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))

  • $\begingroup$ This resolved I will post answer. $\endgroup$
    – user682141
    Commented Jan 12, 2020 at 12:00

1 Answer 1


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

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 $L\cap E\subset L^{\braket{\sigma}}=K$.



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .