= 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 determines
$$
\mathfrak p=\{x\in\mathcal O_K:|x|<1\},
$$
a <prime ideal> satisfying $\mathfrak p\cap\mathbb Z=(p)$. Conversely, each prime $\mathfrak p$ above $p$ defines the normalized absolute value
$$
|x|_{\mathfrak p}=p^{-v_{\mathfrak p}(x)/e_{\mathfrak p}},
$$
where $e_{\mathfrak p}=v_{\mathfrak p}(p)$. 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 irreducibles
$$
f=f_1\cdots f_r,
$$
indexed by the primes $\mathfrak p\mid p$. The <Chinese remainder theorem> gives
$$
K\otimes_{\mathbb Q}\mathbb Q_p
\cong\mathbb Q_p[X]/(f)
\cong\prod_{i=1}^r\mathbb Q_p[X]/(f_i).
$$
The $i$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>
$$
K\otimes_{\mathbb Q}\mathbb Q_p\cong\prod_{\mathfrak p\mid p}K_{\mathfrak p}.
$$
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