As the title explains, I'm trying to solve a question which asks me to determine which are the irreducible elements of the ring of numbers of the form $2^ab,$ where $a$ and $b$ are integers (with the usual addition and multiplication).
I have no idea to do this, so I'd really appreciate any help you could give.
On
In a totally elementary fashion, note first that the integer prime $2$ has become a unit in this ring $R$.
Now show that every element of $R$, which is not zero nor a unit, can be written, up to unit factors, as the product of odd integer primes. (Just note that in the representation $2^{a} b$, you may take $b$ to be odd.)
And then, show that the odd integer primes remain prime in $R$. (Let $p$ be such a prime, and suppose it divides $2^{a_{1}} b_{1} \cdot 2^{a_{2}} b_{2}$ in $R$ (thus $a_{i}, b_{i}$ are integers). Then it divides $2^{a} b_{1} b_{2}$ in $\mathbb{Z}$ for some $a \ge 0$, and thus it divides either $b_{1}$ or $b_{2}$.)
On
We write $b$ into the form $b = 2^ic$, while c is an odd integer. And $2^ab=2^a(2^ic)$
If $c=1$, $2^ab=2^a2^i$ which is a unit
If $c$ is a composite number, for example $c=xy$, then $2^ab=x2^a*y2^i$, which is reducible
Only if $c$ is an odd prime, $2^ab$ is irreducible
In conclusion, the irreducible elements can be written in the form $2^ab$, while $b = 2^i*p$
($p$ is an odd prime)
This is the localisation of $\mathbb Z$ at two. Thus, it is also a UFD. Hence, the irreducible elements are precisely the primes, and by the correspondence theorem, these are the primes $p \neq 2$ multiplied by some power of $2$.