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A finite separable extension is solvable when its normal closure has a solvable group as Galois group. In characteristic zero it is contained in a radical tower after adjoining suitable roots of unity; containment need not mean it is itself a tower of radical adjunctions.

Let a separable polynomial $f$ of degree four be irreducible over a field $F$. If all three pair sums $r_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$, $r_2={\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4$, and $r_3={\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3$ lie in $F$, its Galois group is the Klein four-group. Indeed $r_1-r_2=({\alpha }_1-{\alpha }_4)({\alpha }_2-{\alpha }_3)\ne 0$, and similarly the other pair sums differ. Every Galois automorphism therefore fixes each of the three partitions into pairs. The kernel of this action of $S_4$ on three partitions is $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Irreducibility gives a transitive group action on the four roots, so the orbit-stabilizer theorem forces $4$ to divide the group order. Hence the Galois group equals $V_4$. The argument also works in characteristic two.

The polynomial $X^3-X-1$ is irreducible over $\mathbb{Q}$ and has discriminant $-23$, so its splitting field has Galois group $S_3$. Its unique quadratic subfield is the fixed field of $A_3$, namely $\mathbb{Q}(\sqrt{-23})$.

In a Galois extension, the prime ideals above a fixed unramified prime have a common ramification index and residue degree, and the Galois group acts transitively on them.

For a separable polynomial $f\in F[T]$ in characteristic two, the Vandermonde determinant of its roots is fixed by every Galois automorphism: root permutations change it only by a sign, and $-1=1$ in $F$. It therefore lies in $F$. Its square, multiplied by the square of the appropriate leading-coefficient power for a nonmonic polynomial, is the polynomial discriminant. Thus being a square does not detect containment of the Galois group in the alternating group. For example $T^2+T+1$ over ${\mathbb{F}}_2$ has discriminant one and Galois group $C_2$, while $A_2$ is trivial. The discriminant criterion for an alternating Galois group requires characteristic different from two.

The Cyclotomic character of a finite cyclotomic extension and the Chinese remainder theorem give the displayed Galois group. It has three order-two subgroups and three order-four subgroups; the latter are kernels of its three nonzero group homomorphisms to $C_2$. By the Fundamental theorem of Galois theory, its proper intermediate fields strictly above $\mathbb{Q}$ are the three quadratic fieldsand the three quartic fieldsThe first quartic has cyclic Galois group of order four and unique quadratic subfield $\mathbb{Q}(\sqrt{5})$; the second is a biquadratic extension, so they differ. The last is real whereas the others are not, by the maximal real subfield of a cyclotomic field formula. Also $\sqrt{-3}+\sqrt{5}$ generates the second quartic: if $s=\sqrt{-3}+\sqrt{5}$, then $\sqrt{-3}=(s-8/s)/2$ and $\sqrt{5}=(s+8/s)/2$.

For a finite Galois extension of number fields, choose a nonzero residue field generator at $\mathfrak{q}$ and lift it to an integral element $a$ whose residues at the other primes above $\mathfrak{p}$ are zero, using the Chinese remainder theorem. The orbit polynomial ${\prod }_{\sigma \in G}(T-\sigma a)$ has coefficients in the base ring of integers. Modulo $\mathfrak{q}$, its nonzero roots are precisely the orbit of $\overline{a}$ under the decomposition group. Its coefficients belong to $k_{\mathfrak{p}}$, so finite-field Frobenius automorphism permutes these roots. Consequently some decomposition-group element acts as Frobenius on the generator, and therefore on the entire residue field. Frobenius generates the finite-field Galois group, proving surjectivity.

If $L/K$ is a finite Galois extension of number fields, its Galois group acts transitively on the prime ideals of ${\mathcal{O}}_L$ lying above a fixed prime ideal of ${\mathcal{O}}_K$.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.

The splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{-59})$. The extension over that subfield is abelian of degree three and unramified everywhere. The only possible rational ramified prime is $59$; at that prime a simple residue root lifts, so the local splitting field is quadratic and coincides with the quadratic subfield's completion. It follows from the Hilbert class field degree that $3\mid h_{\mathbb{Q}(\sqrt{-59})}$.

Suppose an irreducible monic integral cubic has discriminant $-p$ for a prime $p>3$, and let $L$ be its splitting field. Its Galois group is $S_3$ and its quadratic subfield is $\mathbb{Q}(\sqrt{-p})$. The extension over that subfield is cyclic cubic and everywhere unramified. Away from $p$, separable reduction and Hensel lemma put the roots in an unramified local extension. At $p$, discriminant valuation one gives a double residue root and a separate simple root; lift the simple root, and the remaining quadratic's discriminant has square class $-p$. Its splitting field is already the quadratic subfield's completion. The imaginary quadratic base has no real places.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

For a Zp-extension, let $L_{\infty }$ be the maximal unramified abelian pro-$p$ extension of $F_{\infty }$. Its Galois group is a compact Iwasawa module, identifiable by Artin reciprocity with the inverse limit of the $p$-primary ideal class groups of finite layers under norms.

Let $R$ be an integrally closed domain and $f\in R[X]$ monic, with discriminant outside a prime ideal $P$. The roots stay distinct after reduction. In the splitting algebra choose a prime above $P$. Its decomposition group maps onto the Galois group of the residue splitting field; its inertia group fixes every reduced root and is trivial because reduction bijects the distinct root sets. Thus the residue Galois group embeds in the generic group as a permutation group. For finite residue fields, irreducible factor degrees give the Frobenius cycle type.

For a normal subgroup $H$ of a finite local Galois group, the upper ramification numbering on its quotient is obtained by projecting the upper groups. The Herbrand function is the change of variable that ensures this compatibility. In contrast, the lower ramification numbering restricts directly to subgroups.

The upper numbering is obtained by applying the inverse Herbrand function to the index in the lower ramification numbering. It is compatible with quotient Galois groups, whereas lower numbering is compatible with subgroups.

For a non-Archimedean local field $K$, the Weil group is the inverse image of the infinite cyclic subgroup generated by Frobenius under $\mathrm{Gal}(\overline{K}/K)\to \mathrm{Gal}(\overline{k}/k)\cong \hat{\mathbb{Z}}$. It is dense in the absolute Galois group with its profinite topology.

A Zp-extension is an infinite Galois extension of a number field whose Galois group is topologically isomorphic to the additive p-adic integers. There is a unique intermediate field of degree $p^n$ for every $n\ge 0$, corresponding to $p^n{\mathbb{Z}}_p$. The finite layers form a tower with cyclic successive degree $p$ extensions.