-1

Let $p$ be prime and $n = p^2$. Show that the polynomial $1+px$ has a multiplicative inverse in $\mathbb{Z}_n[x]$. Is this polynomial invertible if $n = p^k$ for arbitrary $k>0$?

user26857
  • 1
  • 13
  • 62
  • 125
kt046172
  • 505
  • 2
  • 15

2 Answers2

2

$(1+px)(1-px)=1-p^2x^2=1$ and so $1+px$ has a multiplicative inverse.

If $n=p^k$ the same idea works. The inverse is $$1-px+(px)^2-(px)^3+...$$

0

Well we need $(1+px)P(x) \equiv 1$ so if $P(x) = \sum_{k=0}^m a_k x^k$ we must have

$(1+px)\sum_{k=0}^m a_k x^k = \sum_{k=0}^m a_k x^k + \sum_{k=0}^m pa_k x^{k+1}= a_0 + \sum_{k=1}^m (pa_{k-1}+ a_k)x^k + pa_m x^{m+1}$ were

$a_0 \equiv 1 \pmod n$ for $k=1,....m$ that $pa_{k-1}\equiv -a_{k}\pmod n$ and that $pa_m \equiv 0 \mod n$

So $pa_0 \equiv p \equiv -a_1\pmod n$ so $a_1 \equiv -p\pmod n$. And $p*a_1 = -p^2\equiv 0\equiv -a_2$ so $a_2 \equiv 0\pmod n$.

So $P(x) = 1 - px$ and $(1 + px)(1-px) = 1-p^2x^2 \equiv 1\pmod n$.

That's it.

fleablood
  • 1
  • 5
  • 39
  • 125