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 satisfies $a \in \mathrm{F}$ and $\sqrt[p]{a} \notin \mathrm{F}$. Consider the absolute galoisGalois group of $\mathrm{F}$, denoted by $G_F$, and its open sub-groupsubgroup $G_{F[\sqrt[p]{a}]}$ of index $p$ which(which is the absolute galoisGalois group of the extension $F[\sqrt[p]{a}]$). There is a well-known map between the following galoisGalois cohomology groups, which is the corestrictionco-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 wheterwhether the corestrictionco-restriction map and the norm map are actually the same thing, once looking onvia these identifications? In other words, does the corresponding diagram commutescommute? I couldn't find any referencereferences for thatthis.
capitalized names (Galois, Kummer); hyphenated co-restriction (otherwise it looks like core-striction); some syntax, and spelling.
David Handelman
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