Prime-power cyclotomic ring, discriminant support and p factor

$\mathrm{\Phi }(\zeta )=0$ by [F1], so $\zeta$ is a root of the degree-$e$ monic $\mathrm{\Phi }$, which is irreducible by [F2]; hence $K$ has degree $e$ over $\mathbb{Q}$, the powers $1,\zeta ,\dots ,{\zeta }^{e-1}$ form a $\mathbb{Z}$-basis of $R$, and $R$ is an order in ${\mathcal{O}}_K$ with finite index $m$ satisfying $m{\mathcal{O}}_K\subseteq R$.

For every integer $j$ coprime to $p$ the element ${\zeta }^j$ is a root of $\mathrm{\Phi }$: its order is $p^a$, so $({\zeta }^j{)}^{p^a}=1$ while $({\zeta }^j{)}^{p^{a-1}}\mathrm{\ne }1$, and [F1] then forces $\mathrm{\Phi }({\zeta }^j)=0$; the $e$ elements ${\zeta }^j$ with $j\in (\mathbb{Z}\mathrm{/}{p}^a{)}^{\times }$ are pairwise distinct, so comparison with the monic degree-$e$ polynomial $\mathrm{\Phi }$ gives $\mathrm{\Phi }(X)={\prod }_{j\in (\mathbb{Z}\mathrm{/}{p}^a{)}^{\times }}(X-{\zeta }^j)$ and $p=\mathrm{\Phi }(1)={\prod }_{j\in (\mathbb{Z}\mathrm{/}{p}^a{)}^{\times }}(1-{\zeta }^j)$.

Each factor of the product in step 1.2 is a unit multiple of $\lambda =1-\zeta$ inside $R$: after replacing $j$ by its least positive residue modulo $p^a$ the quotient $(1-{\zeta }^j)\mathrm{/}(1-\zeta )=1+\zeta +\cdots +{\zeta }^{j-1}$ lies in $R$, and if $s$ satisfies $js\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{p}^a)$ then $\zeta =({\zeta }^j{)}^s$ and $(1-\zeta )\mathrm{/}(1-{\zeta }^j)=1+{\zeta }^j+\cdots +({\zeta }^j{)}^{s-1}\in \mathbb{Z}[{\zeta }^j]\subseteq R$; the two quotients are inverse to each other, so $(1-{\zeta }^j)\mathrm{/}(1-\zeta )\in R^{\times }$. Hence $p=u{\lambda }^e$ for some $u\in R^{\times }$, and consequently ${\lambda }^e{\mathcal{O}}_K=p{\mathcal{O}}_K$.

Taking the product formula of [F5] over the Galois group identified in [F3] and substituting step 1.2 gives $N_{K\mathrm{/}\mathbb{Q}}(1-\zeta )={\prod }_{\sigma }\sigma (1-\zeta )={\prod }_{j\in (\mathbb{Z}\mathrm{/}{p}^a{)}^{\times }}(1-{\zeta }^j)=\mathrm{\Phi }(1)=p$; moreover $N_{K\mathrm{/}\mathbb{Q}}(\zeta )\in \mathbb{Z}$ by [F5] and $N_{K\mathrm{/}\mathbb{Q}}(\zeta {)}^{p^a}=N_{K\mathrm{/}\mathbb{Q}}({\zeta }^{p^a})=N_{K\mathrm{/}\mathbb{Q}}(1)=1$, so $N_{K\mathrm{/}\mathbb{Q}}(\zeta )=\pm 1$, since the only integers whose $p^a$-th power is $1$ are $\pm 1$.

For $0\le s\le a-1$ one has $N_{K\mathrm{/}\mathbb{Q}}(1-{\zeta }^{p^s})=p^{p^s}$: the element ${\zeta }_s:={\zeta }^{p^s}$ is a primitive $p^{a-s}$-th root of unity in $L_s:=\mathbb{Q}({\zeta }_s)$, the computation of step 2.2 with $a$ replaced by $a-s$ gives $N_{L_s\mathrm{/}\mathbb{Q}}(1-{\zeta }_s)=p$, and the embedding formula of [F5] applied to the tower $\mathbb{Q}\subseteq L_s\subseteq K$ gives $N_{K\mathrm{/}\mathbb{Q}}(1-{\zeta }^{p^s})=N_{L_s\mathrm{/}\mathbb{Q}}(1-{\zeta }_s{)}^{[K:L_s]}=p^{p^s}$, since $[K:L_s]=\phi (p^a)\mathrm{/}\phi (p^{a-s})=p^s$ by [F2].

$\mathbb{Z}\cap \lambda {\mathcal{O}}_K=p\mathbb{Z}$: step 2.1 gives $p=u{\lambda }^e$ with $u\in R^{\times }$, so $p\in \lambda {\mathcal{O}}_K$ and $p\mathbb{Z}\subseteq \mathbb{Z}\cap \lambda {\mathcal{O}}_K$; conversely, if $c\in \mathbb{Z}\cap \lambda {\mathcal{O}}_K$, then $c^e\in \mathbb{Z}\cap {\lambda }^e{\mathcal{O}}_K=\mathbb{Z}\cap p{\mathcal{O}}_K$ by step 2.1, say $c^e=p\beta$ with $\beta =c^e\mathrm{/}p\in {\mathcal{O}}_K$, and $\beta \in {\mathcal{O}}_K\cap \mathbb{Q}=\mathbb{Z}$ by [F7], so $p$ divides $c^e$ in $\mathbb{Z}$ and hence $p$ divides $c$, that is $c\in p\mathbb{Z}$.

$(p{\mathcal{O}}_K)\cap R=pR$: the inclusion $\supseteq$ is clear; for $\subseteq$ let $\alpha \in p{\mathcal{O}}_K\cap R$ and expand $\alpha =c_0+c_1\lambda +\cdots +c_{e-1}{\lambda }^{e-1}$ with $c_i\in \mathbb{Z}$, which is possible because $\zeta =1-\lambda$ makes $1,\lambda ,\dots ,{\lambda }^{e-1}$ a $\mathbb{Z}$-basis of $R$ as well. We show $c_i\in p\mathbb{Z}$ for all $i$ by induction: if $\alpha ={\sum }_{i\ge i_0}{c}_i{\lambda }^i\in p{\mathcal{O}}_K$ with $0\le i_0<e$, then $i_0+1\le e$ and $p{\mathcal{O}}_K={\lambda }^e{\mathcal{O}}_K$ by step 2.1, so $\alpha \in {\lambda }^{i_0+1}{\mathcal{O}}_K$ and $\alpha \mathrm{/}{\lambda }^{i_0}\in \lambda {\mathcal{O}}_K$; on the other hand $\alpha \mathrm{/}{\lambda }^{i_0}=c_{i_0}+\lambda {\sum }_{i>i_0}{c}_i{\lambda }^{i-i_0-1}\in c_{i_0}+\lambda {\mathcal{O}}_K$, the bracket lying in $R\subseteq {\mathcal{O}}_K$. Hence $c_{i_0}\in \mathbb{Z}\cap \lambda {\mathcal{O}}_K=p\mathbb{Z}$ by step 3.2, and subtracting $c_{i_0}{\lambda }^{i_0}\in pR\subseteq p{\mathcal{O}}_K$ from $\alpha$ leaves ${\sum }_{i>i_0}{c}_i{\lambda }^i\in p{\mathcal{O}}_K\cap R$ for the next index. Thus all coefficients are multiples of $p$ and $\alpha \in pR$.

Differentiating the identity $t^{p^a}-1=(t^{p^{a-1}}-1)\mathrm{\Phi }(t)$ of [F1] gives ${\mathrm{\Phi }}^{\mathrm{\prime }}(t)(t^{p^{a-1}}-1)+\mathrm{\Phi }(t)p^{a-1}{t}^{p^{a-1}-1}=p^a{t}^{p^a-1}$, and evaluating at $t=\zeta$, where $\mathrm{\Phi }(\zeta )=0$ and ${\zeta }^{p^a}=1$, yields ${\mathrm{\Phi }}^{\mathrm{\prime }}(\zeta )=p^a{\zeta }^{p^a-1}\mathrm{/}({\zeta }^{p^{a-1}}-1)$. Taking norms with the product formula of [F5] and using multiplicativity of the norm together with step 3.1 at $s=a-1$ and $N_{K\mathrm{/}\mathbb{Q}}(\zeta )=\pm 1$ from step 2.2 gives $N_{K\mathrm{/}\mathbb{Q}}({\mathrm{\Phi }}^{\mathrm{\prime }}(\zeta ))=(p^a{)}^e{N}_{K\mathrm{/}\mathbb{Q}}(\zeta {)}^{p^a-1}\mathrm{/}{N}_{K\mathrm{/}\mathbb{Q}}({\zeta }^{p^{a-1}}-1)=\pm p^{ae-p^{a-1}}=\pm p^{p^{a-1}(a(p-1)-1)}$, so [F6] gives $\mathrm{disc}(R)=\pm p^N$ with $N:=p^{a-1}(a(p-1)-1)$.

Since $\mathrm{disc}(R)=m^2{d}_K$ by [F6] and both $\mathrm{disc}(R)$ and $d_K$ are nonzero integers, $m^2$ divides $\mathrm{disc}(R)=\pm p^N$ in $\mathbb{Z}$, so the positive index is $m=p^{\nu }$ for some integer $\nu \ge 0$.

${\mathcal{O}}_K\cap p^{-1}R=R$: if $\beta \in {\mathcal{O}}_K$ and $p\beta \in R$, then $p\beta \in p{\mathcal{O}}_K\cap R=pR$ by step 4.1, say $p\beta =p\rho$ with $\rho \in R$, and cancelling $p$ in the domain $K$ gives $\beta =\rho \in R$; the reverse inclusion is trivial.

By induction on $j\ge 0$ one has ${\mathcal{O}}_K\cap p^{-j}R=R$: the case $j=0$ is trivial and the case $j=1$ is step 5.2; if $j\ge 1$ and $x\in {\mathcal{O}}_K$ satisfies $p^{j}x\in R$, then $p^{j-1}(px)\in R$ with $px\in {\mathcal{O}}_K$, so the induction hypothesis gives $px\in R$, and then step 5.2 gives $x\in R$.

By step 5.1 write the finite index as $m=p^{\nu }$ with $\nu \ge 0$; then $m{\mathcal{O}}_K=p^{\nu }{\mathcal{O}}_K\subseteq R$ by step 1.1, so for any $x\in {\mathcal{O}}_K$ we get $p^{\nu }x\in R$ and step 6.1 with $j=\nu$ gives $x\in R$. Hence ${\mathcal{O}}_K=R=\mathbb{Z}[\zeta ]$, and the equality $p=u{\lambda }^e$ with $u\in R^{\times }$ from step 2.1 is an equality of principal ideals $(p)=(\lambda {)}^e$ in ${\mathcal{O}}_K$; in particular the power basis $1,\zeta ,\dots ,{\zeta }^{e-1}$ is an integral basis of ${\mathcal{O}}_K$ whose discriminant, by step 4.2, equals $\pm p^{p^{a-1}(a(p-1)-1)}$. For $p^a=2$ one has $e=1$, $\lambda =2$, $K=\mathbb{Q}$ and exponent $0$, in agreement with the general computation.