Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 18G (Galois Theory)

Because $L/K$ is a finite Galois extension,

Suppose the irreducible quadratic $f$ became reducible over $L$. It would then have a root $\alpha \in L$, and $K(\alpha )/K$ would be a quadratic intermediate extension. By the Fundamental theorem of Galois theory, $\mathrm{Gal}(L/K(\alpha ))$ would be a subgroup of index two in $A_{10}$ and hence a normal subgroup by the index-two subgroup is normal result. But the Simplicity of alternating groups says that $A_{10}$ is simple, while this subgroup would be proper and nontrivial. This contradiction proves that $f$ remains irreducible in $L[x]$.

The eleventh cyclotomic field $L=\mathbb{Q}({\xi }_{11})$ hasThe cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The Galois correspondence therefore gives exactly four subfields:The Quadratic Gaussian period in the eleventh cyclotomic field satisfies $\eta =(-1+\sqrt{-11})/2$, while $\theta$ generates the maximal real subfield of the eleventh cyclotomic field. Every subgroup of the abelian Galois group is normal, so all four subfields are Galois over $\mathbb{Q}$.

Each nontrivial proper subfield has prime degree, two or five. If $a\in M\setminus \mathbb{Q}$, then $\mathbb{Q}⊊\mathbb{Q}(a)\subseteq M$, and the tower law forces $\mathbb{Q}(a)=M$. Thus every such element is a primitive element. The displayed $\eta$ and $\theta$ are the requested examples. Finally, ${\xi }_{11}$ itself is a primitive element for $L/\mathbb{Q}$.