Is there a nice explicit description for the group $K^*/Nm_{L/K}(L^*)$ for a finite field extension $L/K$?

What if for example, $L$ is obtained from $K$ by ajoining an n-th root of some $\alpha \in K$ (and assuming that $K$ contains the n-th root of unity)? I don't see a nice answer even for the case $n=2$.

Thank you.

EDIT: Thanks for the answers! Is that correct that in the case of a cyclic extension this group is isomorphic to $Br(L/K)$, since both these groups are identified with $H^2(Gal(L/K), L^*)$?

  • $\begingroup$ If $L/K$ is cyclic with generator $\sigma$, then Hilbert 90 tells us that $a \mapsto a \sigma(a)^{-1}$ yields an isomorphism $L^* / K^* \cong ker(N)$. But I don't see any connection with $coker(N)$. $\endgroup$ – Martin Brandenburg Sep 27 '10 at 21:52
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    $\begingroup$ The meaning of "nice explicit" is unclear; hard to know what you have in mind to do with an answer. In the cyclic Galois case for number fields one can give a pseudo-answer using class field theory and general considerations with cohomology of tori (see case $r=3$ on p. 199 of Cassels-Frohlich, and use double-periodicity of cyclic cohomology to shift to $r=1$), but not really "explicit", is it? With general fields it seems doubtful that anything useful can be said. $\endgroup$ – BCnrd Sep 27 '10 at 22:44
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    $\begingroup$ There are very simple cases where "explicit" is really explicit. The simplest being the trivial case of a quadratic extension of the rationals: the group is isomorphic to the product (strict class group)x(infinitely many copies of $C_2$), a copy for each inert prime. A generalization works for Kummer extensions as well. So not all is lost. $\endgroup$ – Dror Speiser Sep 27 '10 at 23:10
  • $\begingroup$ @Evgeny Regarding your edit, yes, when $L/K$ is a finite cyclic extension with group $G$, $H^0_T(G,L^\times)$ ($0$-dimensional Tate cohomology), your norm residue group, is isomorphic to $H^2(G,L^\times)$ via cup product with a generator of $H^0(G,\mathbb{Z})$, which is cyclic of order $\vert G\vert$ (this isomorphism depends on the choice of generator). $\endgroup$ – Keenan Kidwell Sep 28 '10 at 18:52

It really depends on what kinds of field $K$ and $L$ are. For example, if they are finite fields, then the quotient vanishes, since the norm is always surjective.

If $K$ is a local field with finite residue field and $L/K$ is abelian, then class field theory says that $K^\times/{\rm Norm}_{L/K}(L^\times)$ is isomorphic to the Galois group of $L/K$ (more generally, if $L/K$ is an arbitrary finite extension of local fields, then the quotient is isomorphic to the Galois group of the maximal abelian subextensions of $K$). This isomorphism is reasonably explicit and is described in any good exposition of class field theory. If $K$ is perfect, but not necessarily finite, or a local field with perfect residue field, then one can still say something reasonably explicit (see e.g. Serre's book "Corps Locaux").

If $K$ is a global field, then the situation is more complicated and less explicit, because the "right" object to look at, from the point of view of class field theory, is not the norm quotient you have written down, but the same with the multiplicative groups of the fields replaced by idèle class groups.

You need to tell us more about the fields in question to get a better answer.

Edit: to slightly generalise what I have said above, you can replace "finite field" by "quasi-finite field" throughout, and the statements will still hold.


In addition to Alex's answer I would like to point out that studying the group $K^\times/N(L^\times)$ is a perfectly fine goal (related of course to the validity of the Hasse norm principle in extensions of number fields), which should not simply be dismissed as a "wrong question". A good place to start is the work by Leonid Stern, the most recent article being On the norm groups of Galois $2\frak n$-extensions of algebraic number fields, J. Number Theory 129, No. 5, 1191-1204 (2009).


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