We consider the following sub-ring of $ \mathbb{Q} $:
$$ \mathbb {Z} _2: = \left\{\frac {a} {2 ^k}; a \in \mathbb{Z}, k \in \mathbb {N} \right\} $$
the set of units of $\mathbb {Z}_2^× = \{ ±2^k, k \in \mathbb {Z}\}$
Problem
What are the irreducible elements of $ \mathbb {Z}_2 $
My effort:
We have $2$ and $−2 $ were units of $ \mathbb {Z}_2 $. Let $ p $ be a prime number other than 2. (The case $-p $ is treated in the same way). If $ p = \frac {a}{2 ^ l} \frac {b} {2 ^ k} = \frac {ab} {2 ^{k + l}} $ then $ ab = 2 ^{k + l } p $. Hence, $ a = p2 ^m $ and $ b = 2 ^{k + l − m} $ (or the other way around). In this case, $ \frac {b} {2 ^ k} $ is a unit and $ p $ remains irreducible in $ \mathbb {Z} _2 $.
Question:
can we say that: The irreducible elements of $ \mathbb {Z} _2 $ are therefore the $ p $ and $ −p $ for $ p$ prime other than 2.
Best Answer
Yes, almost – since $\mathbf Z_2$ is the localisation of $\mathbf Z$ w.r.t. the multiplicative set $S=\{2^k\mid k\in \mathbf N\}$, therefore it is a P.I.D. and its irreducible elements generate prime ideals, which correspond bijectively to the prime ideals of $\mathbf Z$ which do not meet $S$, i.e. to the irreducible elements of $\mathbf Z$ different from $2$.
As irreducible elements are defined up to a unit factor, the associated elements of an irreducible element $\frac p1\in\mathbf Z_2$ are all $\pm \frac p{2^k}$, where $p$ is an odd prime of $\mathbf Z$ and $k$ is an integer.