Cyclotomic Arithmetic and Reciprocity via Frobenius

$\mathrm{\Phi }(\zeta )=0$ 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.

By [F1] the given bases are $\mathbb{Q}$-bases of $K$ and $L$, and [F2] gives $[KL:L]=m$ and $[KL:K]=n$.

Every embedding $\sigma :K\to \mathbb{C}$ extends to an embedding $\tilde{\sigma }:KL\to \mathbb{C}$ with $\tilde{\sigma }{\mathrm{\mid }}_L={\mathrm{i}\mathrm{d}}_L$. Indeed, by [F9] write $K=\mathbb{Q}(\alpha )$ with minimal polynomial $f$ of degree $m$; then $KL=L(\alpha )$, and $[L(\alpha ):L]=m=\mathrm{deg}f$ shows that $f$ has no factor of intermediate degree over $L$, so $f$ is irreducible over $L$ and $KL\cong L[t]\mathrm{/}(f)$ as $L$-algebras. The evaluation homomorphism $L[t]\to \mathbb{C}$, $t\mapsto \sigma (\alpha )$, kills $f$ because $\sigma$ fixes $\mathbb{Q}$ and $f(\alpha )=0$; it therefore induces the required $\tilde{\sigma }$ under the isomorphism $L[t]\mathrm{/}(f)\cong KL$.

Put $V:={\sum }_{i=1}^{m}L{\alpha }_i\subseteq KL$. Since ${\alpha }_1,\dots ,{\alpha }_m$ are a $\mathbb{Q}$-basis of $K$, each product ${\alpha }_i{\alpha }_j$ is a $\mathbb{Q}$-linear combination of them; the same structure constants lie in $L$, so $V$ is closed under multiplication. Also $1\in V$, and $V$ contains both $L$ and $K$. It is finite-dimensional over $L$. For each nonzero $x\in V$, multiplication by $x$ is an injective $L$-linear map $V\to V$, because $V$ lies in the field $KL$; finite dimensionality makes the map surjective, so some $y\in V$ satisfies $xy=1$. Thus $V$ is a field containing $K$ and $L$, hence $V=KL$. By [F2] it has $L$-dimension $m$. Its $m$ elements ${\alpha }_i$ span $V$; if they were dependent, a nonzero coefficient in a dependence relation could be inverted in $L$ to express one ${\alpha }_i$ in terms of the others, so the remaining $m-1$ vectors would still span $V$. But an $m$-element basis of $V$ is an independent subset of a space with that $(m-1)$-element spanning set, contradicting [F7]. Thus the ${\alpha }_i$ are independent and spanning, hence an $L$-basis by [F7]. The symmetric argument with $W:={\sum }_{j=1}^{n}K{\beta }_j$ shows that ${\beta }_1,\dots ,{\beta }_n$ form a $K$-basis of $KL$ by the same finite argument.

Applying [F6] to the tower $\mathbb{Q}\subseteq L\subseteq KL$ with the $\mathbb{Q}$-basis ${\beta }_1,\dots ,{\beta }_n$ of $L$ and the $L$-basis ${\alpha }_1,\dots ,{\alpha }_m$ of $KL$, the products ${\alpha }_i{\beta }_j$ form a $\mathbb{Q}$-basis of $KL$; in particular they are $\mathbb{Q}$-linearly independent and there are $mn$ of them.

Let $\gamma \in {\mathcal{O}}_{KL}$. By step 2.1 there are unique $x_1,\dots ,x_m\in L$ with $\gamma ={\sum }_i{x}_i{\alpha }_i$. Each $x_i$ lies in $L=\mathrm{Frac}({\mathcal{O}}_L)$, so there are integers $a_{ij}$ and a positive integer $r$ with $x_i={\sum }_j(a_{ij}\mathrm{/}r){\beta }_j$; dividing out common factors, we may assume that no prime divides $r$ and all $a_{ij}$ simultaneously.

Let ${\sigma }_1,\dots ,{\sigma }_m$ be the distinct embeddings of $K$ into $\mathbb{C}$ and let ${\tilde{\sigma }}_k$ extend ${\sigma }_k$ as in step 1.2. Applying ${\tilde{\sigma }}_k$ to $\gamma ={\sum }_i{x}_i{\alpha }_i$ and using ${\tilde{\sigma }}_k{\mathrm{\mid }}_L=\mathrm{i}\mathrm{d}$ together with $x_i\in L$ gives the linear system ${\sum }_i{\sigma }_k({\alpha }_i)x_i={\tilde{\sigma }}_k(\gamma )$, $k=1,\dots ,m$. Its coefficient matrix $M:=({\sigma }_k({\alpha }_i))$ has determinant $D$ with $D^2=d_K\mathrm{\ne }0$ by [F5]; the entries ${\sigma }_k({\alpha }_i)$ of $M$ are algebraic integers, and so are the entries ${\tilde{\sigma }}_k(\gamma )$, by [F3] applied to the integral elements ${\alpha }_i\in {\mathcal{O}}_K$ and $\gamma \in {\mathcal{O}}_{KL}$.

Cramer's rule applied to the system of step 4.1 gives $D{x}_i=D_i$, where $D_i$ is the determinant of a matrix with algebraic-integer entries; hence $D$ and every $D_i$ are algebraic integers by [F3], and $\mathrm{\Delta }{x}_i=D\cdot D_i$ is an algebraic integer, where $\mathrm{\Delta }:=D^2=d_K$ is a nonzero integer.

By step 3.2, $\mathrm{\Delta }{x}_i={\sum }_j(\mathrm{\Delta }{a}_{ij}\mathrm{/}r){\beta }_j$ lies in $L$, so $\mathrm{\Delta }{x}_i\in L\cap \bar{\mathbb{Z}}={\mathcal{O}}_L$ by step 5.1 and [F4]. As ${\beta }_1,\dots ,{\beta }_n$ is an integral basis, the rational numbers $\mathrm{\Delta }{a}_{ij}\mathrm{/}r$ are integers; equivalently $r\mid \mathrm{\Delta }{a}_{ij}=d_K{a}_{ij}$ for all $i,j$.

The denominator was reduced in step 3.2, so no prime divides $r$ and every $a_{ij}$; therefore their common gcd is $1$. By [F8], choose integers $c,b_{ij}$ with $cr+{\sum }_{i,j}{b}_{ij}{a}_{ij}=1$. Multiplying by $d_K$ and using $r\mid d_K{a}_{ij}$ from step 6.1 shows that every term on the left is divisible by $r$, hence $r\mid d_K$.

Regroup the same expansion from step 3.2 as $\gamma ={\sum }_j{y}_j{\beta }_j$, where $y_j={\sum }_i(a_{ij}\mathrm{/}r){\alpha }_i\in K$. The symmetric trace-matrix argument, now using the $K$-basis ${\alpha }_i$ and the $L$-basis ${\beta }_j$ from step 2.1, gives $r\mid d_L{a}_{ij}$ for every same coefficient $a_{ij}$. The same reduced-denominator identity of [F8] therefore gives $r\mid d_L$. Together with step 7.1 and $\gcd (d_K,d_L)=1$, this implies $r=1$. Thus $\gamma ={\sum }_{i,j}{a}_{ij}{\alpha }_i{\beta }_j\in {\mathcal{O}}_K{\mathcal{O}}_L$ for every $\gamma \in {\mathcal{O}}_{KL}$, proving ${\mathcal{O}}_{KL}\subseteq {\mathcal{O}}_K{\mathcal{O}}_L$.

Conversely ${\mathcal{O}}_K{\mathcal{O}}_L\subseteq {\mathcal{O}}_{KL}$: products and sums of integral elements are integral by [F3], and they lie in $KL$. With step 8.1 this gives ${\mathcal{O}}_{KL}={\mathcal{O}}_K{\mathcal{O}}_L$, and since the products ${\alpha }_i{\beta }_j$ are $\mathbb{Q}$-linearly independent by step 3.1 and span ${\mathcal{O}}_K{\mathcal{O}}_L$ over $\mathbb{Z}$ by construction, they form an integral basis of ${\mathcal{O}}_{KL}$.

The restrictions of the $mn$ distinct embeddings of $KL$ into $\mathbb{C}$ to $K$ and to $L$ define a map $\mathrm{Emb}(KL)\to \mathrm{Emb}(K)\times \mathrm{Emb}(L)$; it is injective, because an embedding of $KL$ is determined by its restrictions to $K$ and to $L$, which together generate $KL$ over $\mathbb{Q}$. Both sides have $mn$ elements by [F5], so the map is a bijection. Order rows by pairs $(k,l)$ and columns by pairs $(i,j)$, with the first index slow in each order. The embedding matrix of the integral basis ${\alpha }_i{\beta }_j$ of step 9.1 then has entries ${\sigma }_k({\alpha }_i){\tau }_l({\beta }_j)$.

The embedding matrix of step 10.1 is $K_1{K}_2$: summing over the intermediate pair $(a,b)$ gives ${\sum }_{a,b}{\sigma }_k({\alpha }_a){\delta }_{lb}{\delta }_{ai}{\tau }_b({\beta }_j)={\sigma }_k({\alpha }_i){\tau }_l({\beta }_j)$. Here $(K_1{)}_{(k,l),(i,j)}={\sigma }_k({\alpha }_i){\delta }_{lj}$ and $(K_2{)}_{(k,l),(i,j)}={\delta }_{ki}{\tau }_l({\beta }_j)$. By [F10], $\det (K_1{K}_2)=\det (K_1)\det (K_2)$. In the first-index-slow ordering of step 10.1, $K_2$ is block diagonal with $m$ blocks $N=({\tau }_l({\beta }_j))$, so $\det (K_2)=\det (N{)}^m$. For $K_1$, reorder both its rows and its columns by the same perfect shuffle from $(k,l)$ to $(l,k)$ and from $(i,j)$ to $(j,i)$. These are the same permutation of $mn$ positions, so their determinant signs multiply to $1$; in the resulting ordering $K_1$ is block diagonal with $n$ blocks $M=({\sigma }_k({\alpha }_i))$. Hence $\det (K_1)=\det (M{)}^n$, using [F10] and the Leibniz formula for block diagonal matrices.

By [F5], $d_{KL}=\det ({\sigma }_k({\alpha }_i){\tau }_l({\beta }_j){)}^2=(\det M{)}^{2n}(\det N{)}^{2m}=d_K^n{d}_L^m=d_K^{[L:\mathbb{Q}]}{d}_L^{[K:\mathbb{Q}]}$, using $\det (M{)}^2=d_K$ and $\det (N{)}^2=d_L$ from [F5] and $m=[K:\mathbb{Q}]$, $n=[L:\mathbb{Q}]$. This completes the proof.