A question about isomorphism and congruence in a cyclotomic extension.

Question

If $p$ is an odd prime and $\zeta=\zeta_p$ a primitive $p$-root of unity, let's consider $\mathbb{Z}(\zeta)$ and the element $\lambda=1-\zeta$. I have proved that $\mathbb{Z}(\zeta)/\lambda\mathbb{Z}(\zeta)\simeq \mathbb{Z}/p\mathbb{Z}$. Is this enough to say that every element of $\mathbb{Z}(\zeta)$ is congruent to a rational integer mod $\lambda$? I have read some papers using this, but is not really clear to me: in my opinion the quotient group $\mathbb{Z}(\zeta)/\lambda\mathbb{Z}(\zeta)$ isn't forced to contain just rational integers...

Answer 1

I try to give a constructive proof, basically by making the isomorphism $\mathbb{Z}(\zeta)/\lambda\mathbb{Z}(\zeta)\simeq \mathbb{Z}/p\mathbb{Z}$ explicit.

The minimal polynomial of $\zeta=\zeta_p$ is $\Phi_p(X)=(X^p-1)/(X-1)=X^{p-1}+\dots+X+1$. So $\Phi_p(X)=(X-\zeta)(X-\zeta^2)\dots(X-\zeta^{p-1})=\prod_\sigma (X-\sigma\zeta)$, where $\sigma$ walks in the Galois group of the field extension $\Bbb Q(\zeta):\Bbb Q$.

$\lambda=1-\zeta$ is a root of $F(X)=\Phi_p(1-X)= p\pm \dots-X^{p-1}$ (for an odd $p$), and we can write $F(X)=\prod_\sigma (\sigma\lambda-X)$. So the norm of $\lambda$ is $p$, the free coefficient above.

Now we start, and consider an arbitrary element $f(\zeta)$ in $\Bbb Z[\zeta]$, $f=f(X)\in \Bbb Z[\zeta]$. Let $g(X)=f(1-X)\in \Bbb Z[\zeta]$, so $g(1-X)=f(1-(1-X))=f(X)$.

So $f(\zeta)=g(1-\zeta)=g(\lambda)$. Working modulo $\lambda$, we just pick the free coefficient of $g$, which is an integer, so $f(\zeta)$ is congruent to an integer modulo $\lambda$. We can also "move" this integer modulo $p$ in the range $0,1,\dots,(p-1)$ (or some other representative set) by using $$ p=\operatorname{Norm}(\lambda) =\lambda\cdot \prod_{\sigma\ne \text{id}}\sigma\lambda \equiv 0\mod(\lambda)\ . $$ This gives an explicit construction for the given isomorphism. Having it, the question may dissolve itself.