Show that a certain set of elements is a basis of the free module $\mathbf{Z}[\xi]$ Let $\xi$ be a $p$-th root of unity for $p$ a prime. It is well-known that $\mathbf{Z}[\xi]$ is a free $\mathbb Z$-module. 

Now I'd like to show that $1, (1-\xi)^2, ..., (1-\xi)^{p-1}$ is a basis of $\mathbf{Z}[\xi]$. 

Since I know that the rank of $\mathbf{Z}[\xi]$ is $p-1$, we only need to show that $1, (1-\xi)^2, ..., (1-\xi)^{p-1}$ is a set of linearly independent elements (over $\mathbf{Z}$). This is where I'm stuck and need help.
However, I do not know if this is even a basis, it is only a guess. 
Thanks for any help!
 A: It is not sufficient to show that a set is linearly independent to show that it is an integral basis. For instance, consider $ \mathbf Z[\sqrt{2}] $ which is a free $ \mathbf Z $-module of rank $ 2 $ - the set $ \{ 1, 2 \sqrt{2} \} $ is linearly independent over $ \mathbf Z $, but it is not an integral basis of $ \mathbf Z[\sqrt{2}] $, for instance, $ \sqrt{2} $ does not lie in its integral span.
Unfortunately, your set is not an integral basis - the easiest way to see this is to show that the change of basis matrix from an integral basis we know to this basis has determinant $ \neq \pm 1 $. Clearly $ \mathbf Z[\xi] = \mathbf Z[1 - \xi] $, so the set $ \{ 1, 1 - \xi, (1 - \xi)^2, \ldots, (1 - \xi)^{p-2} \} $ is an integral basis of $ \mathbf Z[\xi] $. Write the change of basis matrix from this basis to the linearly independent set in question, and use expansion along the second row to compute its determinant. You will see that the determinant turns out to be $ p(p-1)/2 $ up to a sign, so in fact the integral span of your set is a submodule of index $ p(p-1)/2 $ in $ \mathbf Z[\xi] $.
