Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 4 / c

The polynomial $X^3-2$ is Eisenstein over ${\mathbb{Q}}_2$, hence irreducible. Its discriminant is $-108=36(-3)$, and $-3\equiv 5(\mathrm{mod}8)$ is not a square in ${\mathbb{Q}}_2$. Therefore its splitting field has Galois group $S_3\cong D_6$.

Solved by gpt-5.6-sol high.

Suppose $G\cong D_{10}$. For residue characteristic two, $G_1$ is a normal $2$-group, $G_0/G_1$ embeds in $k_L^{\ast }$, and $G/G_0\cong \mathrm{Gal}(k_L/{\mathbb{F}}_2)$ is cyclic. The only proper nontrivial normal subgroup of $D_{10}$ is its rotation subgroup $C_5$, and it has no nontrivial normal $2$-subgroup.

Thus either $G_0=G$ or $G_0=C_5$. In the first case $G_1=1$, so part (b) would embed the noncyclic group $D_{10}$ in the cyclic group $k_L^{\ast }$, impossible. In the second case the residue degree is $|G/G_0|=2$, so $k_L={\mathbb{F}}_4$ and $k_L^{\ast }$ has order three; it cannot contain the injected group $G_0/G_1=C_5$. Both cases contradict the ramification constraints, so no such extension exists.

Solved by gpt-5.6-sol high.