# Why is the residue field of $\mathbb{Q}_p$ isomorphic to $\mathbb{F}_p$?

$$\mathcal{O}=\{x\in \mathbb{Q}_p:v(x)\geq0\}$$ 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?

• 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 – reuns Feb 21 at 0:23
• How you come to see the truth of this claim may depend on which definition of $\Bbb Q_p$ you’re using. – Lubin Feb 21 at 5:12

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 $$n=1$$.