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The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_{p^n})/{\mathbb{Q}}_p$ is generated by a primitive $p^n$th root of unity. It is totally ramified and has Galois group $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$.

A cyclotomic field is generated by a root of unity. If $K\subseteq \mathbb{C}$ and ${\zeta }_n$ is primitive, every $K$-conjugate of ${\zeta }_n$ is a power ${\zeta }_n^a$ and already lies in $K({\zeta }_n)$. Thus $K({\zeta }_n)/K$ is Galois, and its Galois group embeds in $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$, so it is abelian.

For odd $p$, remove the finite torsion subgroup ${\mu }_{p-1}$ from the Galois group of $\mathbb{Q}({\mu }_{p^{\infty }})/\mathbb{Q}$ by taking its fixed field. For $p=2$, take the fixed field of $\{\pm 1\}$ instead. The remaining Galois group is ${\mathbb{Z}}_p$, using the p-adic logarithm on $1+p{\mathbb{Z}}_p$ or $1+4{\mathbb{Z}}_2$. For a number field $K$, the cyclotomic Zp-extension is the compositum with this rational tower.

Let $E$ be a finite normal closure of the number field $K$. The permutation representation of its Galois group on the embeddings of $K$ into $E$ has one trivial constituent, because the action is transitive. At a prime, the inertia group invariant basis consists of sums over inertia orbits of embeddings; Frobenius automorphism permutes them in cycles of lengths equal to the residue degrees of the primes of $K$. A cycle of length $f$ contributes $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Hence the local factors equal those of the Dedekind zeta function, including at ramified primes. If nontrivial irreducible Artin L-functions are holomorphic and nonzero at one, this factorization and the simple pole of the Riemann zeta function show that ${\zeta }_K$ has a simple pole there.

If $r$ cyclically sends $\alpha \mapsto \beta \mapsto -\alpha \mapsto -\beta$ and $s$ fixes $\alpha$ while negating $\beta$, then $r^4=s^2=1$ and $srs=r^{-1}$. These automorphisms realize $D_8$ as the Galois group of the corresponding quartic splitting field.

For a monic separable polynomial over a field of characteristic other than two, its Galois group is contained in the alternating group exactly when its discriminant is a square in the base field. The Galois action multiplies the Vandermonde product by the sign of the induced root permutation.

The Galois group of $\mathbb{Q}({\zeta }_{11})/\mathbb{Q}$ isIts only subfield degrees over $\mathbb{Q}$ are $1,2,5,10$, and every subfield is Galois because this group is abelian.

Finite places of a number field above $p$ correspond to absolute Galois group orbits of embeddings of a number field into a p-adic algebraic closure. The absolute value is invariant under the group. Conversely, embeddings inducing the same place extend to embeddings of the same local completion and are conjugate.

Finite abelian extensions of $K$ correspond contravariantly to congruence subgroups $H$ with $P_K(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$, and the Artin map identifies $I_K(\mathfrak{m})/H$ with the Galois group.

A field automorphism is a bijective ring homomorphism from a field to itself. The automorphisms fixing a specified subfield form a Galois group when the extension is Galois.

A finite extension of the complete discretely valued field $\mathbb{C}((t))$ has residue field $\mathbb{C}$, hence is totally ramified. If $\pi$ is its uniformizer, write $t=u{\pi }^n$. A root of the residue of $u$ lifts by Hensel lemma to an $n$th root $v$ of $u$, since the residue characteristic is zero. Then $s=v\pi$ satisfies $s^n=t$. Distinct valuations modulo $n$ make $1,s,\dots ,s^{n-1}$ linearly independent over the base, so $L=\mathbb{C}((t))(s)$. Expanding in $s$ with constant coefficients identifies it with $\mathbb{C}((s))$. Its automorphisms are $s\mapsto {\zeta }_{n}s$, giving a cyclic Galois group.

Every degree-$n$ extension of ${\mathbb{F}}_q$ is isomorphic to ${\mathbb{F}}_{q^n}$.

Realize a finite p-group $P$ as the Galois group of a totally ramified extension $E/{\mathbb{F}}_p((t))$, using Artin-Schreier embedding problems over a characteristic-p field. A uniformizer $u$ of $E$ gives $E\cong {\mathbb{F}}_p((u))$: every element has a unique convergent Laurent expansion with coefficients in the residue field. An automorphism has $u\mapsto a_{1}u+a_2{u}^2+\cdots$, and $a_1$ defines a homomorphism $P\to {\mathbb{F}}_p^{\times }$. Its image is trivial because $|P|$ is a $p$-power and $p\nmid |{\mathbb{F}}_p^{\times }|$. Thus the faithful Galois action lies in the Nottingham group, embedding $P$.

For a finite residue field $k$ of cardinality $q$, arithmetic Frobenius is $x\mapsto x^q$. Its lift generates the Galois group of a finite unramified extension; geometric Frobenius is its inverse.

A continuous one-cocycle for a Galois group acting on an abelian module $M$ is a map satisfying the displayed identity. Cocycles modulo maps $\sigma \mapsto \sigma u-u$ form $H^1(K,M)$ in Galois cohomology. For an elliptic curve, choosing $mQ=P$ gives the cocycle $\sigma \mapsto \sigma Q-Q$ in $E[m]$, which represents the Kummer map of an elliptic curve.

Galois cohomology is the group cohomology of a Galois group acting on a module. For a discrete $G$-module $M$, the first group is the quotient of crossed homomorphisms $c(\sigma \tau )=c(\sigma )+\sigma c(\tau )$ by maps $c(\sigma )=\sigma m-m$.

For a Finite Galois extension $L/K$ and a finite-dimensional $L$-space $V$ with a semilinear action of its Galois group, the map $L{\otimes }_K{V}^G\to V$ is an isomorphism. The vectors ${\sum }_{\sigma }\sigma (a)\sigma (v)$ are invariant and span $V$ by the Artin independence theorem. Choosing an invariant $L$-basis then identifies the invariant space with its $K$-span. The proof does not divide by the extension degree.

The Galois group $\mathrm{Gal}(L/K)$ consists of the field automorphisms of $L$ that fix $K$ pointwise. For a finite Galois extension, its order is $[L:K]$.

An irreducible cubic over a field of characteristic other than two has Galois group $A_3$ when its discriminant is a square and $S_3$ otherwise.

Let $g(X)=X^4+b{X}^2+c\in \mathbb{Q}[X]$ be irreducible, let $\alpha$ be a root, and put $\delta =\sqrt{c}$. Its roots areso its splitting field is $K=\mathbb{Q}(\alpha ,\delta )$. If $\delta \notin \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=8$ and the Galois group is $D_8$. If $\delta \in \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=4$ and the group is $V_4$ when $\delta \in \mathbb{Q}$, and $C_4$ otherwise.