# 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 I wanted to show now, is that because every polynomial from $\mathbb{Z}[a]$ can be written as a polynomial from $\mathbb{Z}[a+1]$ (this requires solving a system of linear equations for the coefficients), it has the same basis. Is this argument correct?

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Yes it is. Indeed, $\mathbb Z[a]$ and $\mathbb Z[a+1]$ are the same thing (as subrings of $\mathbb C$, for example) –  Mariano Suárez-Alvarez May 27 '12 at 2:05

As Mariano already noted, you are indeed correct $\mathbb{Z}[a]=\mathbb{Z}[a+1]$ (e.g. as subrings of $\mathbb{C}$).