# Congruence modulo prime power

In the book "A Classical Introduction to Modern Number Theory", I saw the following theorem (p. 43):

If $p\neq 2$, and $p\nmid a$ then $p^{l-1}$ is the order of $(1+ap)$ mod $p^l.$

i.e. $(1+ap)^{p^{l-1}}\equiv 1(\text{mod } p^{l})$, and $(1+ap)^{p^k}\not\equiv 1(\text{mod } p^l)$ for $0< k< p^{l-1}.$

The question is:

Suppose we consider the finite field $\mathbb{F}_{p^l}$ , $p\neq 2$, and take $a=1$. Then by above theorem, the multiplicative order of $(1+p)$ in $\mathbb{F}_{p^l}$ is $p^{l-1}$. But, as $1+p$ is non-zero element of the Field $\mathbb{F}_{p^l}$, so it is an element of the cyclic group $\mathbb{F}^*_{p^l}$ of order $p^l-1$, so its multiplicative order should divide $p^l-1$, and hence can not be prime power, hence the order can not be $p^{l-1}.$

What's going wrong here?

$\mathbb{F}_{p^l}$ has nothing to do with $\mathbb{Z}/p^l\mathbb{Z}$. When you do stuff mod $p^l$, you do stuff in the ring $\mathbb{Z}/p^l\mathbb{Z}$, whose group of invertible elements, $(\mathbb{Z}/p^l\mathbb{Z})^*$, has $\phi(p^l) = p^{l-1}(p-1)$ elements, and is not necessarily cyclic.
• It might be added that although the multiplicative group of $\mathbb{Z}/p^l\mathbb{Z}$ is not always cyclic, it is almost always so. It is cyclic whenever $p$ is odd or $l \leq 2$ – Tobias Kildetoft Apr 11 '11 at 13:19
The finite field $\mathbb{F}_{p^\ell}$ is not the same thing as the ring of integers modulo $p^{\ell}$! (Except in the case $\ell=1$, of course.)
To construct $\mathbb{F}_{p^\ell}$, take a root $\alpha$ of an irreducible degree $\ell$ polynomial over $\mathbb{F}_p$; then $\mathbb{F}_{p^\ell} \simeq \mathbb{F}_p(\alpha)$.
The integers modulo $p^\ell$ form a ring, not a field. Another big difference is that $\mathbb{F}_{p^\ell}$ has characteristic $p$ (so $p \alpha = 0$), but of course there are many integers $a$ such that $pa \not \equiv 0 \pmod{p^{\ell}}$.