Exercise 2.2.14* (Generalization of Problem 2.1.45)

Question

Show that $\binom{p^{\alpha}}{k} \equiv 0 \pmod p$ for $0 < k < p^\alpha$.

Richard Ganaye · Accepted Answer

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ewcommand{\N}{\mathbb{N}}

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\begin{proof} Here $p$ is a prime number.

By Exercise 2.1.45, $\binom{p}{k} \equiv 0 \pmod p$ for $0<k<p$. This imply that the polynomial $f(x) = (1 + x)^p \in \F_p[x]$ (where $x$ is an variable) satisfies the equality in $\F_p[x]$
$$(1 +x)^p = 1 +x^p.$$
If we substitute $x^r$ to $x$ (formal composition), we obtain $(1 + x^r)^p = 1 + x^{rp}$.
Reasoning by induction suppose that $(1+x)^{p^k} = 1 + x^{p^k}$. Then $$(1+x)^{p^{k+1}} = \left[(1 +  x)^{p^k}ight] ^p= (1 + x^{p^k})^p = 1+x^{ p^k p} = 1 + x^{p^{k+1}},$$ and the induction is done. Therefore, for all $\alpha \in \N$, the equality
$$(1+x)^{p^\alpha}  = 1 + x^{p^\alpha}$$
is true in $\F_p[x]$.
Since $(1+x)^{p^\alpha} = \sum_{k = 0}^\alpha \binom{p^\alpha}{k} x^k$, the comparison of coefficients of $x^k$ for $0 <  k < p^\alpha$ gives $\left[\binom{p^\alpha}{k}ight]_{p} = [0]_p$, that is
$$\binom{p^\alpha}{k} \equiv 0 \pmod p, \qquad 0 < k < p^\alpha.$$
\end{proof}

Exercise 2.2.14* (Generalization of Problem 2.1.45)

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Exercise 2.2.14* (Generalization of Problem 2.1.45)

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