# Field theoretic problem in a proof of Artin reciprocity law

Let $$L/K$$ be a cyclic extension of number fields of degree $$n$$, $$\zeta_m$$ a primitive $$m$$ th root of unity.

Artin's lemma Let $$S$$ a finite set of primes of $$\mathbb{Z}$$, $$\mathfrak{p}$$ a prime of $$\mathcal{O}_K$$. Then there is some $$m\in \mathbb{Z}_{>0}$$, prime to the elements of $$S$$ and to $$\mathfrak{p}$$, and an extension $$E/K$$ such that

(1) $$L\cap E=K$$

(2) $$L(\zeta_m)=E(\zeta_m)$$

(3) $$L\cap K(\zeta_m)=K$$

(4) $$\mathfrak{p}$$ splits completely in $$E/K$$.

(proof is in here for example)

argument Let $$\mathfrak{p}_1,\dots ,\mathfrak{p}_r$$ be prime ideals. Then applying Artin's Lemma to $$\mathfrak{p}_1,\dots ,\mathfrak{p}_r$$ in succession, we get integers $$m_1,\dots ,m_r$$ and fields $$E_1,\dots, E_r$$.

We may assume that $$m_1,\dots ,m_r$$ are pairwise relatively prime (enlarge $$S$$ each time we apply Artin's Lemma) and that each is prime to all the primes that ramify in $$K/\mathbb{Q}$$ and to $$\mathfrak{p}_1,\dots ,\mathfrak{p}_r$$ (again by enlarging $$S$$).

Question Then why $$L\cap E=K$$ (where $$E:=E_1\cdots E_r$$) ?

(This argument appears in a proof of Artin reciprocity law in Nancy Childress "Class Field Theory"(p.121))

• This resolved I will post answer. Commented Jan 12, 2020 at 12:00

$$\require{begingroup} \begingroup$$ $$\def\gal{\operatorname{Gal}}$$ $$\def\res{\operatorname{Res}}$$ $$\newcommand{\inv}[1]{#1^{-1}}$$ $$\def\coloneqq{\colon=}$$ $$\def\Q{\mathbb{Q}}$$ $$\def\R{\mathbb{R}}$$ $$\def\Z{\mathbb{Z}}$$ $$\def\braket#1{\mathinner{\left\langle{#1}\right\rangle}}$$ $$\def\dprod{\displaystyle\prod}$$

First, we have to improve Artin's lemma as follows. You can obtain this reading the proof of Artin's lemma.

Improved Artin's Lemma. Let $$L/K$$ be a cyclic extension of number fields with Galois group $$G=\gal(L / K)=\langle\sigma\rangle$$ and degree $$n=[L: K] .$$ Let$$S$$ be a finite set of prime number and let $$\mathfrak{p}$$ be a prime ideal of $$K$$. Then (by applying $$K,S'\coloneqq \left\{\text{p :prime \mid (p) ramifies in K/\Q}\right\}\cup S,\mathfrak{p},n$$ to lemma2.7 in the book) there exists positive integer $$m\notin \mathfrak{p}$$ so that

(a). $$(K(\zeta_m)/K,\mathfrak{p})$$ has order divisible by $$n$$

(b). restriction onto each factor gives an isomorphism $$\res\colon\gal(L(\zeta_m)/K)\xrightarrow{\sim} G\times\gal(K(\zeta_m)/K)%\xrightarrow{\sim} \braket{\sigma}\times(\Z/m\Z)^{\times}$$ (c). there exists $$\tau\in\gal(K(\zeta_m)/K)$$ independent of $$(K(\zeta_m)/K,\mathfrak{p})$$ and with order also divisible by $$n$$.

If we let $$H'\coloneqq \braket{(\sigma,\tau),((L/K,\mathfrak{p}),(K(\zeta_m)/K,\mathfrak{p}))}$$, $$H\coloneqq \inv{\res}(H') \subset \gal(L(\zeta_m)/K)$$ and we set $$E\coloneqq L(\zeta_m)^H$$.

Then positive integer $$m$$ and field extension $$E/K$$ satisfies following condition:

(0)$$p\nmid m$$ for all $$p\in S$$.

(1) $$L\cap K(\zeta_m)=K$$.

(2) $$\mathfrak{p}$$ splits completely in $$E/K$$.

(3) $$L\cap E=K$$

(4) $$L(\zeta_m)=E(\zeta_m)$$.

We write $$H_1\dots H_r,H_1'\dots H_r'$$ as corresponding groups $$H,H'$$ for each of Improved Artin's Lemma in the argument.

Then we get $$L\cap E=K$$ as follows:

Fix for any $$i=1\dots r$$ and let $$F\coloneqq K(\zeta_{m_1})\cdots K(\zeta_{m_r})$$.

Now we consider a commutative diagram.

$$\require{AMScd}$$ $$\begin{CD} \gal(LF/K) @>{r_1}>> \gal(L/K)\times \dprod_{j=1}^r \gal(K(\zeta_{m_j})/K) \\ @VVV @V{p}VV \\ \gal(L(\zeta_{m_i})/K) @>{r_2}>> \gal(L/K)\times \gal(K(\zeta_{m_i})/K) \end{CD}$$

Where $$p$$ is a projection, others are restrictions. $$r_1,r_2$$ are surjective.

Let $$\rho\in\gal(LF/K)$$.

\begin{align*} & \rho\in\gal(LF/E_i)=\left\{\text{fixing group of E_i in LF/K}\right\}\\ \iff& \rho|_{L(\zeta_{m_i})}\in\left\{\text{fixing group of E_i in L(\zeta_{m_i})/K}\right\}=H_i\\ \iff& r_2(\rho|_{L(\zeta_{m_i})})\in r_2(H_i)=H_i'\\ \iff& p(r_1(\rho))\in H_i'\\ \iff& r_1(\rho)\in H_i'\times \dprod_{j\neq i} \gal(K(\zeta_{m_j})/K)\\ \iff& \rho\in \inv{r_1} (H_i'\times \dprod_{j\neq i} \gal(K(\zeta_{m_j})/K))=: H_i''. \end{align*}

Therefore $$H_i''=\gal(LF/E_i)$$.

$$\epsilon\coloneqq \inv{r_1}((1,\tau_1,\dots,\tau_r))$$ fixes $$L$$. And by definition of $$H_i$$, $$\lambda\coloneqq \inv{r_1}((\sigma,\tau_1,\dots,\tau_r))\in H_i''=\gal(LF/E_i)$$ (for each $$j=1\cdots r$$). So $$\lambda$$ fixes $$E$$. So $$\lambda\inv{\epsilon}$$ fixes $$L\cap E$$.

Now since $$r_1(\lambda\inv{\epsilon})=(\sigma,1,\dots,1)$$, $$\sigma$$ fixes $$L\cap E$$. Therefore $$L\cap E\subset L^{\braket{\sigma}}=K$$.

$$\endgroup$$