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Write $K=\mathbb{Q}({\mu }_m)$, $L=\mathbb{Q}({\mu }_n)$. For coprime $m,n$, Bézout's identity and powers of a primitive $mn$-th root give $KL=\mathbb{Q}({\mu }_{mn})$. The degrees of these cyclotomic fields are $\phi (m)$, $\phi (n)$ and $\phi (mn)=\phi (m)\phi (n)$. The natural restriction mapis injective and, by these orders, bijective. Under this product description the subgroups fixing $K$ and $L$ are the two complementary factors. Together they generate the whole group, so their common fixed field is $\mathbb{Q}$. By the Fundamental theorem of Galois theory, $K\cap L=\mathbb{Q}$. This also includes $m=1$ or $n=1$.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.

Let $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The splitting field is $L=\mathbb{Q}(\alpha ,\zeta )$. A splitting field in characteristic zero is normal and separable, hence Galois. Eisenstein criterion at $2$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic polynomial. Thus $[L:\mathbb{Q}]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The subgroup fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition subgroups, so it is not normal.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.

The Fundamental theorem of Galois theory applies to a finite Galois extension $L/K$ with group $G$. It gives inclusion-reversing inverse bijectionsThe fixed field $L^H$ consists of elements fixed by every member of $H$. Moreover $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois iff $H=\mathrm{Gal}(L/M)$ is normal, in which case restriction gives $\mathrm{Gal}(M/K)\cong G/H$. All intermediate $L/M$ are themselves Galois. Finiteness and normality/separability of $L/K$ are essential for this stated form.

Let $\sigma (\alpha )={\xi }_d\alpha$ with $\alpha \ne 0$. Thenand these $d$ elements are distinct because ${\xi }_d$ is a primitive root of unity. Thus the stabilizer of $\alpha$ in the cyclic Galois group is trivial. The orbit-stabilizer form of the Fundamental theorem of Galois theory givessoMoreover,The fixed field of the whole Galois group is $K$, and thereforeThis is the basic eigenvector mechanism behind a cyclic Kummer extension.

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 $L[x]$.