Place (number field)

Remarks

Equivalent absolute values define the same topology on $K$. A place is infinite (archimedean) if the values $\abs{n \cdot 1}$ for $n \in \N$ are unbounded, and finite (non-archimedean) otherwise.

Let $K$ be a number field with $r_1$ real embeddings and $r_2$ pairs of complex embeddings. Each embedding $\sigma \colon K \to \C$ gives the absolute value $\abs{x}_\sigma = \abs{\sigma(x)}$, and conjugate embeddings give the same one. These give the $r_1 + r_2$ infinite places. Each nonzero prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ gives $\abs{x}_{\mathfrak{p}} = N(\mathfrak{p})^{-v_{\mathfrak{p}}(x)}$ for $x \ne 0$. Here $v_{\mathfrak{p}}(x)$ is the exponent of $\mathfrak{p}$ in the factorisation of $x\mathcal{O}_K$, and $N(\mathfrak{p})$ is the norm. These give all finite places. Every place arises in one of these two ways.

Example: on $\Q$, the p-adic absolute value $\abs{x}_p = p^{-v_p(x)}$ satisfies the three clauses, even with $\abs{x + y}_p \le \max(\abs{x}_p, \abs{y}_p)$, and it is nontrivial since $\abs{p}_p = 1/p$; so it defines a place of $\Q$. By Ostrowski's theorem, the places of $\Q$ are these and the place of the usual absolute value.

Non-example: on $\Q$, the map $x \mapsto \abs{x}^2$ fails clause 3, since $\abs{1 + 1}^2 = 4 > 1^2 + 1^2$. So it is not an absolute value in this sense, although it is a power of one.