3
votes
1answer
56 views

Free module, $\mathbb{Z}[a]$ over $\mathbb{Z}[(a+1)^2]$ for transcendental number a

I'm trying to prove that for a transcendental number $a$ the module $\mathbb{Z}[a]$ over $\mathbb{Z}[(a+1)^2]$ is free. For $\mathbb{Z}[a+1]$ over $\mathbb{Z}[(a+1)^2]$, the basis is $\{1,a+1\}$. What ...