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$\mathcal{O}=\{x\in \mathbb{Q}_p:v(x)\geq0\}$ is a valuation ring.

$\mathfrak{m}=\{x\in \mathbb{Q}_p: v(x)>0\}$ is the maximal ideal of $\mathcal{O}$.

Why is $K=\mathcal{O}/\mathfrak{m}$ isomorphic to $\mathbb{F}_p$, the finite field with p elements?

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  • $\begingroup$ Because $O = \mathbb{Z}_p, \mathfrak{M} = \bigcup_{a=0}^{p-1} a+p\mathbb{Z}_p$. If you define $\mathbb{Z}_p$ as the completion of $\mathbb{Z}$ for $|x|_p = p^{-v(x)}$ then that $O/\mathfrak{M} = (O \cap \mathbb{Z})/(\mathfrak{M}\cap \mathbb{Z}) = \mathbb{Z}/p\mathbb{Z}$ is a consequence of that $v$ is a discrete valuation $\endgroup$ – reuns Feb 21 at 0:23
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    $\begingroup$ How you come to see the truth of this claim may depend on which definition of $\Bbb Q_p$ you’re using. $\endgroup$ – Lubin Feb 21 at 5:12
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We have the following exact sequence $$ 0\rightarrow \mathbb{Z}_p\rightarrow \mathbb{Z}_p\rightarrow \mathbb{Z}/p^n\mathbb{Z}\rightarrow 0, $$ where the first map is multiplication by $p^n$ and the second sends $x=(x_i)\in \mathbb{Z}_p=\lim_{\leftarrow}\mathbb{Z}/p^n\mathbb{Z}$ to its $n$th term. Thus $ \mathbb{Z}_p/p^n \mathbb{Z}_p\cong \mathbb{Z}/p^n\mathbb{Z}$, so take $n=1$.

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