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I'm new in this forum. I have a question about the proof of Lucas' theorem on Wikipedia (link).

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The proof is completed by looking at the coefficients of $X^n$ on both sides, ${m \choose n}$ and $\prod{m_i \choose n_i}$ respectively. And it comes out that the coefficients are congruent $\bmod p$.

But how can we say this? As I know if two polynomials of the same degree are congruent $\bmod p$, that doesn't imply that corresponding coefficients are also congruent $\bmod p$. Then how is Lucas' theorem proved?

Zev Chonoles
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shota kobakhidze
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    I think the key is in your last paragraph. You are simply mistaken. When the text says that two polynomials are congruent modulo $p$ it means exactly that their corresponding coefficients are congruent modulo $p$. Why do you think that would not be the case? – Jyrki Lahtonen Aug 28 '16 at 11:13
  • i have come across some articles which were against it and also this page with a counter example: http://math.stackexchange.com/questions/163490/congruent-polynomials – shota kobakhidze Aug 28 '16 at 11:21
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    You (and possibly also some posters in that thread) are confusing congruence in the polynomial ring with congruence of the values of two polynomials at integer points. In other words, you need to be careful with the distinction between formal polynomials and polynomial functions. The difference is that two formal polynomials $p$ and $q$ are equal modulo $n$, iff their coefficients are so. But they give rise to the same function if $p(k)\equiv q(k)\pmod n$ for all integers $k$. In this proof the focus is on the formal polynomials. – Jyrki Lahtonen Aug 28 '16 at 11:45
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    Observe that the argument does work with formal polynomial (or their coefficients). For example when $p=3$ we do have $$(1+x)^3=1+3x+3x^2+x^3\equiv1+x^3\pmod3.$$ – Jyrki Lahtonen Aug 28 '16 at 11:49
  • how can we say that there are formal polynomials in this proof? – shota kobakhidze Aug 28 '16 at 12:09
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    In abstract algebra contexts formal polynomials are the default. It takes a while to get used to that. When I was an undergrad I had not fully internalized it within a year to getting exposed to the idea. Basically because the difference is important only in relatively few areas of math. Probably whoever wrote that piece assumed that their readers either know this or can figure it out. Or (possibly more likely), they did not want to spend time explaining this. – Jyrki Lahtonen Aug 28 '16 at 12:17
  • does "formal" in this case mean that $X$ isn't required to be an integer? – shota kobakhidze Aug 28 '16 at 15:44
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    That's one way of thinking about it. It is just a variable, so thinking of it as any kind of number is "wrong". With formal polynomials we never really give $X$ any kind of value. Sorry about not bringing that up right away. – Jyrki Lahtonen Aug 28 '16 at 16:00

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There are two polynomials under discussion, which I will name $f(X)$ and $g(X)$: $$f(X)=\sum_{n=0}^m\binom{m}{n}X^n\qquad \qquad g(X)=\sum_{n=0}^m\left(\prod_{i=0}^k \binom{m_i}{n_i}\right)X^n$$ Note that $f(X),g(X)\in\mathbb{Z}[X]$. That is, they have integer coefficients.

For any two polynomials $a(X),b(X)\in\mathbb{Z}[X]$, if we say that $$a(X)\equiv b(X)\bmod p$$ then that is intended to mean that $a(X)$ and $b(X)$ represent the same element of $(\mathbb{Z}/p\mathbb{Z})[X]$, which is equivalent to each of their coefficients being equivalent mod $p$. For example, if $$a(X)=4-3X+9X^2+4X^8,\quad b(X)=6+15X-X^2-10X^7\in\mathbb{Z}[X]$$ then $$a(X)\equiv b(X)\bmod 2$$ since they are both equivalent to the element $X+X^2\in(\mathbb{Z}/2\mathbb{Z})[X]$.

Thus, since the argument shows $f(X)\equiv g(X)\bmod p$, we can conclude $$\binom{m}{n}\equiv\prod_{i=0}^k \binom{m_i}{n_i}\bmod p$$ for each $0\leq n\leq m$.

Zev Chonoles
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