# What are the irreducible elements of $\mathbb {Z}_2$

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.

## 1 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.

• I don't understand "which correspond bijectively to the prime ideals of $\mathbb {Z}$".Prime ideal of $\mathbb {Z}_2$ is $\frac {p}{1}\mathbb {Z}_2$? – Made Feb 24 at 23:14
• That's a general fact about localisations: for any multiplicative subset $S$ of a commutative ring $R$, the prime ideals of $S^{-1}R$ have the form $S^{-1}\mathfrak p$, for exactly one prime ideal $\mathfrak p$ of $R$ such that $\mathfrak p\cap S=\varnothing$. – Bernard Feb 24 at 23:21