Let the monic minimal polynomial of $\alpha$ over $L$ have coefficients $a_0,\dots ,a_{r-1}$, and put $F=K(a_0,\dots ,a_{r-1})$. Each coefficient is algebraic over $K$, so adjoining these finitely many coefficients successively gives $[F:K]<\infty$. The same polynomial makes $\alpha$ algebraic over $F$, giving $[F(\alpha ):F]\le r$. The tower law for field extensions yields $[F(\alpha ):K]<\infty$. Its subfield $K(\alpha )$ therefore also has finite degree over $K$. This holds for every $\alpha \in M$, proving $M/K$ is algebraic.

Let $a=\sqrt[n]{2}$ and $F=\mathbb{Q}(\xi )$. The roots of $X^n-2$ are $a{\xi }^j$, so $L=F(a)$. This splitting extension is separable in characteristic zero, hence Galois over $\mathbb{Q}$. The subextension $F/\mathbb{Q}$ is itself Galois by part (iii), so restriction gives a surjectionwith normal kernel $H=\mathrm{Gal}(L/F)$. Surjectivity follows from the extension theorem for embeddings and normality of $L/\mathbb{Q}$.

For $\sigma \in H$, write $\sigma (a)=a{\xi }^{j(\sigma )}$. This defines an injective homomorphism $H\to ⟨\xi ⟩$: injectivity holds because $L=F(a)$, and multiplicativity holds because $H$ fixes all roots of unity in $F$. A subgroup of a cyclic group is cyclic. The first isomorphism theorem therefore givesThere is no need to assume that $|H|=n$; the intersection with the cyclotomic field can reduce its order.

The cyclotomic polynomial isBy the allowed irreducibility, this is the minimal polynomial of $\xi$ and its degree is $\phi (n)$. The field $\mathbb{Q}(\xi )$ contains every $n$th root of unity and is the splitting field of $X^n-1$. This polynomial has no repeated roots in characteristic zero, so the extension is Galois.

An automorphism sends $\xi$ to another primitive root ${\xi }^a$, where $a$ is a unit modulo $n$. Conversely each such conjugate defines an embedding, and it is an automorphism of $\mathbb{Q}(\xi )$ because $a$ has an inverse modulo $n$. Composition corresponds to multiplication of exponents. Therefore

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection for a finite Galois extension $L/K$ with group $G$:It satisfies $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois precisely when its corresponding subgroup $H$ is normal, in which case restriction induces $\mathrm{Gal}(M/K)\cong G/H$.

The Galois group $\mathrm{Gal}(L/K)$ is the group of all field automorphisms of $L$ fixing every element of $K$, with composition as its operation. A finite extension is a Galois extension if it is both normal and separable; equivalently, it is a splitting extension of separable polynomials, or its group has order $[L:K]$. Normality means that every irreducible polynomial over $K$ with a root in $L$ splits there.

Any extending field homomorphism $j$ is determined by $j(\alpha )$, since every element of $K(\alpha )$ is represented by a polynomial in $\alpha$ with coefficients in $K$. Applying $j$ to the minimal-polynomial equation shows $i(m_{\alpha })(j(\alpha ))=0$. Thus there are at most $m$ choices.

For each listed root ${\beta }_r$, define $K[x]\to L$ by $\sum a_k{x}^k\mapsto \sum i(a_k){\beta }_r^k$. Its kernel contains $(m_{\alpha })$, so it induces a unital map from the field $K[x]/(m_{\alpha })\cong K(\alpha )$ into $L$. A unital map from a field is injective. This gives an extending homomorphism $j_r$ with $j_r(\alpha )={\beta }_r$. Distinct roots give distinct maps, proving exactly $m$ extensions.

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.

The rotations form a cyclic subgroup $R\cong C_n$ of index two, so it is normal. The quotient $D_{2n}/R$ has order two and is cyclic. Hence $1⊴R⊴D_{2n}$ has abelian factors, proving $\boxed{D_{2n}\text{ is solvable}}$. Its commutator subgroup is contained in the abelian rotation subgroup, so its second derived subgroup is already trivial.

Let $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. This is an abelian subgroup, and conjugation preserves its nonidentity cycle type, so it is normal in $S_4$ and hence in $A_4$. The serieshas factors $V_4\cong C_2\times C_2$, $A_4/V_4\cong C_3$, and $S_4/A_4\cong C_2$, all abelian. Thus $\boxed{S_4\text{ is solvable}}$.

A finite group is solvable if its derived series, defined by $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$, reaches the identity after finitely many steps. Equivalently it has a subnormal series with abelian factor groups. The factors need not all have prime order for this definition.

In the distinct-root convention, a polynomial is separable if it has no repeated roots in its splitting field, equivalently $\gcd (f,f^{\prime })=1$. A common factorwise convention instead requires each distinct irreducible factor to have no repeated roots; repeated occurrences of an already separable factor do not affect the splitting field. Under either separability convention, the splitting field is a finite normal separable extension andFor failure in the inseparable case, take $K={\mathbb{F}}_p(t)$ and $f=X^p-t$. It is irreducible, for example by Eisenstein at the prime $t$ in ${\mathbb{F}}_p[t]$. Its splitting field is $L=K(t^{1/p})$ of degree $p$, but the polynomial has just one distinct root. Every $K$-automorphism fixes that root, so the Galois group is trivial rather than having order $p$.

A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. Prove existence together with the factorial divisibility by strong induction on $d$, with degrees zero and one immediate for a nonzero polynomial.

If $f$ is reducible, write $f=gh$ with $1\le e=\mathrm{deg}g<d$, $\mathrm{deg}h=d-e$. By induction first split $g$ in $K_1/K$ with degree dividing $e!$, then split $h$ over $K_1$ with degree dividing $(d-e)!$. The resulting field is generated over $K$ by all roots of $f$, hence is a splitting field. Its degree divides $e!(d-e)!$, which divides $d!$ because the binomial coefficient $d!/[e!(d-e)!]$ is an integer.

If $f$ is irreducible, adjoin a root $\alpha$, obtaining degree $[K(\alpha ):K]=d$. Over this field divide out $X-\alpha$ and split the degree-$(d-1)$ quotient by induction. That splitting field together with the base already contains every root and has degree over $K$ dividing $d(d-1)!=d!$. ThusThe reducible/irreducible distinction proves divisibility, not just an upper bound on the degree, and works without assuming separability.

Set $Y=X+1$. The polynomial becomes $Y^4+Y+1$, and translating all roots leaves their pairwise differences unchanged. Part (b) givesIt is nonzero and lies strictly between $15^2=225$ and $16^2=256$. An integer which is a rational square is an integer square, so this is not a rational square. The roots are distinct because the discriminant is nonzero. By (c), $\boxed{G_f⊈A_4}$.