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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?

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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 $r$.

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    $\begingroup$ For any ring $A$, the unique homomorphism $\mathbb{Z} \to A$ allows any integer to be viewed as an element of $A$. $\endgroup$ – user14972 Jun 3 '16 at 1:36
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    $\begingroup$ @Hurkyl. the answer makes it clear that it also considers rings without 1. $\endgroup$ – Carsten S Jun 3 '16 at 11:14
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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).

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    $\begingroup$ So, in $F_p$, it would be $\binom{n}{k} \pmod{p}$? $\endgroup$ – Zubin Mukerjee Jun 2 '16 at 20:08
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    $\begingroup$ Yes. To interpret an integer $n$ in a ring, you add together $n$ copies of the multiplicative identity of the ring. $\endgroup$ – Eric Wofsey Jun 2 '16 at 20:09
  • $\begingroup$ That makes sense :) $\endgroup$ – Zubin Mukerjee Jun 2 '16 at 20:09
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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).

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