Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 32 / 1 / iii / Solution

Put $\alpha =44^{1/11}$ and $L=\mathbb{Q}(\alpha )$. The Eisenstein criterion at $11$ shows that $X^{11}-44$ is an irreducible polynomial, so $[L:\mathbb{Q}]=11$. Its derivative is $11X^{10}$, and the polynomial discriminant isThe discriminant-index formula for an integral lattice implies that the field discriminant has no prime divisor outside $\{2,11\}$. By the discriminant obstruction to ramification, every other prime is unramified.

To prove that neither candidate disappears through an index correction, let $\mathfrak{p}$ lie above $p\in \{2,11\}$ and normalize its valuation to have value group $\mathbb{Z}$. If $e=e_{\mathfrak{p}/p}$, the relation ${\alpha }^{11}=44$ givesIn both cases $11\mid e$. The fundamental identity for prime decomposition, ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}/p}{f}_{\mathfrak{p}/p}=11$, now forces exactly one prime ideal above each candidate, with $e=11$ and $f=1$. This is ramification forced by a coprime root valuation; at $2$ it avoids an invalid direct application of the Eisenstein criterion, since $4\mid 44$. Hence