 Solution

The Ostrowski theorem says that every nontrivial absolute value on a field defined on $\mathbb{Q}$ is equivalent either to the usual absolute value or to $|\cdot {|}_p$ for a unique prime $p$.

Let an absolute value on the number field $K$ extend $|\cdot {|}_p$. Its valuation ring determinesa prime ideal satisfying $\mathfrak{p}\cap \mathbb{Z}=(p)$. Conversely, each prime $\mathfrak{p}$ above $p$ defines the normalized absolute valuewhere $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$. It restricts to $|\cdot {|}_p$. The correspondence between extensions and primes follows either from the valuation ring or from local factorization and extended absolute values; distinct primes give inequivalent valuations. Thus these $|\cdot {|}_{\mathfrak{p}}$ are exactly the extensions, up to equivalence.

For the tensor-product assertion, choose a primitive element of a field extension $\alpha$ for $K/\mathbb{Q}$, with minimal polynomial $f$. Because number fields are separable, over ${\mathbb{Q}}_p$ it factors into distinct irreduciblesindexed by the primes $\mathfrak{p}\mid p$. The Chinese remainder theorem givesThe $i$th factor is the completion of a number field at a prime ideal $K_{{\mathfrak{p}}_i}$. Under these identifications the isomorphism is the natural diagonal map $x\otimes a\mapsto (ax{)}_{\mathfrak{p}}$, proving the p-adic tensor decomposition of a number field