Existence of inverses in $ℤ/p^nℤ$

Question

I am working through text on number theory by Borevich and Shafarevish, section on p-adic numbers. I can not grasp the idea of part of the proof, which can be very easily formulated.

Imagine $a \neq0, a \in ℤ/pℤ$, $p$ is prime. Then $a$ has an inverse in $ℤ/pℤ$, because $ℤ/pℤ$ is a field. It is well known fact and I understand it and its proof. What I do not understand is the following:

They conclude, that $a$ has an inverse in $ℤ/p^nℤ$ for any natural $n$. I am frustrated because I can not understand why it is true. $ℤ/p^nℤ$ is not a field. I tried to construct inverses explicitly in $ℤ/p^nℤ$, but failed. I also failed to find a counterexample.

Please advice. Thanks for you help!

Answer 1

Recall that an integer $a$ is invertible in $\mathbb{Z}/m\mathbb{Z}$ if and only if $a$ and $m$ are relatively prime.

In your case, since $a\neq 0$ in $\mathbb{Z}/p\mathbb{Z}$ it follows that $a$ is not divisible by $p$. This means that $a$ and $p^n$ are relatively prime for all natural numbers $n$, hence $a$ is invertible in $\mathbb{Z}/p^n\mathbb{Z}$.

Answer 2

I don't have the book that you're referring to, but it's not true that every nonzero $a \in \mathbb{Z}/p^n\mathbb{Z}$ has an inverse in $\mathbb{Z}/p^n\mathbb{Z}$. For example, 6 has no inverse in $\mathbb{Z}/8 \mathbb{Z}$, because every integer congruent to $1 \bmod 8$ is odd, but every multiple of 6 is even. In general, no multiple of $p$ has an inverse mod $p^n$, for the same reason.

If $a \in \mathbb{Z}$ and $p^n$ are relatively prime, though, then $a$ has an inverse mod $p^n$: Bézout's identity guarantees that there are some integers $r, s$ such that $ar + p^ns = 1$, so $a$ and $r$ are modular inverses. $a$ and $p^n$ are relatively prime iff $a$ is not divisible by $p$, which is equivalent to saying that $a$ corresponds to a nonzero element in $\mathbb{Z}/p\mathbb{Z}$—perhaps this is what the book means?