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Suppose that $L/\mathbb{Q}$ were a Finite Galois extension with Galois group $S_3$ and unramified away from $7$. Its quadratic subfield is the fixed field of the alternating subgroup $A_3$. A quadratic field ramified only at $7$ must be $\mathbb{Q}(\sqrt{-7})$: the classification by fundamental discriminants shows that $-7$ is the only nontrivial quadratic discriminant supported at $7$. In particular, $7$ ramifies in this quadratic subfield.

Because $7$ does not divide $|S_3|=6$, all ramification at $7$ is tame ramification. Its inertia group is therefore cyclic. Its image in $S_3/A_3\simeq C_2$ is nontrivial because the quadratic subfield is ramified, so the inertia group is generated by a transposition and has order two; it cannot have order six because it is cyclic.

The discriminant exponent of a tame Galois extension is thereforeThere is no other finite ramification, so $|d_L|=7^3=343$.

On the other hand, the Minkowski bound for ideal classes implies the Minkowski lower bound for a number-field discriminantSince $0\le r_2\le 3$, the right side is smallest at $r_2=3$ and is greater than $980$. This contradicts $|d_L|=343$, so no such extension exists.

The Artin–Wedderburn theorem says that every central simple algebra $A$ over $k$ has the form $M_r(D)$ for a finite-dimensional central division algebra $D$, uniquely up to the evident data. If $A$ and $B$ are central simple, extend scalars to an algebraic closure $\bar{k}$. Both become full matrix algebras, hencefor suitable $a,b$. Any nonzero proper ideal of $A{\otimes }_{k}B$ would extend to one in this simple matrix algebra, and faithful flatness prevents it from vanishing or becoming the whole algebra. The same scalar-extension argument shows that the center is $k$. This proves the tensor product of central simple algebras theorem.

The Brauer group $\mathrm{Br}(k)$ consists of Morita equivalence classes of central simple $k$-algebras. Its product is $[A][B]=[A{\otimes }_{k}B]$, its identity is $[k]$, and $[A{]}^{-1}=[A^{\mathrm{op}}]$ because $A{\otimes }_k{A}^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ is a full matrix algebra.

Let $L/k$ be a Finite Galois extension with Galois group $G_L$, and let $\varphi :G_L\times G_L\to L^{\times }$ be a normalized two-cocycle. The crossed-product algebra of a Galois extension has underlying left $L$-vector spaceand multiplicationThe cocycle identity is exactly associativity. After scalar extension to $L$, the algebra acts by the twisted regular representation and becomes $M_{|G_L|}(L)$; Galois descent shows that it is central simple over $k$. If $\varphi$ is multiplied by the coboundary of a one-cochain $b_{\sigma }$, rescaling $u_{\sigma }$ by $b_{\sigma }$ gives an isomorphic algebra. Hence the cohomological construction of a Brauer class gives a well-defined map

It remains to show that every Brauer class is torsion. For a finite group $G$, restriction and corestriction on normalized bar cochains satisfyon cohomology: the first equality follows by summing the translated cochain over coset representatives, and each $g^{\ast }$ is the identity because an inner automorphism is cochain-homotopic to the identity. Restriction to the trivial subgroup is zero in positive degree, so this proves that finite-group cohomology is annihilated by the group order. In multiplicative notation, every $\alpha \in H^2(G_L,L^{\times })$ therefore satisfies ${\alpha }^{|G_L|}=1$.

By the permitted assumption, $[A]$ is the image of such an $\alpha$ for some $L$. Consequently $[A{]}^{|G_L|}=1$ in $\mathrm{Br}(k)$. By the definition of Brauer equivalence, this says that for some $n$,as required.

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.

A finite extension is Galois when it is both normal and separable; over $\mathbb{Q}$ separability is automatic, so it is equivalently a splitting field of a polynomial with distinct roots. Its Galois group consists of the field automorphisms fixing $\mathbb{Q}$. If $G=\mathrm{Gal}(L/\mathbb{Q})$, the Fundamental theorem of Galois theory gives mutually inverse inclusion-reversing bijectionsThey satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb{Q}]=[G:H]$. The intermediate extension $K/\mathbb{Q}$ is Galois exactly when $H=\mathrm{Gal}(L/K)$ is normal, and then $\boxed{\mathrm{Gal}(K/\mathbb{Q})\cong G/H}$ via restriction.

Every permutation of the four roots permutes the three unordered partitions into two pairs and hence permutes the three ${\beta }_j$. Thus each coefficient of $g(X)={\prod }_{j=1}^3(X-{\beta }_j)$ is a symmetric polynomial in the four roots. The Fundamental theorem of symmetric polynomials expresses it in the elementary symmetric functions, which are rational coefficients of $f$. Hence $g\in \mathbb{Q}[X]$. Each product ${\alpha }_i{\alpha }_j$, $i<j$, occurs exactly twice in ${\beta }_1+{\beta }_2+{\beta }_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.

For the given cubic write $g(X)=(X-2{)}^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its Galois group is transitive on three roots, so its order is divisible by three; complex conjugation is a transposition. A subgroup of $S_3$ with both a three-cycle and a transposition is all of $S_3$. HenceLet $G=\mathrm{Gal}(L/\mathbb{Q})\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic splitting field $K$ lies inside $L$ by the definitions of the ${\beta }_j$, and restriction onto its Galois group gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a normal subgroup of $A_4$ of index six, contrary to the other supplied result. Thus

For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

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 $p$.

Eisenstein's criterion at two makes $X^5-4X+2$ irreducible: all nonleading coefficients are divisible by two and the constant is not divisible by four. Its Galois group is therefore transitive on five roots, and its order is divisible by five, so Cauchy's theorem for finite groups supplies a five-cycle.

The derivative is $5X^4-4$, with real critical points $\pm c$, $c=(4/5{)}^{1/4}$. The values are $f(-c)=2+16c/5>0$ and $f(c)=2-16c/5<0$. The derivative is positive outside these two points and negative between them, so there are exactly three real roots and one nonreal conjugate pair. Complex conjugation therefore acts as a single transposition.

A five-cycle together with any transposition generates $S_5$: conjugating that transposition by powers of the cycle gives edges joining labels with a fixed nonzero difference modulo five. These edges form a connected graph, and transpositions on a connected graph generate the symmetric group. HenceThe equality of degree and group order uses separability in characteristic zero.

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.

The polynomial is irreducible by Eisenstein's criterion at $2$. Its Galois group therefore acts transitively on five roots, so its order is divisible by $5$ by the orbit-stabilizer theorem.

To count the real roots, differentiate:It is positive on $(-\infty ,0)$, negative on $(0,\sqrt[3]{4})$, and positive thereafter. The local maximum is $f(0)=2>0$ and the local minimum is $2-6(\sqrt[3]{4}{)}^2<0$. The limits at infinity have opposite signs, so there are exactly three real roots and one conjugate pair of nonreal roots. Complex conjugation fixes the three real roots and interchanges the other two; its permutation is a transposition. The prime-degree subgroup result supplied in the paper now gives

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 $A_3$.

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

Let $M_{\infty }$ be the maximal abelian pro-$p$ extension of $F_{\infty }$ unramified away from primes over $p$. Its Galois group is the p-ramified Iwasawa module. In the cyclotomic tower of $\mathbb{Q}({\mu }_p)$ for odd $p$, its Iwasawa-module rank is $(p-1)/2$. It is therefore quite different from the torsion unramified Iwasawa module.

If a separable cubic has polynomial discriminant $\Delta$, then adjoining $\sqrt{\Delta }$ gives the subfield fixed by the alternating part of its Galois group. When the cubic Galois group is $S_3$, this is its unique quadratic subfield.

For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ is the displayed field. A quadratic Gauss sum squares to $(-1{)}^{(\ell -1)/2}\ell$, and the cyclic Galois group has a unique subgroup of index two.

In characteristic different from two, a field generated by finitely many nonzero square roots is the splitting field of a collection of separable quadratic polynomials. Its Galois group embeds into the sign-vector group $(\mathbb{Z}/2{)}^n$. Every index-two subgroup is the kernel of a nontrivial sign character. Coordinate sign characters span the dual of this subgroup of $(\mathbb{Z}/2{)}^n$, by extending a linear functional to the ambient vector space. A product of the corresponding roots transforms by that character and generates precisely the quadratic fixed field.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

A resolvent is an auxiliary polynomial built from expressions in the roots of a polynomial. The action of its Galois group on those expressions gives information about solvability and intermediate field extensions. A Lagrange resolvent for a cubic uses Fourier combinations of the three roots.

If an irreducible quartic over a field of characteristic zero has a cubic resolvent of a quartic whose Galois group of a polynomial is $S_3$, then the quartic has Galois group $S_4$. Irreducibility places the quartic group among the transitive subgroups of the symmetric group on four points, while the action on its three root pairings is surjective onto $S_3$. Of the possibilities $C_4,V_4,D_8,A_4,S_4$, only $S_4$ has this image.

For an extension $L/K$ with Galois group $G$, a semilinear action on an $L$-space is an additive group action satisfying $\sigma (av)=\sigma (a)\sigma (v)$. Scalar coefficients are transformed along with vectors. Invariants form a $K$-space, and Galois descent of vector spaces reconstructs the original space from it.

A polynomial is solvable by radicals when its roots lie in a tower obtained by adjoining successive roots; equivalently, its Galois group is solvable.