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I want to solve the following exercise from Dummit & Foote's Abstract Algebra:

Let $p$ be an odd prime and let $n$ be a positive integer. Use the Binomial Theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$ but $(1+p)^{p^{n-2}} \not \equiv 1 \pmod{p^n}$. Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $\left( \mathbb{Z}/p^n \mathbb{Z} \right)^\times$.

I have managed to show that $$(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n} $$ for all primes $p$ (even $p=2$), and for all $n \in \mathbb{Z}^+$.

However, I got stuck while trying to prove that $$(1+p)^{p^{n-2}} \not \equiv 1 \pmod{p^n} $$ (obviously, this statement makes sense only for $n \geq 2$, and here the requirement that $p$ is odd is essential). I've expanded the LHS using the binomial theorem, and I counted the powers of $p$ in each term. My proof will be complete provided that the following holds for all odd primes $p$, integers $n \geq 3$ and $2 \leq k \leq p^{n-2}$:

$$\sum_{m=1}^\infty \left\lfloor \frac{k}{p^m} \right\rfloor+\left\lfloor \frac{p^{n-2}-k}{p^m} \right\rfloor=\frac{p^{n-2}-1}{p-1}+2-n $$

Can anyone help me prove this? Alternatively, can anyone provide me with another solution of the problem?

Thanks.

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Your identity does not always hold. For example, for $p = 5, n = 5$ and $k = 5$, the right hand side is

$$\frac{5^3-1}{4} - 3 = 31-3 = 28,$$

but the left is

$$\left\lfloor \frac{5}{5}\right\rfloor + \left\lfloor \frac{120}{5}\right\rfloor + \left\lfloor \frac{120}{25}\right\rfloor = 1 + 24 + 4 = 29.$$

Suggestion for an alternative proof: For an odd prime $p$ and $k\in\mathbb{N}$ we have

$$(1+p)^{p^k} \equiv 1 + p^{k+1} \pmod{p^{k+2}}.$$

That is easily proved by induction.

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  • $\begingroup$ Where does the oddness of $p$ come in? $\endgroup$ – Guacho Perez May 15 '17 at 16:36
  • $\begingroup$ @GuachoPerez For $p = 2$ and $k \geqslant 1$, we have $(1+p)^{p^k} \equiv 1 + p^{k+2} \pmod{p^{k+3}}$, which comes because in the binomial expansion of $(1 + 2)^2$ we have two terms $\equiv 2^2 \pmod{2^3}$ and those add to yield $(1+2)^2 = 1 + 2^3 \not\equiv 1 + 2^2 \pmod{2^3}$, while for $p > 2$ there is only one term divisible by $p$ but not by $p^3$ in the expansion of $(1+p)^p$, namely $\binom{p}{1}p=p^2$. Thus for $p = 2$ the exponent is one greater than for $p>2$. Hence for $n \geqslant 3$ the group $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ is not cyclic (it's $\cong C_2\times C_{2^{n-2}}$). $\endgroup$ – Daniel Fischer May 15 '17 at 17:50

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