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.