show that $pO_K=(p,(1-\zeta_p)^{p-1}.$

Question

$\text{Totally Ramified primes in cyclotomic fields:}$

Suppose we have $$ \Bbb F_p[x]/\overline{\Phi(x)} \cong O_k/pO_K.$$ If $ \ \overline{\Phi(x)}=\bar x^{p-1}+\cdots+\bar x+1=\large \frac{\bar x^p-1}{\bar x-1}=(x-1)^{p}$, where $ \large \bar x \equiv x \ (\mod \ p)$,

then how it implies $$pO_K=(p,(1-\zeta_p)^{p-1}.$$

when you have Eisenstein's criterion, you have a totally ramified prime. — Angina Seng, Nov 09 '19 at 20:48
@LordSharktheUnknown, I want to understand , how $(p,(1-\zeta_p))^{p-1}$ is equal to $pO_K$? Why $(p, (1-\zeta_p))$ is prime ideal? This expression is new to me — MAS, Nov 09 '19 at 21:21

score 6 · Accepted Answer · answered Nov 12 '19 at 18:54

The result you're looking for follows from the following well-known theorem of Dedekind (see here or here for a discussion):

Let $\mathcal{O}_K$ be the ring of integers of a number field $K,$ and assume $K$ is generated by adjoining the algebraic integer $\alpha$ to $\Bbb Q.$ Let $f\in\Bbb{Z}[x]$ be the minimal polynomial of $\alpha.$ Suppose $p$ does not divide the index $[\mathcal{O}_K:\Bbb{Z}[\alpha]],$ and write $$ f(x)\equiv \pi_1(x)^{e_i}\cdots\pi_r(x)^{e_r}\pmod{p}, $$ where the $\pi_i$ are distinct monic irreducible polynomials in $\Bbb{F}_p[x].$ Then $p\mathcal{O}_K$ factors into prime ideals as $$ p\mathcal{O}_K = (p,\tilde{\pi}_1(\alpha))^{e_1}\cdots (p,\tilde{\pi}_r(\alpha))^{e_r}, $$ where $\tilde{\pi}_i$ is a lift of $\pi_i$ to a polynomial in $\mathcal{O}_K[x].$

Now, if $K = \Bbb Q(\zeta_p),$ we know that $\mathcal{O}_K = \Bbb Z[\zeta_p],$ so that $[\mathcal{O}_K : \Bbb Z[\zeta_p]] = 1.$ Moreover, the minimal polynomial of $\zeta_p$ is $\Phi_p(x) = \frac{x^p - 1}{x-1},$ and modulo $p$ we have \begin{align*} \Phi_p(x) &\equiv \frac{x^p - 1}{x-1}\pmod{p}\\ &\equiv\frac{(x - 1)^p}{x-1}\pmod{p}\\ &\equiv (x-1)^{p-1}\pmod{p}. \end{align*}

Since $p\nmid 1,$ the theorem implies that $p\mathcal{O}_K = (p,1 - \zeta_p)^{p-1}.$

show that $pO_K=(p,(1-\zeta_p)^{p-1}.$

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