Exercise 9.1.1

Proof. If $[i]$ has a multiplicative inverse in the ring $\mathbb{Z}∕\mathit{n\mathbb{Z}}$, then there exists $[j]\in \mathbb{Z}∕\mathit{n\mathbb{Z}}$ such that $[i][j]=[\mathit{ij}]=1$, so $\mathit{ij}\equiv 1[n]$. Thus there exists $k\in \mathbb{Z}$ such that $\mathit{ij}-\mathit{kn}=1$. This Bézout’s relation between $i$ and $n$ shows that $i\wedge n=1$.

Conversely, if $i\wedge n=1$, by Bézout’s Theorem, there exist integers $j,k$ such that $\mathit{ij}-\mathit{kn}=1$, so $[i][j]=[1]$, and $[i]$ has a multiplicative inverse $\mathbb{Z}∕\mathit{n\mathbb{Z}}$.

The mapping

obtained by restriction of the bijection $\text{[[}0,n\text{[[}\to \mathbb{Z}∕\mathit{n\mathbb{Z}},i\mapsto [i]$, is well defined, and this is a bijection.

Therefore

If $n=1$, the ring $\mathbb{Z}∕1\mathbb{Z}$ is the trivial ring $\{[0]\}$, where $[0]=[1]$, so the multiplicative group ${(\mathbb{Z}∕1\mathbb{Z})}^{\text{∗}}=\{[1]\}$ has one element, and the set of integers $i$ such that $0\le i<1=n$ is reduced to $\{0\}$, which satisfies $0\wedge 1=1$, so $\varphi (1)=1$. □