Basis of $\mathbb{Z}[\zeta_p]/p\mathbb{Z}[\zeta_p]$ over $\mathbb{Z}_p$.

Question

Let $\zeta_p$ be a primitive $p$-root of unity. We know that $\{1,\zeta_p,\ldots,\zeta_p^{p-2}\}$ is a integral basis of $\mathbb{Z}[\zeta_p]$ over $\mathbb{Z}$. Is true that if we quotient both with the ideal generated by $p$, and we see $\mathbb{Z}[\zeta_p]/p\mathbb{Z}[\zeta_p]$ as vectorial space over $\mathbb{Z}_p$, we keep the same basis? Is this a fact true in general, or just in some cases?

Answer 1

I’m rather a duffer at this, but if you start with a free $R$-module $M$, of rank $m$ and tensor with an $R$-algebra $S$ to get $M\otimes_RS$, then the result is still free over $S$ with the corresponding basis. The important thing is freeness of $M$ over $R$.

That’s the fancy talk. You’re taking a ring that’s free over $\Bbb Z$ and tensoring with $\Bbb Z/p\Bbb Z$, so the result will be free over the field, with the “same” basis.

For my part, I understand this particular situation better by taking the basis $\{1,\pi,\pi^2,\cdots,\pi^{p-2}\}$, where $\pi$ is the local prime element $\pi=\zeta_p-1$. This is root of a $p$-Eisenstein polynomial, as you can check, so that $\pi^{p-1}$ is divisible by $p$. All becomes clear now (to me).