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For a monic polynomial with roots $r_i$, its discriminant is ${\prod }_{i<j}(r_i-r_j{)}^2$ and vanishes exactly for a repeated root.

For a finite Galois extension, subgroups correspond in reverse order to intermediate fields by taking fixed fields and field-fixing subgroups.

If a finite group $G$ acts faithfully by automorphisms on a field $L$, then $L/L^G$ is Galois, has degree $|G|$, and has Galois group $G$.

This theorem constructs a Galois extension from a finite group of field automorphisms; it is not an alternative name for the extension itself.

For a group $G$ of automorphisms of a field $L$, the fixed field isFor one automorphism $\sigma$, it is also written $L^{\sigma }$.

A finite real extension $K/\mathbb{Q}$ is constructible when it is contained in a towerEquivalently, every element of $K$ can be obtained from rational numbers by field operations and successive square roots.

${\Phi }_n$ is the monic polynomial whose roots are the primitive $n$th roots of unity. It belongs to $\mathbb{Z}[X]$ and is irreducible over $\mathbb{Q}$.

At an unramified prime, the degrees of the irreducible factors of a polynomial modulo that prime give the cycle type of an element of the Galois group.

Over ${\mathbb{F}}_q$, Frobenius acts on roots of an irreducible degree-$n$ polynomial as an $n$-cycle. Factor degrees therefore give its cycle type.

Let $g\in K[T]$ be irreducible and separable in characteristic $p$, and put $f(T)=g(T^p)$. Every monic irreducible factor $h$ of $f$ is either $f$ itself or separable. Indeed, if $h$ were inseparable, then $h(T)=q(T^p)$; divisibility of $q(T^p)$ into $g(T^p)$ forces $q\mid g$, hence $q=g$ and $h=f$.

The Galois group of a separable polynomial acts faithfully on its roots; this action is transitive exactly when the polynomial is irreducible.

If a field $k$ has characteristic prime to $m$ and contains all $m$th roots of unity, then $k(X)/k(X^m)$ is a cyclic Galois extension of degree $m$, with automorphisms $X\mapsto \zeta X$.

The multiplicative group ${\mathbb{F}}_q^{\times }$ is cyclic of order $q-1$, so it contains all $m$th roots of unity when $m$ divides $q-1$ and the characteristic does not divide $m$.

A finite extension $L/K$ is Galois when it has $[L:K]$ automorphisms fixing $K$.

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.

The splitting field of a polynomial over $K$ is the smallest field extension of $K$ in which the polynomial is a product of linear factors.

For finite extensions $K\subseteq E\subseteq L$, the degrees multiply:If $(u_i)$ is a $K$-basis of $E$ and $(v_j)$ is an $E$-basis of $L$, then the products $(u_i{v}_j)$ form a $K$-basis of $L$: spanning follows by expanding first over $E$ and then over $K$, while $K$-linear independence follows by grouping coefficients of each $v_j$ and using both basis properties.

A field embedding $L\to \Omega$ is an injective field homomorphism. A $K$-embedding fixes every element of the base field $K$.

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.

Up to conjugacy, the transitive subgroups of $S_4$ are $C_4$, $V_4$, $D_8$, $A_4$, and $S_4$.

If $a,b,ab$ are nonsquares in a field $K$ of characteristic other than two, thenis a degree-four Galois extension with group $C_2\times C_2$. Its three quadratic intermediate fields are $K(\sqrt{a})$, $K(\sqrt{b})$, and $K(\sqrt{ab})$.