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Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.

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

By local factorization and extended absolute values, extensions of $|\cdot {|}_p$ to the number field $L=\mathbb{Q}(\alpha )$ correspond to the irreducible factors of $X^3-2$ over ${\mathbb{Q}}_p$.

For $p=2$, the polynomial is Eisenstein, hence irreducible, so there is one extension. For $p=3$, a root would be a unit $x$ with $x\equiv -1(\mathrm{mod}3)$, but then $x^3\equiv -1\equiv 8(\mathrm{mod}9)$, not $2(\mathrm{mod}9)$. A reducible cubic has a root, so the polynomial is again irreducible and there is one extension.

For $p=5$, reduction givesThe factors are coprime, and the quadratic has discriminant $3$, a nonsquare modulo $5$. Hensel lemma lifts this as one linear and one irreducible quadratic factor over ${\mathbb{Q}}_5$, giving two extensions. The requested numbers are thereforefor $p=2,3,5$, respectively.