## reciprocity maps and norm maps in class field theory

we consider number fields embedded in a fixed algebraic closure $k$ of $\mathbb{Q}$, or simply consider number fields in $\mathbb{C}$. for $L$ a number field, we write $G_L$ for the galois group $Gal(k/L)$.

Take $F$, $F_1$, $F_2$ and $E$ number fields such that $F=F_1\cap F_2$, $E=F_1F_2$, and $F_1$ Galois over $F$. Then $G_E=G_{F_1}\cap G_{F_2}$ and $G_F=G_{F_1}G_{F_2}$ (as we have assumed that $F_1$ is galois over $F$).

Then what happens with the abelianization? the norm maps give us a diagram

$$\xymatrix{ G_E^{ab} \ar[d]^{\subset} \ar[r]^{\subset} & G_{F_1}^{ab} \ar[d]^\subset \ G_{F_2}^{ab} \ar[r]^\subset & G_F^{ab}}$$

Is there chance that this diagram is cartesian?

We can also write $T^L$ for the torus $Res_{L/\mathbb{Q}}\mathbb{G}_\mathrm{m}$. Then the diagram above can be produced from the one below by passing to connected components of idele classes

$$\xymatrix{T^E \ar[d] \ar[r] & T^{F_1} \ar[d] \ T^{F_2} \ar[r] & \T^F}$$

However by dimension arguments one cannot expect this lower diagram to be cartesian. What happens if we only take connected components of the idele class groups?

thanks.

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Your xymatrix isn't rendering for me, which makes this question very hard to read. But if I understand correctly, your diagram is not right in general because there are several $\subset$s that are not inclusions once you pass to abelianisations: if $H$ is a subgroup of $G$ then there's no reason to expect $H^{ab}$ to be a subgroup of $G^{ab}$; there's a map but why should it be injective? The diagram could be cartesian (e.g. if $E=F_1=F_2=F$) but it could not be (for example if all the fields are totally real and satisfy Leopoldt then the $Z_p$-ranks don't add up in general). – Kevin Buzzard Nov 4 2011 at 21:47