular, you can use the recurrence relation $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$ with the starting conditions $$\binom{n}{0}=\binom{n}{n}=1$$ as a definition (which of course is just the mathematical way of presenting the construction of Pascal's triangle).
A: In the formula, $\binom{n}{k}$ is an ordinary integer. It need not be viewed as an element of our  ring $A$, and in general commutative rings, which may not have a multiplicative unit, it cannot be so viewed.
If $r$ is a ring element, then $\binom{n}{k}r$ is interpreted as the sum in the ring of $\binom{n}{k}$ copies of $r$.
A: Although the expression ${n \choose k} = \frac{n!}{(n-k)!k!}$ involves division, it actually always evaluates to be an integer (namely, the cardinality of the set of $k$-element subsets of a set with $n$ elements).  So in a general ring, ${n\choose k}$ just refers to that integer in that ring (even though the formula $\frac{n!}{(n-k)!k!}$ may not make sense).
