Complete $\mathbb{Q}$ at the prime $(p)$ to obtain $\mathbb{Q}_p = \{x = \sum_{n=n_0}^\infty a_n p^n : a_n \in \{0, 1, \ldots, p-1\}\}$.

Arithmetic: Addition and multiplication defined term-by-term (with carrying). The valuation ring $\mathbb{Z}_p = \{x : v_p(x) \geq 0\}$ consists of $p$-adic integers.

For $p = 2$: $-1 = 1 + 2 + 2^2 + 2^3 + \cdots$ (geometric series) For $p = 5$: $1/3 = 2 + 1 \cdot 5 + 3 \cdot 5^2 + 1 \cdot 5^3 + 3 \cdot 5^4 + \cdots$

Construct $\mathbb{Q}_p(\sqrt{d})$:

1. If $d$ is a square in $\mathbb{Q}_p$: extension is trivial
2. If $d$ is not a square: get degree 2 extension, either unramified or ramified

Criterion: $d \in (\mathbb{Q}_p^*)^2$ if and only if $v_p(d)$ is even and $d/p^{v_p(d)} \equiv x^2 \pmod{p}$ for some $x$.

For $\mathbb{Q}_3(\sqrt{-1})$: Since $-1 \not\equiv x^2 \pmod{3}$, this is unramified with $e = 1, f = 2$.

For $\mathbb{Q}_p(\zeta_{p^n})$ where $\zeta_{p^n}$ is a primitive $p^n$-th root of unity: $[\mathbb{Q}_p(\zeta_{p^n}) : \mathbb{Q}_p] = \varphi(p^n) = p^{n-1}(p-1)$

This is totally ramified when $n \geq 1$. The extension $\mathbb{Q}_p(\zeta_{p^\infty})/\mathbb{Q}_p$ (union of all cyclotomic $p$-power extensions) is the starting point of Lubin-Tate theory.

For each local field $K$, there exist Lubin-Tate formal groups $F$ over $\mathcal{O}_K$ providing explicit totally ramified abelian extensions.

The $\pi$-division points (where $\pi$ is a uniformizer) generate abelian extensions. For $K = \mathbb{Q}_p$ and $F = \hat{\mathbb{G}}_m$ (multiplicative group), division points are roots of unity, recovering cyclotomic extensions.

This constructs all abelian extensions explicitly, proving local class field theory constructively.

Consider $x^3 + 117x^2 - 97 = 0$ over $\mathbb{Q}_p$ for various $p$.

For $p = 2$: Check $\bar{f}(x) \equiv x^3 + x^2 + 1 \pmod{2}$ has root $\bar{x} = 1$. Since $\bar{f}'(1) \equiv 1 \pmod{2} \neq 0$, Hensel's lemma gives a $2$-adic root.

For $p = 3$: Check $\bar{f}(x) \equiv x^3 + 2 \pmod{3}$. This has no roots mod 3, so no $3$-adic solutions exist.

For $p = 5$: Similar analysis determines existence of $5$-adic roots.

To solve $N_{K/\mathbb{Q}}(\alpha) = m$ locally: check if $m$ is a norm from $K_p$ for all primes $p$.

For $K = \mathbb{Q}(i)$ and $m = 5$: We have $5 = |2+i|^2 = N(2+i)$, so 5 is a norm globally. Locally, check $\mathbb{Q}_p(i)/\mathbb{Q}_p$ for each $p$ using local norms.

The local-global principle for norms (Hasse norm theorem) states: $m$ is a global norm if and only if it's a local norm everywhere.