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Let $d \geq 2$ be an integer, $K$ a number field containing the $d$-th roots of unity $\mu_d(\mathbb{C})$ and $\mathfrak{p}$ a prime ideal of $K$ not dividing $d$. Let $\mathbb{F}_q$ be the residue field of $\mathfrak{p}$. I have seen in several places a multiplicative character $$ \chi: \mathbb{F}_q^\times \to \mu_d(\mathbb{C}) $$ defined by $$ \chi(x)=Teich_{\mathfrak{p}}(x^{(q-1)/d}) $$ where $Teich_{\mathfrak{p}}: \mu_d(\mathbb{F}_q) \to \mu_d(\mathbb{C})$ is the inverse of reduction mod $\mathfrak{p}$.

For this to make sense, we need that reduction mod $\mathfrak{p}$ of roots of unity gives an isomorphism (then it follows that $d$ divides $q-1$). But why is this true?

Also, is there any relation between the Teichmüller character here and the one for $p$-adic numbers?

Thanks for your help!

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    $\begingroup$ if $\mathfrak p|t$, $t\neq0$, and $k$ is maximal such that $\mathfrak p^k|t$, then $(1+t)^d=1+s$ and $\mathfrak p^{k+1}\not|s$, so $s\neq 0$, and thus $1+t$ is not a $d$-th root of $1$. This shows that the only $d$-th root of $1$ which reduces to $1$ mod $\mathfrak p$ is $1$, i.e. injectivity, hence $d|(q-1)$. $\endgroup$ – user8268 Jan 5 '15 at 18:13

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