= Solution
By the <Fundamental theorem of finitely generated abelian groups>, write
$$
A\cong\mathbb Z^r\oplus F
$$
with $F$ finite. Taking <profinite completions> gives
$$
\widehat A\cong\widehat{\mathbb Z}^{,r}\oplus F.
$$
If $A\cong\mathbb Z$, this is plainly $\widehat{\mathbb Z}$.
Conversely, suppose $\widehat A\cong\widehat{\mathbb Z}$. Reduction modulo a prime $p$ gives
$$
A/pA\cong(\mathbb Z/p\mathbb Z)^r\oplus F/pF,
$$
whereas $\widehat{\mathbb Z}/p\widehat{\mathbb Z}\cong\mathbb Z/p\mathbb Z$. Choosing $p\nmid|F|$ first shows $r=1$. If $F\ne0$, choosing a prime divisor $p$ of $|F|$ makes $F/pF\ne0$, a contradiction. Thus $F=0$ and $A\cong\mathbb Z$.
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