Prime factorisation in a cyclotomic field

Assume $a\ge 1$. Since $\gcd (\mathrm{\ell },m)=1$, the divisors of ${\mathrm{\ell }}^{a}m$ are exactly the ${\mathrm{\ell }}^j{m}^{\mathrm{\prime }}$ with $0\le j\le a$ and $m^{\mathrm{\prime }}\mid m$; reducing the product identity of [F4] at $n={\mathrm{\ell }}^{a}m$ and at $n={\mathrm{\ell }}^{a-1}m$ modulo $\mathrm{\ell }$ and using $(X^m{)}^{{\mathrm{\ell }}^j}-1=(X^m-1{)}^{{\mathrm{\ell }}^j}$ in ${\mathbb{F}}_{\mathrm{\ell }}[t]$ therefore gives ${\prod }_{j=0}^a{\prod }_{m^{\mathrm{\prime }}\mid m}{\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^j{m}^{\mathrm{\prime }}}=(X^m-1{)}^{{\mathrm{\ell }}^a}$ and ${\prod }_{j=0}^{a-1}{\prod }_{m^{\mathrm{\prime }}\mid m}{\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^j{m}^{\mathrm{\prime }}}=(X^m-1{)}^{{\mathrm{\ell }}^{a-1}}$. The second product is the part $j\le a-1$ of the first, and $(X^m-1{)}^{{\mathrm{\ell }}^{a-1}}\mathrm{\ne }0$, so cancelling this common factor in the domain ${\mathbb{F}}_{\mathrm{\ell }}[t]$ gives ${\prod }_{m^{\mathrm{\prime }}\mid m}{\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^a{m}^{\mathrm{\prime }}}=(X^m-1{)}^{{\mathrm{\ell }}^a-{\mathrm{\ell }}^{a-1}}=(X^m-1{)}^{\phi ({\mathrm{\ell }}^a)}$.

Applied to $n=m$, which is coprime to $\mathrm{\ell }$, [F5] gives ${\bar{\mathrm{\Phi }}}_m={\prod }_{i=1}^g{h}_i$ with $g=\phi (m)\mathrm{/}d$, where the $h_i$ are pairwise distinct monic irreducible elements of ${\mathbb{F}}_{\mathrm{\ell }}[t]$ of degree $d={\mathrm{ord}}_m(\mathrm{\ell })$; in particular ${\bar{\mathrm{\Phi }}}_m\mathrm{\ne }0$.

Claim: for our fixed $a\ge 1$, ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^{a}n}={\bar{\mathrm{\Phi }}}_n^{\phi ({\mathrm{\ell }}^a)}$ in ${\mathbb{F}}_{\mathrm{\ell }}[t]$ for every $n\ge 1$ with $\gcd (n,\mathrm{\ell })=1$. This is proved by strong induction on $n$. For $n=1$, step 1.1 with $m=1$ gives ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^a}=(X-1{)}^{\phi ({\mathrm{\ell }}^a)}$, while ${\bar{\mathrm{\Phi }}}_1=X-1$ by [F4], so ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^a}={\bar{\mathrm{\Phi }}}_1^{\phi ({\mathrm{\ell }}^a)}$. For $n>1$, assume the claim for every proper divisor $m^{\mathrm{\prime }}\mid n$, $m^{\mathrm{\prime }}\mathrm{\ne }n$; step 1.1 with $m=n$ gives ${\prod }_{m^{\mathrm{\prime }}\mid n}{\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^a{m}^{\mathrm{\prime }}}=(X^n-1{)}^{\phi ({\mathrm{\ell }}^a)}$, and [F4] gives ${\prod }_{m^{\mathrm{\prime }}\mid n}{\bar{\mathrm{\Phi }}}_{m^{\mathrm{\prime }}}=X^n-1$, so substituting ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^a{m}^{\mathrm{\prime }}}={\bar{\mathrm{\Phi }}}_{m^{\mathrm{\prime }}}^{\phi ({\mathrm{\ell }}^a)}$ for the proper divisors yields ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^{a}n}\cdot {\prod }_{m^{\mathrm{\prime }}\mid n,m^{\mathrm{\prime }}<n}{\bar{\mathrm{\Phi }}}_{m^{\mathrm{\prime }}}^{\phi ({\mathrm{\ell }}^a)}={\bar{\mathrm{\Phi }}}_n^{\phi ({\mathrm{\ell }}^a)}\cdot {\prod }_{m^{\mathrm{\prime }}\mid n,m^{\mathrm{\prime }}<n}{\bar{\mathrm{\Phi }}}_{m^{\mathrm{\prime }}}^{\phi ({\mathrm{\ell }}^a)}$; the common factor is a nonzero product of nonzero monic polynomials, so cancellation in the domain ${\mathbb{F}}_{\mathrm{\ell }}[t]$ gives ${\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^{a}n}={\bar{\mathrm{\Phi }}}_n^{\phi ({\mathrm{\ell }}^a)}$.

Since $\gcd ({\mathrm{\ell }}^a,m)=1$, [F6] gives $\phi (f)=\phi ({\mathrm{\ell }}^{a}m)=\phi ({\mathrm{\ell }}^a)\phi (m)$, so $edg=\phi ({\mathrm{\ell }}^a)\cdot d\cdot (\phi (m)\mathrm{/}d)=\phi ({\mathrm{\ell }}^a)\phi (m)=\phi (f)$.

In all cases $a\ge 0$ one has ${\bar{\mathrm{\Phi }}}_f={\bar{\mathrm{\Phi }}}_{{\mathrm{\ell }}^{a}m}={\prod }_{i=1}^g{h}_i^e$ in ${\mathbb{F}}_{\mathrm{\ell }}[t]$, a product of pairwise distinct monic irreducibles of degree $d$ with common multiplicity $e$: if $a\ge 1$ this is step 2.1 at $n=m$ combined with step 1.2, and if $a=0$ then $e=\phi (1)=1$ and $m=f$, so the same formula is step 1.2 itself.

By [F1], [F2] the element $\alpha :=\zeta$ has ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$ and monic minimal polynomial ${\mathrm{\Phi }}_f$, so the monogenic factorisation [F3] applies with $p=\mathrm{\ell }$ to the factorisation of step 3.1: $\mathrm{\ell }{\mathcal{O}}_K={\prod }_{i=1}^g{P}_i^e$ with pairwise distinct primes $P_i=(\mathrm{\ell },{\tilde{h}}_i(\zeta ))$ of residue degree $\mathrm{deg}{h}_i=d$, where ${\tilde{h}}_i$ is the coefficientwise integer lift modulo $\mathrm{\ell }$, and ${\prod }_{i=1}^g{P}_i^e=(P_1\cdots P_g{)}^e$.

Edge cases. If $f=1$ then $a=0$, $m=1$, $e=d=g=1$ and ${\bar{\mathrm{\Phi }}}_1=X-1$, so steps 3.1 and 4.1 give $\mathrm{\ell }\mathbb{Z}=(\mathrm{\ell })$, and step 2.2 gives $edg=1=\phi (1)$; if $m=1$ and $a\ge 1>0$ then $g=d=1$ and $P_1=(\mathrm{\ell },\zeta -1)$, so $\mathrm{\ell }{\mathcal{O}}_K=P_1^{\phi ({\mathrm{\ell }}^a)}$, while [F8] with $p=\mathrm{\ell }$, $b=a$ gives $\mathrm{\ell }{\mathcal{O}}_K=(\lambda {)}^{\phi ({\mathrm{\ell }}^a)}$ with $\lambda =1-\zeta$ the unique prime above $\mathrm{\ell }$; the two descriptions agree by uniqueness of the prime above $\mathrm{\ell }$.