Let $p$ be a prime number. Define the $p$-adic valuation on $\mathbb{Z}$ as $v_p(p^kx) = p^{-k}$ where $x$ is not divisible by the prime $p$. Let $\mathbb{Z}_p$ (the $p$-adic integers) be a completion of $\mathbb{Z}$ under this valuation.

The problem is to prove that if an integer $z$ is not divisible by $p$, then it is invertible in the ring $\mathbb{Z}_p$.

My thoughts by now are the following:

For any such $z$ there are integers $q \neq 0$ and $-p<r < p$ such that $ z = qp + r $. Then we claim existence of $p$-adic integer $a = \sum^\infty_{n=0}a_np^n$ with the properties:

  1. $b_0 =\mathbf{a_0}(r + qp) + rp\mathbf{a_1} = 1 $
  2. $\forall k : \mathrm{even} . b_k = qa_{k-1} + \mathbf{a_k}(r + qp) + rp\mathbf{a_{k+1}} =0$

If such $a$ exists then $za = 1$ as $za = \sum^\infty_{n=0}b_{2n}p^n$.

We will try to derive coefficients of $a$ from this conditions starting from $a_0$ and $a_1$. If $r+qp$ and $rp$ are coprime then we are done. This will be true in case $\gcd(r,q) = 1$, otherwise we will have problem with solving for $a_1, a_0$ in first condition. However, if $m = gcd(r,q)$ then $m < p$ which gives us representation of form $m = p - (p-m)$ and, as $q$ in this representation is $1$, number $m$ must be divisible in $\mathbb{Z}_p$. Same holds for $m^{-1}z = m^{-1}qp +m^{-1}r$ as $\gcd(m^{-1}q,m^{-1}r) = 1$. Then we may construct $z^{-1} = m^{-1}(m^{-1}qp +m^{-1}r)^{-1}$ which completes the proof.

Is this proof correct?

Do you know better proofs of this fact?


3 Answers 3


One possible slight improvement but still in the style you propose is as follows. Since $z$ is prime to $p$, there are integers $y$ and $k$ such that $yz=1+kp$. We can invert $1+kp$ in $\mathbb Z_p$ by expanding the geometric series $$ {1\over 1+kp} \;=\; 1-kp+(kp)^2-(kp)^3+\ldots $$ This converges very nicely in $\mathbb Z_p$. Then $$ \Big(y\cdot (1-kp+(kp)^2-\ldots)\Big) \cdot z \;=\; 1 $$


I would say:

Since $p$ does not divide $z$, $z$ is coprime to every $p^n$, and therefore it has an inverse $y_n$ modulo $p^n$.

If $n<m$, then $y_mz\equiv 1\pmod{p^m}$ and therefore also modulo $p^n$. Since $z$ can have only one inverse modulo $p^n$ we must have $y_m\equiv y_n\pmod{p^n}$, so the $p$-adic distance between $y_n$ and $y_m$ is at most $p^{-n}$. Thus, the sequence of $y_n$s is Cauchy and converges in $\mathbb Z_p$.

Furthermore the sequence $(y_nz-1)_n$ clearly converges to $0$, so the limit of the $y_n$s is inverse to $z$.


See Jean-Pierre Serre, A course in arithmetic, chapter II, section 2, "Properties of $Z_p$", Proposition 2, for a slightly more general result. To extend his proof to the case of $Z_p$, you just need to see that if $p$ doesn't divide an element $x=(x_n)\in Z_p$, then it doesn't divide any $x_n$ (for example, if $p$ divides $x_1$, using the morphism $\varphi_2$ you see that $p$ divides $x_2$, then $x_3$, then ...). So every $x_n$ is invertible in $Z/p^nZ$.


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