5
$\begingroup$

Let $\mathrm{F}$ be a field that contains a root of unity of order $p$, where $p$ is a prime number. Fix an element $a$ such that $a \in \mathrm{F}$ and $\sqrt[p]{a} \notin \mathrm{F}$. Consider the absolute Galois group of $\mathrm{F}$, denoted by $G_F$, and its open subgroup $G_{F[\sqrt[p]{a}]}$ of index $p$ (which is the absolute Galois group of the extension $F[\sqrt[p]{a}]$). There is a well-known map between the following Galois cohomology groups, the co-restriction: $$Cor:H^1(G_{F[\sqrt[p]{a}]}, \mathbb{Z}/p) \rightarrow H^1(G_F, \mathbb{Z}/p)$$ In addition, there is a map between the abelian groups $F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p$ and $F^*/{F^*}^p$ which is the field norm: $$ N_{F[\sqrt[p]{a}] / F} : F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p \rightarrow F^*/{F^*}^p $$ Kummer theory yields the isomorphisms $H^1(G_F, \mathbb{Z}/p) \cong F^*/{F^*}^p$ and $H^1(G_{F[\sqrt[p]{a}]}, \mathbb{Z}/p) \cong F[\sqrt[p]{a}]^*/{F[\sqrt[p]{a}]^*}^p$, so the natural question to be asked is whether the co-restriction map and the norm map are actually the same thing, via these identifications? In other words, does the corresponding diagram commute? I couldn't find any references for this.

$\endgroup$
7
$\begingroup$

The corestriction map on cohomology is indeed the norm in degree zero (see Tate's notes on Galois cohomology for example). By a dimension shifting argument, it then easily follows that the corestriction in degree one also corresponds to taking norms (under your isomorphisms).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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