Write $s=\sqrt{-33}$. The discriminant is $-132$, and the bound of part (a) isEvery ideal class therefore has an integral representative of norm at most $7$. The Kummer-Dedekind theorem givesModulo $5$, $x^2+33=x^2+3$ is irreducible, so there is no ideal of norm $5$. Modulo $7$ it has roots $\pm 3$, giving $(7)=R\bar{R}$, where $R=(7,s+3)$. The ideal of $3+s$ has norm $42$, and reduction in the three residue fields shows that it lies in $P,Q,R$. Equality of norms therefore givesIt follows that $[R]=[P][Q]$, because $[P{]}^2=[Q{]}^2=1$. The other prime over $7$ has the inverse class, which is the same class here.

By unique factorization of nonzero ideals, all norms at most $7$ now give only the classes $1,[P],[Q],[PQ]$: norm $4$ gives $P^2=(2)$, norm $6$ gives $PQ$, and norm $7$ was handled above. None of $P,Q,PQ$ is principal, since their norms are $2,3,6$ andAlso $[P]\ne [Q]$, since equality would make $PQ$ principal. Thus there are exactly four classes, all of exponent two:

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one obtains such an ideal by multiplying the fractional ideal $A^{-1}$ by a suitable nonzero integer. The assumed Minkowski bound for ideal classes gives $0\ne \beta \in B$ withBecause $(\beta )\subseteq B$ and nonzero ideals of the ring of integers are invertible,It is an integral ideal in class $[B{]}^{-1}=[A]$. By multiplicativity of the ideal norm,The same argument applies if the starting $A$ is a fractional ideal, since its class is unchanged by clearing denominators.

Put $s=\sqrt{-5}$ and $P=(2,1+s)$. The polynomial $x^2+5$ reduces to $(x+1{)}^2$ modulo $2$, so the Kummer-Dedekind theorem givesIf $P=(a+bs)$, its norm would imply $a^2+5b^2=2$, impossible for integers $a,b$. Thus $P$ represents the nontrivial ideal class. Since the class number is $2$, every nonprincipal integral ideal $I$ has $[I]=[P]$, and $[IP]=1$. Write $IP=(\gamma )$ with $\gamma \in {\mathcal{O}}_M$; integrality follows from $IP\subseteq {\mathcal{O}}_M$. Multiplying by $P$ givesNorms give $4N(I)=2N(\gamma )$, hence $N(\gamma )=2N(I)$. If $N(I)=p_3\equiv 1(\mathrm{mod}20)$, writing $\gamma =a+bs$ yieldscontradicting the list $0,1,4$ of squares modulo $5$. Therefore no nonprincipal ideal has norm $p_3$.

As $p_3\equiv 1(\mathrm{mod}4)$ and $p_3\equiv 1(\mathrm{mod}5)$, quadratic reciprocity gives $(-5/p_3)=1$. Thus $x^2+5$ has two distinct roots modulo $p_3$ andThe norm obstruction shows that $R$ must be principal, say $R=(x+ys)$. Taking its norm proves

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

Let $\theta =\sqrt[3]{3}$. The power-basis discriminant isIf $\mathbb{Z}[\theta ]$ were a proper sublattice of ${\mathcal{O}}_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb{Z}[\theta ]$ already belongs to $\mathbb{Z}[\theta ]$. This is a contradiction. Hence

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

A finite solvable group is a group admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chainhas quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every subgroup of a solvable group is solvable, because intersecting the chain with the subgroup gives factors isomorphic to subgroups of the original abelian factors. A Galois group of a polynomial of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the polynomial is reducible or has repeated roots.

For the substitution, the characteristic-zero solvability by radicals theorem says that a polynomial has solvable Galois group if and only if its splitting field is contained in a finite tower formed by adjoining successive roots of elements of the preceding field. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th root of unity and, for every nonzero $\alpha$, an $m$th root ${\beta }_{\alpha }$. These are radical adjunctions, and all roots of $x^m-\alpha$ are ${\beta }_{\alpha }{\zeta }_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the splitting field of $f(x^m)$. The theorem implies

The Fundamental theorem of symmetric polynomials states that every symmetric polynomial in $n$ variables over a field is a polynomial in the elementary symmetric polynomials, uniquely so. For the three roots, Vieta's formulas give $e_1=0$, $e_2=p$, $e_3=-q$.

The squared product of root differences is symmetric. Expansion in elementary symmetric polynomials givesOne can verify the displayed identity simply by substituting the definitions of $e_1,e_2,e_3$ and expanding; it is the cubic polynomial discriminant.

If the cubic is irreducible, its roots are distinct in characteristic zero, so $\delta \ne 0$. Its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$. A permutation of the roots multiplies $\delta$ by its sign. The fixed-field assertion of the Fundamental theorem of Galois theory therefore gives $\delta \in \mathbb{Q}$ exactly when all group elements are even. ThusFor example, $f=x^3-3x-1$ has no rational root: neither $1$ nor $-1$ is a root, and these are the only possibilities by the rational root theorem. Its discriminant is $108-27=81$, so $\delta =\pm 9\in \mathbb{Q}$ and its group is $A_3$.

The Fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with group $G$, the maps $H\mapsto L^H$ and $E\mapsto \mathrm{Gal}(L/E)$ are mutually inverse inclusion-reversing correspondences between subgroups and intermediate fields. They satisfy $[L:L^H]=|H|$ and $[L^H:F]=[G:H]$. An intermediate extension $E/F$ is Galois exactly when its subgroup is normal, in which case its group is the quotient by that subgroup.

Let $a=\sqrt{2}$, $b=\sqrt{3}$, $c=i$. The real field $\mathbb{Q}(a,b)$ has degree $4$: if $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$, squaring forces $uv=0$ and either $u^2=3$ or $v^2=3/2$, both impossible. Adjoining the nonreal $i$ doubles the degree. All three square-root signs can be changed independently, givingWrite ${\sigma }_{uvw}$ for the automorphism sending $(a,b,c)$ to $((-1{)}^{u}a,(-1{)}^{v}b,(-1{)}^{w}c)$, where $(u,v,w)\in {\mathbb{F}}_2^3$. Every nonzero vector generates an order-two subgroup, so there are $7$ such subgroups. Their fixed fields are:

Each listed field has degree $4$ and is fixed by the indicated sign change, which proves equality with the fixed field.

There are also $7$ order-four subgroups: they are the kernels of the seven nonzero linear functionals on ${\mathbb{F}}_2^3$. For $e=(e_1,e_2,e_3)\ne 0$, define $H_e=\{{\sigma }_{uvw}:e_{1}u+e_{2}v+e_{3}w=0\}$. ThenThus the seven quadratic fields have radicands $2,3,-1,6,-2,-3,-6$. Their square classes are distinct, so all are different. Finally the trivial subgroup fixes $L$, and all of $G$ fixes $\mathbb{Q}$. This accounts for every subgroup and intermediate field.

The fixed field isIt is a subfield because every field automorphism preserves sums, products and inverses.

Set $F=K^G$. If $[K:F]$ is infinite the desired finite lower bound is immediate. Otherwise let $b_1,\dots ,b_d$ be an $F$-basis of $K$. If $|G|>d$, the homogeneous system over $K$has a nonzero solution. For $a={\sum }_j{a}_j{b}_j$, with $a_j\in F$, every $\sigma$ fixes $a_j$, so the same coefficients give ${\sum }_{\sigma }{c}_{\sigma }\sigma (a)=0$. This contradicts linear independence of distinct field embeddings. Consequently

A splitting field of a nonconstant polynomial $f$ over a field $K$ is an field extension $L$ in which $f$ factors into linear factors and which is generated over $K$ by those roots. We prove existence and the divisibility simultaneously by induction on $n=\mathrm{deg}f$; constant nonzero polynomials have splitting field $K$.

If $f$ is reducible, write $f=gh$ with positive degrees $r,s<n$. By induction first split $g$, obtaining $E/K$ of degree dividing $r!$, then split $h$ over $E$, obtaining $L/E$ of degree dividing $s!$. The tower law for field extensions gives $[L:K]\mid r!s!\mid (r+s)!$. All adjoined elements are roots of $f$, so $L$ has the required minimality.

If $f$ is irreducible, take $E=K[x]/(f)$ and let $\alpha$ be the image of $x$. Then $[E:K]=n$ and $f=(x-\alpha )h$ in $E[x]$. Apply induction to $h$, of degree $n-1$, to obtain $[L:E]\mid (n-1)!$. ThereforeThis proof also covers inseparable polynomials; it does not incorrectly identify extension degree with automorphism count in that case.

Identify $\mathbb{C}$ with the Euclidean plane and start with the points $0,1$. A complex number is a constructible number if a finite sequence of drawing lines through known points and circles with known centres and known distances as radii, and selecting their intersections, produces that point.

Line-line intersections solve linear equations in the current coordinate field. Line-circle and circle-circle intersections require at most one quadratic equation after eliminating a coordinate. Hence all coordinates obtained lie in a finite tower of real quadratic extensions of $\mathbb{Q}$. If $\alpha =x+iy$ is constructible, choose a common such real field $F$ containing $x,y$. Adjoining $i$ if needed still gives an extension $F(i)/\mathbb{Q}$ of power-of-two degree. By the tower law for field extensions,A regular heptagon centred at $0$ with one vertex $1$ would have $\xi =e^{2\pi i/7}$ as another constructible vertex. But its cyclotomic degree is $\phi (7)=6$, which is not a power of two. Thus the regular heptagon is not constructible.