0

Let $p$ be an odd prime number and $n$ be a positive integer. Use the binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \mod p^n$ but $(1+p)^{p^{n-2}} \ne 1 \mod p^n$ Deduce that $(1+p)$ is an element with order $p^{n-1}$ of the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$

I started with $\sum_{k=0}^{p^{n-1}} {p^{n-1} \choose k} 1^k p^{p^{n-1}-k}$

Writing out terms looks like $(1)p^{p^{n-1}-0} + (1)p^{p^{n-1}-1} + (1)p^{p^{n-1}-2} + ... + (1)p^{p^{n-1} - p^{n-1}}$ is this the correct way to write this sum?

$p^{p^{n-1}} + p^{p^{n-1}}p^{-1} + p^{p^{n-1}}p^{-2} + ... + p^{p^{n-1}}p^{-p^{n-1}}$ reduces to $p^{p^{n-1}}(p^{-1} + p^{-2} + ... + p^{-p^{n-1}})$ somehow evaluates to $1 \mod p^n$

Any help appreciated, thanks!

Obliv
  • 407
  • What happened to the binomial coeffcients? – quid Jul 11 '16 at 17:16
  • @quid Someone told me I was missing coefficients, too. I don't see how. I'm simply writing out the terms of the sum. Maybe I'm doing it wrong? – Obliv Jul 11 '16 at 17:21
  • In the sum you have binomial coefficents. Why do you not write those? Differently each summand has 3 terms. You then write 2. – quid Jul 11 '16 at 17:22
  • sorry @quid but can you show me how to do this then? $(1+p)^5 = \sum_{k=0}^{5} {5 \choose k} 1^{k}p^{5-k} = ? $ is this correct set-up? Then don't I have terms $1p^5 + 1p^4 + 1p^3 + ... + 1$? – Obliv Jul 11 '16 at 17:31
  • Obliv, if you look up the binomial theorem and write out your coefficients carefully, I think your first set of problems will solve itself. – The Count Jul 11 '16 at 17:43
  • A cleaner way to write the sum is

    $$\sum_{k=0}^{p^{n-1}} {p^{n-1} \choose k}p^{k}={p^{n-1} \choose 1}p^0+{p^{n-1} \choose 2}p^1+\cdots+{p^{n-1} \choose p^{n-1}}p^{p^{n-1}}$$

     – pancini Jul 11 '16 at 17:47
  • Ohhh.. the coefficient would be ${p^{n-1} \choose k}$ right? How is that evaluated as $k$ increments? – Obliv Jul 11 '16 at 17:48
  • What you need to do is figure out which power of $p$ divides the binomial coefficients. – quid Jul 11 '16 at 17:55

1 Answers1

1

We have

$$(a+b)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}ka^{p^{n-1}-k}b^k\implies(1+p)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}kp^k$$

Now prove (something very similar to this appears in a standard proof of Sylow's first theorem)

Lemma: For $\;m\in\Bbb N\;,\;\;0\le m\le n-1\;$ and $\;0< k< p^{n-1}$ , we have that

$$\;p^m\,\mid\,\left(p^{n-1}-k\right)\;\iff p^m\,\mid\,k$$

Assuming the above, we have that

$$\binom{p^{n-1}}k=\frac{\left(p^{n-1}-(k-1)\right)\left(p^{n-1}-(k-2)\right)\cdot\ldots\cdot p^{n-1}}{k(k-1)(k-2)\cdot\ldots\cdot1}\implies p^{n-1}\,\mid\,\binom{p^{n-1}}k$$

and it follows that $\;\binom{p^{n-1}}k p^k=0\pmod{p^n}\;$ , so we get

$$(1+p)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}kp^k=\left(1+p^{p^{n-1}}\right)\pmod{p^n }=1\pmod{p^n}$$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • I'm alright until the evaluation of ${p^{n-1} \choose k} = ...$ I was under the impression it was $\frac{p^{n-1}!}{k!(p^{n-1} - k)!}$ not $\frac{(p^{n-1}-k)!}{k!}$ – Obliv Jul 11 '16 at 19:01
  • @Obliv And your impression is correct. I just cancelled terms... – DonAntonio Jul 11 '16 at 19:28