Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 18I / c / Solution

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

Solved by gpt-5.6-sol high.