2
$\begingroup$

Let $E/F$ be a quadratic extension of global fields. Denote by $J_F$ the idèle group of $F$, by $P_F$ the subgroup of principal idèles and by $N_{E/F} : J_E \to J_F$ the norm map between idèles. In Timothy O'Meara's "Introduction to Quadratic Forms" it is shown that $[J_F : P_FN_{E/F}J_E] = 2$ (65:21), which is then used to prove a 'converse' to the Hilbert Reciprocity Law.

However, O'Meara derives these results under the assumption that the global characteristic is not 2. Does the statement hold if the characteristic is 2? I am interested mainly in this rather explicit converse of the Reciprocity Law and how it could be reformulated if the characteristic is 2, and the index equality seems to be the key lemma.

$\endgroup$

1 Answer 1

5
$\begingroup$

Yes, this is true for separable extensions of global fields of arbitrary characteristic. More generally, Artin reciprocity gives an isomorphism $J_F / P_F {N_{E/F} J_E} \cong \operatorname{Gal}(E/F)^{\text{ab}}$, and in your case the latter is order 2. (In most expositions you'll see the left hand side written as $C_F / N_{E/F} C_E$.)

As pointed out by Daniel Loughran, you do need to assume that E/F is separable. If not then the norm map should be surjective.

$\endgroup$
3
  • 2
    $\begingroup$ Presumably this only works for separable quadratic extensions. $\endgroup$ May 13, 2018 at 13:07
  • $\begingroup$ Thanks! Do you know of a reference with a proof of Artin Reciprocity which does not make any assumptions on the characteristic? I have only come across proofs for algebraic number fields. $\endgroup$
    – Bib-lost
    May 20, 2018 at 14:25
  • $\begingroup$ The standard expositions generally handle number fields and function fields separately, even though the results are uniform. Class Field Theory by Artin-Tate proves both cases, and see also arxiv.org/pdf/1610.07486.pdf for geometric-only proofs. In your case, since the Galois group is order 2, the result you need follows directly from the first and second inequalities without having to do any serious cohomology. $\endgroup$ May 22, 2018 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.