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I would like to know which is the group of units of $\mathbb Z[\frac{1}{3}]=\bigl\{\frac{a}{3^n}\mid a \in \mathbb N, n \in \mathbb N \cup\{0\}\bigr\}$.

Thank you.

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Your ring is a subring of the field of rational numbers, so the inverses will be the same. So your task is to determine, which fractions $m/n$ meet the condition that both $m/n$ and $n/m$ are in your ring. – Jyrki Lahtonen Sep 30 '13 at 19:20
Hint: $\displaystyle\frac{3^n}{a}=\frac{b}{3^m}\implies ?$ – anon Sep 30 '13 at 19:21
I wrote U(Z[1/3])={3^k, where k€Z}. Is it correct? – Blanca Sep 30 '13 at 19:31
up vote 2 down vote accepted

Let's suppose that $\frac{a}{3^m}$ is invertible. Then there exist $b\in\mathbb Z$ and $n\ge 0$ such that $\frac{a}{3^m}\cdot\frac{b}{3^n}=1$ $\Leftrightarrow$ $ab=3^{m+n}$, so $a$ and $b$ are of powers of $3$. If $a=3^t$, then $\frac{a}{3^m}=\frac{1}{3^{m-t}}$. It follows that $U(\mathbb Z[1/3])=\{3^i:i\in\mathbb Z\}$. (This group is obviously cyclic isomorphic to $\mathbb Z$.)

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Thank you. So, it is the solution I gave: U(Z[1/3])={3^k: k∈Z}. – Blanca Sep 30 '13 at 19:55

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