A non-factorisation domain

Question

I have to check if $1+e$ is an irreducible element in $\mathbb{Q}[e,e^{1/2},e^{1/2^2},\ldots]$. I guess the given domain is a non-factorisation domain. I have no idea how to prove it. Also how to check the irreducibility of $1+e$?

Answer 1

Not an answer, but a couple of remarks that may simplify the solution:

• The only special property of $e$ here is that $e$ is transcendental over $\mathbb Q$. We could use $\pi$ instead.

• Since $\mathbb{Q}[e,e^{1/2},e^{1/2^2},\ldots] = \bigcup_n \mathbb{Q}[e,e^{1/2},e^{1/2^2},\ldots, e^{1/2^n}]$, we can work in a fixed $\mathbb{Q}[e,e^{1/2},e^{1/2^2},\ldots, e^{1/2^n}]=\mathbb{Q}[e^{1/2^n}] = \mathbb{Q}[\alpha]$, with $\alpha^{2^n}=e$.