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

Question

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

Answer 1

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 $n=1$.

Answer 2

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 $\mathbb{Z} / p \mathbb{Z}$.