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Does Lucas' Theorem hold for $p$-adic integers? More specifically, does it specifically hold for the case that, given a $p$-adic integer: $$x = x_0 + x_1p + x_2p^2 + \cdots = \sum_{i=0}^\infty x_i p^i$$ where each $x_i \in \{0,1,\ldots,p-1\}$, is it true that for each $n$ a natural number or zero, $$\binom{x}{p^n} \equiv x_n \bmod p\ ?$$ If it is true, does that mean that it holds if we do not put the integer in canonical form?

For example, suppose $a$ is a p-adic integer with a multiplicative inverse and that the canonical representations for $a$, $-a$, and $1/a$, are given by, $$a = \sum_{i=0}^\infty ( a_i p^i ) \ \land \ -a = \sum_{i=0}^\infty ( b_i p^i ) \ \land \ 1/a = \sum_{i=0}^\infty ( c_i p^i )$$ that is, for each $i$, $a_i , b_i , c_i \in \{0,\ldots,p-1\}$. Then is it true that $$ \binom{-a}{p^n} = (-1)^{p^n} \binom{a+p^n-1}{p^n} \equiv b_n \mod p $$ $$ \binom{1/a}{p^n} = \frac{\prod_{i=0}^{p^n-1} (1-a \times i)}{a^{p^n} \times (p^n)!} \equiv c_n \mod p $$ For the case that $p=2$, I've checked the first few digits for the negation operation $-a$ and the first two or three digits for $1/a$ and it seems to hold. But is it true in general?

Clarification concerning interpretation of binomial coefficients: For powers of p in the denominator of binomial coefficients and fractions modulo p, common p factors of the numerator and denominator are to be canceled out before interpreting the fraction mod p so that division by zero will not occur. If common p factors are canceled out and there are still factors of p in the denominator then the fraction is undefined mod p.

Any help or information is very much appreciated. Thank you for your time and have a great day.

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    $\begingroup$ How do you define the binomial coefficient? Recall that $\mathbb{Z}_p$ is just a ring, so that division will be restricted. $\endgroup$ – Martin Brandenburg Oct 19 '14 at 1:21
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    $\begingroup$ Common factors of p are to be canceled before reduction mod p occurs. If there are still factors of p in the denominator after canceling common factors of p, then the fraction is undefined. Thanks for catching that. $\endgroup$ – DAS Oct 19 '14 at 1:47
  • $\begingroup$ This is definitely true. You can write this as a polynomial identity in $\mathbb{Z}/p^n$. Since $\mathbb{Z}/p^n \cong \mathbb{Z}_p / p^n$, it follows that it's equally valid for $\mathbb{Z}_p$. It's a very interesting observation that this gives a neat way of comparing different binomial coefficients. $\endgroup$ – Slade Oct 19 '14 at 2:42
  • $\begingroup$ I don't follow. How is $\mathbb{Z}/p^{n} \cong \mathbb{Z}_p/p^{n}$? You're saying they are isomorphic? But isn't $p^{n-1}$ identically zero in $\mathbb{Z}_p/p^{n}$ while it is not identically zero in $\mathbb{Z}/p^{n}$? $\endgroup$ – DAS Oct 19 '14 at 3:27
  • $\begingroup$ @DAS: Nope: $p^{n-1}$ isn't a multiple of $p^n$ in $\mathbf{Z}_p$, so it can't be zero in $\mathbf{Z}_p / p^n$. $\endgroup$ – Hurkyl Oct 19 '14 at 9:51

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