0
$\begingroup$

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.

$\endgroup$
  • 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
5
$\begingroup$

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 $(X-1)$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.