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I've been learning about local fields for some time now and I wanna prove the following:

Let $p$ be a prime number and $m \in \mathbb{N}$ such that $(m,p) = 1$. Pick a primitive $m$-th root of unity $\zeta_m$ in an algebraic closure of $\mathbb{Q}_p$ and let $K = \mathbb{Q}_p(\zeta_m)$. Then the residue class field $k$ of $K$ is $\mathbb{F}_p(\zeta_m)$.

My attempt of a proof: Since $(m,p) = 1$, the extension $K/\mathbb{Q}_p$ is unramified and $k$ contains a primitive $m$-th root of unity, hence $\mathbb{F}_p(\zeta_m) \subset k$. Let $L$ be the unique corresponding unramified intermediate extension of $K/\mathbb{Q}_p$ such that the residue field $l$ of $L$ is exactly $\mathbb{F}_p(\zeta_m)$. Let $N$ be the order of $p$ in $(\mathbb{Z}/m\mathbb{Z})^{\times}$ (i.e. the degree of $\mathbb{F}_p(\zeta_m)/\mathbb{F}_p$). Then $L = \mathbb{Q}_p(\zeta_r)$, where $r = p^N -1$ (since $L/\mathbb{Q}_p$ is unramified). Hence $m$ divides $r$. Hence $K \subset L$ and thus $l = k$.

Any corrections are greatly appreciated!

Shuster
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  • I feel like in your proof you are using some facts about unramified extensions which in my mind lie a little deeper than the one you want to prove. Are you sure you could back each of them up without a circular argument? --- Also, at the end, do we not rather get that $r$ divides $m$? – Torsten Schoeneberg Apr 25 '22 at 06:04

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A proof goes as follows. I'll indicate parts that might have to be looked up.

The extension $K/\mathbb{Q}_p$ is unramified as $(p,m)=1$. The degree of the extension is the multiplicative order of $p$ modulo $m$ and the extension of residue class fields has the same degree as the extension is unramified. The extension $\mathbb{F}_p[\zeta_m]/\mathbb{F}_p$ has the same degree. Now $k$ certainly contains $\mathbb{F}_p[\zeta_m]$, so they must coincide as the have the same degree over $\mathbb{F}_p$.

nilsw
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