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Let $\mathbb{F}_q$ denote a finite field. Let $n\geq 1$ be an integer such that $n\mid q-1$. Hence the $n$-th roots of unity $\boldsymbol{\mu}_n$ are contained in $\mathbb{F}_q$. Why is the map $$\mathbb{F}_q^{*}/(\mathbb{F}_q^{*})^n\rightarrow \boldsymbol{\mu}_n: \overline{x}\mapsto x^{(q-1)/n} $$ an isomorphism of groups?
It is clearly well defined. But why is it a (iso)morphism ?

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What is its kernel? – Neal Mar 3 '12 at 17:40
$\mathbb{F}_q^\times$ is a cyclic group. Now, apply what you know about cyclic groups. – Hurkyl Mar 3 '12 at 18:39
up vote 2 down vote accepted

There are a few ways to see this. I think I learned this particular argument from Serre's A Course in Arithmetic (He needs the statement for $n = 2$ to prove quadratic reciprocity). First define $\mathbf F_q^* \to \mu_n$ by $x \mapsto x^{(q - 1)/n}$. Fix $x \in \mathbf F_q^*$, and let $y$ be an element in some extension of $\mathbf F_q$ such that $y^n = x$. For $x$ to be an $n$-th power in $\mathbf F_q$ we need $y \in \mathbf F_q$, i.e. $y^{q - 1} = x^{(q - 1)/n} = 1$. So the kernel of our map really is $\mathbf F_q^{*n}$. Now note that $\mathbf F_q^{*n}$ has order $(q - 1)/n$ because $\mathbf F_q^*$ is a cyclic group.

Edit. Is the question simply, "Why is this a homomorphism of groups?" The multiplicative group is commutative, and hence $(xy)^k = x^ky^k$ for all $k$, and in particular for $k = (q - 1)/n$.

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