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.



Is co-restriction in Galois cohomology in fact the norm map via Kummer isomorphism?
The corestriction map on cohomology is indeed the norm in degree zero (see [Tate's notes on Galois cohomology](http://wstein.org/edu/2010/582e/refs/tate-galois_cohomology.pdf) 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).


