Let $\mathrm{F}$ be a field that contains a root of unity of order $p$, where $p$ is a prime number. Fix an element $a$ that satisfies $a \in \mathrm{F}$ and $\sqrt[p]{a} \notin \mathrm{F}$. Consider the absolute galois group of $\mathrm{F}$, denoted by $G_F$, and its **open** sub-group $G_{F[\sqrt[p]{a}]}$ 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, which is the corestriction:
$$Cor:H^1(G_{F[\sqrt[p]{a}]}, \mathbb{Z}/p) \rightarrow H^1(G_F, \mathbb{Z}/p)$$
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 wheter the **corestriction** map and the **norm** map are actually the same thing, once looking on these identifications? In other words, does the corresponding diagram commutes? I couldn't find any reference for that.