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For a Galton-Watson process with one ancestor and offspring probability generating function $F$, the branching extinction probability $q$ is the smallest fixed point of $F$ in $[0,1]$. A direct way to find it is to compute extinction by generation $n$:Zero is an absorbing state, so extinction by generation $n$ is exactly the event $Z_n=0$. ^[ref]

For $F(s)=1/4+s/4+s^2/2$, the equation isBoth $1/2$ and $1$ are fixed points, but iteration from zero selects $q=1/2$. The first values are $q_1=1/4$, $q_2=11/32$, $q_3=809/2048\approx 0.395020$, and $q_6\approx 0.462186$. Iteration started at $1$ would stay at $1$ forever and would give the wrong extinction answer for this offspring distribution.

The vertical steps in the figure apply the offspring probability generating function; horizontal steps return to the diagonal to start the next iteration. This gives a useful numerical calculation even when solving $F(s)=s$ explicitly is difficult. For the displayed law, every finite iteration is a lower bound for the eventual branching extinction probability; at generation six, about $0.037814$ additional extinction probability remains beyond the current empty families.

Why the smallest fixed point? The ancestor's children's descendant families are independent; given $k$ children, all are empty $n$ generations later with probability $q_n^k$. Averaging gives $q_{n+1}={\sum }_k{p}_k{q}_n^k=F(q_n)$. The increasing extinction events give the limit $q$, and continuity of the probability generating function yields $F(q)=q$. For any other fixed point $r\in [0,1]$, $q_0\le r$ and monotonicity of $F$ imply $q_n\le r$ at every step; hence $q\le r$. ^[ref]

Let $x_1,x_2,x_3$ be the roots of the separable polynomial $T^3+aT+b$ in characteristic two. The cyclic sumsare the distinct roots ofEven root permutations fix $r,s$, while odd permutations interchange them. The cubic Galois group is therefore contained in $A_3$ exactly when this quadratic splits over the base field.

A separable 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.

Over a field of characteristic zero, a polynomial is solvable by radicals when its roots lie in a tower obtained by adjoining successive roots; equivalently, its Galois group is solvable.

For roots ${\alpha }_1,{\alpha }_2,{\alpha }_3$ of a depressed cubic $X^3+aX+b$ over a field of characteristic zero, choose a primitive cube root of unity $\omega$. Its Lagrange resolvent for a cubic consists of the Fourier combinationsCyclically permuting the roots multiplies these combinations by cube roots of unity, so their cubes are fixed by the cyclic subgroup. Direct expansion gives $uv=-3a$ and $u^3+v^3=-27b$. Their cubes are therefore the roots of the cubic resolvent $Y^2+27bY-27a^3$.

Since the roots sum to zero, ${\alpha }_1=(u+v)/3$. The quadratic equation for the cubes, followed by compatible cube roots, gives the mechanism of the Cardano formula. A primitive root is required: choosing $\omega =1$ would make both combinations vanish.

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 $\mathbb{Q}$.

Let $f\in \mathbb{Z}[X]$ be monic and separable over $\mathbb{Q}$. If its reduction modulo a prime number $p$ is squarefree, factor it into distinct irreducible polynomials of degrees $d_1,\dots ,d_r$. The Galois group of a polynomial then contains an element with cycle lengths $d_1,\dots ,d_r$: a Frobenius automorphism at a prime of the splitting field acts on the reduced roots in precisely those orbits.

The squarefree-reduction condition, equivalently $p\nmid \mathrm{disc}(f)$, ensures that reduction distinguishes the roots. Merely assuming that $p$ is unramified in the number field generated by a root is insufficient for an arbitrary generator. For example, $X^2-5$ has repeated reduction $(X+1{)}^2$ modulo two even though two is unramified in $\mathbb{Q}(\sqrt{5})$; its power basis has index two. For a good prime, a quadratic times an irreducible cubic therefore gives a permutation with one 2-cycle and one 3-cycle, a useful constraint on 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 monic, 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.