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

$$\mathcal{O}=\{x\in \mathbb{Q}_p:v(x)\geq0\}$$ 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 Feb 21 '19 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. Feb 21 '19 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 $$n=1$$.
How I see this to be true is by using the base-$$p$$ expansions. Now, we know that all the $$p$$-adic elements x can be written as some formal sum $$x = \sum_n a_n p^n, \quad 0 \le a_n \le p-1 \forall n$$ Now, elements in $$\mathcal{O}$$ have nonnegative valuation, so the formal sum has $$a_n = 0$$ for all $$n < 0$$, leaving only the coefficients $$a_n$$ for $$n \ge 0$$
Also, elements in $$\mathfrak{m}$$ have positive valuation, so the formal sum has $$a_n = 0$$ for all $$n \le 0$$, leaving only the coefficients $$a_n$$ for $$n > 0$$.
So intuitively, when you quotient by $$\mathfrak{m}$$, you are basically saying 'ignore all the coefficients $$a_n$$ with $$n > 0$$', and this leaves us only the $$a_0$$ term.
More precisely, for each coset $$x + \mathfrak{m}$$ in $$\mathcal{O} / \mathfrak{m}$$, where $$x$$ is $$\sum_{n \ge 0} a_n p^n$$, you map $$x + \mathfrak{m}$$ to the first coefficient $$a_0$$. This gives us the map from the quotient $$\mathcal{O} / \mathfrak{m}$$ to $$\mathbb{Z} / p \mathbb{Z}$$.