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Consider the field extensions $L\supset K'\supset K$, where both $L|K$ and $K'|K$ are finite and Galois. I want to prove that $$N_{L|K}(L^\ast)\subseteq N_{L|K'}(L^{\ast})$$

Maybe it is very easy, but I wasn't able to find a solution after much time spent on it.

Clearly $N_{L|K}(L^\ast)\subseteq K^\ast$ and $N_{L|K'}(L^{\ast})\subseteq K'^\ast$, also from transitivity of norm we have $N_{L/K}(L^\ast) \subseteq N_{K'/K}(K'^\ast)$, but they obviously doesn't imply anything.


Edit: For completeness I explain from where my question comes from. I'm reading chapter $IV$ of Neukirch's Algebraic number theory book (the chapter is about abstract class field theory) and I saw the following commutative diagram at page $297$: enter image description here

I can't completely understand the vertical arrow on the right. The author says that it is induced by the inclusion $A_K\subset A_{K'}$ (here unfortunately you have to be familiar with Neukirch notation to understand what is $A_K$). So, such a vertical arrow is well defined if I can solve the problem of my question.

I've decided to write my question in the category of fields extensions in order to be as clear as possible. In this way also I avoided a cumbersome introduction about notations.

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  • $\begingroup$ I don't think that the right vertical arrow is the inclusion. $\endgroup$ – manifold Feb 10 '16 at 18:21
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Your map IS indeed induced by the inclusion of K in K' in the following sense. Denote G = Gal(L/K), H = Gal(L/K'), and let $N_G$, $N_H$ etc. be the corresponding norm maps in the group algebras Z[G], Z[H], etc. So, for instance, $N_G$ = $i_{L/K}$ . $N_{L/K}$ (where $i_{L/K}$ is the obvious inclusion) is an endomorphism of L* . The classical "transitivity of the norm" reads : $N_G$ = $N_{G/H}$ . $N_H$ . It shows that $N_G$(L*) is contained in $N_H$(L*), and we are done. Actually, in the presentation of Neukirch ( = the abstract presentation of CFT via "class formations"), the 2 vertical maps have analogous cohomological expressions. The rightmost map is the cohomological restriction $H^0$(G, L*) -> $H^0$(H, L*) (Tate cohomology, with a hat over $H$), whereas the leftmost one (transfer, or Verlagerung) is the restriction $H^{-2}$(G, Z) -> $H^{-2}$(H, Z) .

EDIT. See the comment of @pisco, and my subsequent answer.

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  • $\begingroup$ Sorry for excavating this answer long after. But I have two questions, how do you define $N_{G/H}$, is it $i_{L/K}\circ N_{K'/K}$?. Also, how do you know $$N_G = N_{G/H} \circ N_H \implies N_G(L^\ast) \subset N_H(L^\ast)$$ it seems to me that the inclusion you can derive should instead be $N_{L/K}(L^\ast) \subset N_{K'/K}(L^\ast)$, which is quite different from the question. Thank you very much. $\endgroup$ – pisco Oct 5 '18 at 18:01
  • $\begingroup$ @pisco 1) To a given a finite Galois extension $L/K$ with Galois group $G$, one can associate two types of norms : the arithmetic norm $N_{L/K}$, which is the usual norm map in Galois theory, and the algebraic norm $N_G$ wich is (in additive notation) the sum $\Sigma s$ in the group algebra $\mathbf Z[G]$. They are related by the formula $N_G=i_{L/K} .N_{L/K} $, where $i_{L/K}$ is the extension $K \to L$. Since $i_{L/K}$ is injective, there is no harm in identifying the two norms. In a tower of extensions $K<K'<L$, similar notions hold for $N_{K'/K}$ and $N_{G/H}$ if $K'/K$ is Galois. $\endgroup$ – nguyen quang do Oct 6 '18 at 13:54
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    $\begingroup$ ... But I guess your real concern is about my claim that $N_G=N_{G/H} . N_H$ implies $N_{L/K}(L^*)<N_{L/K'} (L^*)$. You're quite right, I've been careless, this needs an additional argument. In the commutative diagram extracted from Neukirch's book, I think, $A_K , A_K', A_L$ denote three levels in the "class formation", and $r_{L/K} , r_{L/K'}$ the corresponding "reciprocity isomorphisms" of CFT. The left vertical map is the transfer (Verlagerung), and the reciprocity isomorphisms induce the right vertical map. The inclusion $A_K<A_K'$ then implies $N_{L/K}(A_L)<N_{L/K'} (A_L)$. $\endgroup$ – nguyen quang do Oct 6 '18 at 14:37
  • $\begingroup$ For the properties of class formations which were used, I refer to Cassels-Fröhlich, especially chap. VII, §11.3. The "concrete" examples are the multiplicative group $L^*$ in the local case, and the idèle class group $C_L$ in the global case, so the OP general question about $N_{L/K}(L^*)<N_{L/K'}(L^*)$ was misleading. NB. The transfer map - which was used by Furtwängler to show the "principal ideal theorem"- is rather non intuitive. There is a letter of Grothendieck to Serre in which he confessed that he had been "humbled" by his unsuccessful attempt to find a direct proof. $\endgroup$ – nguyen quang do Oct 6 '18 at 14:58
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    $\begingroup$ Yes. In fact, the "inclusion" of $A_K$ in $A_L$ is a lazy way to say that $A_K\cong (A_L)^G$ thanks to the axioms of a class formation. $\endgroup$ – nguyen quang do Oct 7 '18 at 12:32

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