I wondered the relationship between $Cl_K$ and $Cl_F$ where $K/F$ is extension of number field.

Then I found a following short paper: Hiroyuki OSADA "Note on the ideal class group of abelian number fields".

the paper

I have two questions about this:

1.(in the proof of the theorem)

How can we get the map $Gal(\tilde{L} /L)\to Gal(\tilde{K}/K)$? I know it finally equal to composition of norm map and Artin map as wrote below though.

2.(in the proof of the lemma)

Why $f((1-\sigma)x)=0$? I'm also not sure how the action of $G$ defined (it's trivial action as a result though).

  • $\begingroup$ (1) is the translation theorem of Galois theory. Clearly $L \overline{K}$ is a subextension of $\overline{L}/L$. (2) G acts by conjugation with some lift; but the group is abelian, so conjugation is trivial. $\endgroup$
    – user23365
    Commented Oct 28, 2022 at 13:04
  • $\begingroup$ @franz lemmermeyer (1) OK, it's merely restriction map isn't it. (2) Sorry, I still don't figure out yet. Do group cohomology or something relate? $\endgroup$
    – user682141
    Commented Oct 31, 2022 at 13:12

1 Answer 1


Let $L/K/F$ be a tower of Galois extensions. For $\tau \in $ Gal$(K/F)$, let $T$ be a lift to Gal$(L/F)$, i.e., any $F$-automorphism of $L$ that restricts to $\tau$. Then Gal$(K/F)$ acts on Gal$(L/F)$ by conjugation: $\sigma^\tau := T^{-1} \sigma T$. If $L/F$ is abelian, this action is, of course, trivial.

BTW: In such situations, I often find it easier to rewrite the proof in my own words.


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