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 $x\equiv-1\pmod3$, but then $x^3\equiv-1\equiv8\pmod9$, not $2\pmod9$. A reducible cubic has a root, so the polynomial is again irreducible and there is one extension.

For $p=5$, reduction gives
$$
X^3-2=(X-3)(X^2+3X+4)\pmod5.
$$
The 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 therefore
$$
\boxed{1,1,2}
$$
for $p=2,3,5$, respectively.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 4 / b / Solution

If $L=K(\alpha)$ is a finite separable extension and $v$ is an absolute value on $K$, the extensions of $v$ to $L$ correspond to the irreducible factors of the minimal polynomial of $\alpha$ over the completion of $K$ at $v$.


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 Incoming links: Local factorization and extended absolute values