Let consider the ring $\mathbb{Z}_p$ and $\zeta$ be a $p$-th root of unity. Especially $\zeta \not \in \mathbb{Z}_p$. Denote with $\Phi _p(x)$ the cyclotomical polynomial in $p$. Since $p$ is a prime we know that it has the shape $\Phi _p(x)= 1 + x +x^2 +... +x^{p-1}$. This gives rise for the quotient ring

$$ \mathbb{Z}_p[X]/\langle\Phi_p(x)\rangle \cong \mathbb{Z}_p[\zeta] = \mathbb{Z}_p \oplus \zeta \mathbb{Z}_p \oplus \dots \oplus \zeta^{p-2} \mathbb{Z}_p $$

which is obviously a free $\mathbb{Z}_p$-module of rank $p-1$. Denote $g=\zeta -1$.

My question is how to see that $\mathbb{Z}_p[\zeta]$ isa local ring with maximal ideal $g \mathbb{Z}_p[\zeta]$?

I tried to argue in following way: Obviously observation provides $\Phi _p(g +1) =0$ and the formula above gives $$ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $$.

In light of this I can conclude following inclusions:

$p \mathbb{Z}_p[\zeta] \subset g \mathbb{Z}_p[\zeta]$ and $\pi^{p-1} \mathbb{Z}_p[\zeta] \subset p \mathbb{Z}_p[\zeta]$ which imply $(g\mathbb{Z}_p[\zeta]) \cap\mathbb{Z}_p = p\mathbb{Z}_p$.

From here I'm stuck.

  • 3
    $\begingroup$ $\zeta_p \not\in \mathbf{Z}_p$ for $p > 2$. The maximal ideal should be generated by $1-\zeta_p$. Everything is proved in [Neukirch], II.7.13. $\endgroup$ – user19475 Apr 15 '18 at 13:04
  • 3
    $\begingroup$ More generally, if $G$ is a (finite) $p$-group and $R$ a commutative local ring with residue field of characteristic $p$, the group algebra $R[G]$ is local. You can see a proof in Bourbaki's Algèbre 8, §2, exercise 10. $\endgroup$ – abx Apr 15 '18 at 14:08
  • $\begingroup$ You should be able to prove for yourself that if $(\mathfrak o,\mathfrak m)$ is a local ring and $\pi$ is a root of an Eisenstein polynomial over $\mathfrak o$, then $\mathfrak o[\pi]$ is also local, maximal ideal being $\langle\mathfrak m,\pi\rangle$. $\endgroup$ – Lubin Apr 15 '18 at 14:51

Let $m$ be a maximal ideal of $A:=\mathbb{Z}_p[\zeta]$. Then $m\cap\mathbb{Z}_p=p\mathbb{Z}_p$ because $A$ is a finite $\mathbb{Z}_p$-algebra. So the maximal ideals of $A$ are essentially those of $A/pA\cong\mathbb{F}_p[X]/(\Phi_p\bmod p)$. Since $\Phi_p\equiv(X-1)^{p-1}\pmod p$, we see that $A/pA$ is local with maximal ideal $(X-1)$.


Put $A=\mathbb{Z}_p[\zeta]$ and $\pi=\zeta-1$. Then $$ A/\pi=\mathbb{Z}_p[x]/(\Phi_p(x),x-1)= (\mathbb{Z}_p[x]/(x-1))/\Phi_p(x) = \mathbb{Z}_p/\Phi_p(1) = \mathbb{Z}/p. $$ This is a field, so $\pi$ generates a maximal ideal. Now suppose that $u$ lies outside this maximal ideal. Let $v$ be a lift in $\mathbb{Z}_p^\times$ of the image of $u$ in $A/\pi=\mathbb{Z}/p$, so $u=v(1-a\pi)$ for some $a\in A$. As $p$-th powers are additive mod $p$, in $A/p$ we have $\pi^p=\zeta^p-1=0$. This means that $\pi^p$ is divisible by $p$, so the series $\sum_i(a\pi)^i$ converges $p$-adically to an inverse for $1-a\pi$, and we deduce that $u$ is also invertible in $A$. As $A\pi$ is an ideal such that $A\setminus A\pi$ consists of units, we see that it is the unique maximal ideal, so $A$ is local.


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