Hello I try to post but i cannot verify the typesetting because it is completely random (works on a line and not on another)
I was very token by the definition of the conductor of an extension given by S. Lang in Alg. Num. Th.$ \def\U{{\Bbb U}}\def\I{{\Bbb I}}\def\J{{\Bbb J}} \def\fm{{\frak m}}\def\ff{{\frak f}}\def\fp{{\frak p}}\def\fP{{\frak P}} \let\ss\subset\let\fa\forall$
Up here, for me, the conductor was defined through open subgroup of idèles. Such group contain a group $\U_\fm$ of inversible idèles $\equiv 1\mod \fm$; i mean : the $\fp$ component of $\U_m$,
$(\U_\fm)_\fp=U_\fp$ if $\fp\not|\fm$ and
$(\U_\fm)_\fp=1+\widehat\fp^{e(\fp)}$ if $\fm=\prod\fp^{e(\fp)}$ is a $k$-module.
Then one applies this definition to the group $k*N(\J_L)$ where $\J_L$ is the idèles of $L/k$. The conductor of an extension $L/k$ is the gcd of the $\fm$ such that $\U_\fm\ss k*N(\J_l)$.
S. Lang defines the conductor in a completely different way. For him, it's the gcd of the $k$-modules such that $\fa\fp,\fa\fP|\fp$ one have $(\U_\fm)_\fp\ss N(L_\fP)$. With such a definition, his conductor admits only ramified primes as divisor and Lang explain that $1+\widehat\fp^{e(\fp)}$ is the bigest disk centered in 1 and contained in the norm group.
It's clear that the "open-group" conductor (let's note $\ff_G$) divides the Lang conductor, let's say the "normic" conductor and note it $\ff_N$.
I cannot see at all why are they egal ?
More concretely, i cannot see how to evacuate the coefficient in $k^*$ in the definition of $\ff_G$, that is the principal idèles. But it is precisely what happen. In fact $\U_{\ff_G}\ss N(\J_L)$ and i actualy think $\U_{\ff_G}\ss N(\I_L)$ (inversible idèles).
Why do we have $\ff_G=\ff_N$?