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A splitting field of a nonconstant polynomial $f$ over a field $K$ is an field extension $L$ in which $f$ factors into linear factors and which is generated over $K$ by those roots. We prove existence and the divisibility simultaneously by induction on $n=\mathrm{deg}f$; constant nonzero polynomials have splitting field $K$.

If $f$ is reducible, write $f=gh$ with positive degrees $r,s<n$. By induction first split $g$, obtaining $E/K$ of degree dividing $r!$, then split $h$ over $E$, obtaining $L/E$ of degree dividing $s!$. The tower law for field extensions gives $[L:K]\mid r!s!\mid (r+s)!$. All adjoined elements are roots of $f$, so $L$ has the required minimality.

If $f$ is irreducible, take $E=K[x]/(f)$ and let $\alpha$ be the image of $x$. Then $[E:K]=n$ and $f=(x-\alpha )h$ in $E[x]$. Apply induction to $h$, of degree $n-1$, to obtain $[L:E]\mid (n-1)!$. ThereforeThis proof also covers inseparable polynomials; it does not incorrectly identify extension degree with automorphism count in that case.

For a splitting field $L$ of $f$ over $K$, define $\mathrm{Gal}(f)={\mathrm{Aut}}_K(L)$. Each automorphism permutes the distinct roots, since it fixes the coefficients. An automorphism fixing every root fixes the field they generate, so this action is faithful. If there are $r\le n$ distinct roots, it gives an embedding into $S_r$, and hence into $S_n$ by fixing $n-r$ additional symbols.

The Fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with group $G$, the maps $H\mapsto L^H$ and $E\mapsto \mathrm{Gal}(L/E)$ are mutually inverse inclusion-reversing correspondences between subgroups and intermediate fields. They satisfy $[L:L^H]=|H|$ and $[L^H:F]=[G:H]$. An intermediate extension $E/F$ is Galois exactly when its subgroup is normal, in which case its group is the quotient by that subgroup.

Let $a=\sqrt{2}$, $b=\sqrt{3}$, $c=i$. The real field $\mathbb{Q}(a,b)$ has degree $4$: if $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$, squaring forces $uv=0$ and either $u^2=3$ or $v^2=3/2$, both impossible. Adjoining the nonreal $i$ doubles the degree. All three square-root signs can be changed independently, givingWrite ${\sigma }_{uvw}$ for the automorphism sending $(a,b,c)$ to $((-1{)}^{u}a,(-1{)}^{v}b,(-1{)}^{w}c)$, where $(u,v,w)\in {\mathbb{F}}_2^3$. Every nonzero vector generates an order-two subgroup, so there are $7$ such subgroups. Their fixed fields are:

Each listed field has degree $4$ and is fixed by the indicated sign change, which proves equality with the fixed field.

There are also $7$ order-four subgroups: they are the kernels of the seven nonzero linear functionals on ${\mathbb{F}}_2^3$. For $e=(e_1,e_2,e_3)\ne 0$, define $H_e=\{{\sigma }_{uvw}:e_{1}u+e_{2}v+e_{3}w=0\}$. ThenThus the seven quadratic fields have radicands $2,3,-1,6,-2,-3,-6$. Their square classes are distinct, so all are different. Finally the trivial subgroup fixes $L$, and all of $G$ fixes $\mathbb{Q}$. This accounts for every subgroup and intermediate field.

Right multiplication gives a right action $(Hx)\cdot g=Hxg$. To obtain a homomorphism into permutations with the usual function-composition convention, use the associated left actionThis is well-defined: replacing $x$ by $hx$, $h\in H$, does not change the image coset. Moreover $\rho (g)\rho (k)(Hx)=Hx{k}^{-1}{g}^{-1}=\rho (gk)(Hx)$, and each map is bijective with inverse $\rho (g^{-1})$.

An element $g$ is in the kernel precisely when $Hx{g}^{-1}=Hx$ for every $x\in G$. This says $x{g}^{-1}{x}^{-1}\in H$, equivalently $g\in x^{-1}Hx$. HenceThis is the largest normal subgroup of $G$ contained in $H$.

Now the index of $H$ is $q$, so $G/\ker \rho$ embeds into $S_q$. Its order therefore divides $q!$. Since $p>q$ is prime, $p$ does not divide $q!$. From $|G|=pq$ it follows that $p$ divides $|\ker \rho |$. But $\ker \rho \subseteq H$ and $|H|=p$, forcing $\ker \rho =H$. Therefore $H$ is normal.

For the last case let $H=⟨b⟩$. Normality makes conjugation by $a$ an automorphism, so $a^{-1}ba=b^r$ with $r\in \{1,2,3,4\}$. Applying it twice and using $a^2=1$ gives $b=b^{r^2}$, so $r^2\equiv 1(\mathrm{mod}5)$. Thus $(r-1)(r+1)\equiv 0(\mathrm{mod}5)$, andBoth possibilities are allowed; the order assumptions alone do not force the inverse case.

For the given group homomorphism $\phi :H\to \mathrm{Aut}(N)$, the external semidirect product has underlying Cartesian product $N\times H$ and operationThe notation $\mathrm{Aut}(N)$ denotes the automorphism group of $N$. Closure follows because $\phi (h)$ is an automorphism of $N$. To check associativity, calculate both bracketings:They agree because $\phi (hk)=\phi (h)\phi (k)$. The identity is $(e_N,e_H)$, using $\phi (e_H)={\mathrm{id}}_N$, and the two-sided inverse isIndeed multiplication in either order gives the identity, using $\phi (h)\phi (h^{-1})={\mathrm{id}}_N$. This verifies the group axioms.

For the specified group presentation, take $N=C_9=⟨X⟩$ and $H=C_3=⟨Y⟩$. The rule $\alpha (X)=X^4$ defines an automorphism because $4$ is coprime to $9$. Furthermore $4^3\equiv 1(\mathrm{mod}9)$, so $Y\mapsto \alpha$ defines the required group homomorphism $C_3\to \mathrm{Aut}(C_9)$. In the resulting semidirect product $S$, the canonical generators satisfyConsequently the group presentation defines a surjective group homomorphism $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying Cartesian product gives $|S|=27$, whereas part (a) gives $|G|\le 27$. Surjectivity forces equality, so the map is an isomorphism:This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has element order exactly $9$.

Use the fact that $K$ is a normal subgroup to move a factor of $K$ past a factor of $H$. For $h_1,h_2\in H$ and $k_1,k_2\in K$,whileThe identity element belongs to $HK$. Hence $HK$ is a subgroup, by the subgroup test. This proves the product of a subgroup with a normal subgroup construction; $H$ need not be a normal subgroup.

For each $h\in H$, conjugation maps $K$ to itself and respects its multiplication. The inverse of $k\mapsto h^{-1}kh$ is $k\mapsto hk{h}^{-1}$, so both are group automorphisms of $K$.

The printed formula requires a composition convention. For the printed ${\varphi }_h(k)=h^{-1}kh$, ordinary function composition givesIndeed, $(h_1{h}_2{)}^{-1}k(h_1{h}_2)=h_2^{-1}(h_1^{-1}k{h}_1)h_2$. Thus the displayed assignment is a group antihomomorphism with the usual multiplication in the automorphism group. It is a group homomorphism if the automorphisms are written as right operators and multiplied in the corresponding reverse order; this is precisely why right conjugation reverses composition.

With the usual left function composition, the corresponding group homomorphism is insteadDirect expansion gives ${\psi }_{h_1{h}_2}(k)=h_1(h_{2}k{h}_2^{-1})h_1^{-1}=({\psi }_{h_1}\circ {\psi }_{h_2})(k)$. This states the intended conjugation action without silently reversing the order in the printed formula.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

Use letters $x_i$ and their formal inverses. An elementary contraction deletes an adjacent pair $x_i{x}_i^{-1}$ or $x_i^{-1}{x}_i$. A reduced word has no such pair. Repeated contraction terminates because each step reduces length by two, so every word has a reduced descendant.

For uniqueness, two possible first contractions either occur on disjoint pairs, when performing both yields a common descendant, or overlap in three letters $a{a}^{-1}a$, when either contraction produces the same remaining letter. Induct on word length: the shorter descendants of both first contractions therefore have the same eventual reduced word. This proves every contraction sequence from a given word ends at the same reduced word. An insertion or deletion of an inverse pair preserves that reduced result. Since the word model of the free group identifies words exactly under finite sequences of these moves, every group element has a unique reduced representative.

To define the requested homomorphism, substitute $w_i$ for $x_i$ and $w_i^{-1}$ for $x_i^{-1}$ in any word. Inverse-pair insertions or deletions map to inverse words that cancel, so substitution is well defined on group elements and respects concatenation. This proves the universal property of the free group in this case.

For an automorphism of order four, take $\varphi (x_1)=x_2$ and $\varphi (x_2)=x_1^{-1}$. Its iterates on $x_1$ are $x_2,x_1^{-1},x_2^{-1},x_1$; its fourth iterate also fixes $x_2$. Unique reduced representatives show the first three iterates on $x_1$ differ from $x_1$. Hence its order is exactly four.

For the final map, every iterate of $x_2$ is a positive word and contains an $x_2$. Substituting once replaces each occurrence of $x_2$ by $x_{2}w$ and leaves each $x_1$ unchanged. If $\ell =|w|\ge 1$, its length consequently increases by at least $\ell$ on every iteration. No cancellation occurs in a positive word, so $|{\varphi }^n(x_2)|\ge 1+n\ell >1$. It cannot equal $x_2$, proving no positive iterate of this endomorphism is the identity. Surjectivity of this last endomorphism is not assumed.

This uses positive-word growth under a free-group endomorphism.

For $\sigma \in G_i$, the residue of $(\sigma {\pi }_L-{\pi }_L)/{\pi }_L^{i+1}$ gives a homomorphism to the additive residue field with kernel $G_{i+1}$. On composing two automorphisms, their leading displacement coefficients add because each fixes residues and changes the uniformizer only in degree at least two. These successive quotients are elementary abelian groups in residue characteristic $p$. The filtration terminates, so the wild inertia group is a finite $p$-group.

For an abelian group $A$, its $p$-primary torsion subgroup is $A\{p\}=\{a:p^{r}a=0\text{ for some }r\ge 0\}$. It is zero exactly when $A[p]=0$, since a nonzero element of $p$-power order has a multiple of order $p$. Multiplication by an integer prime to $p$ is an automorphism on it.

In a totally ramified Galois extension, the uniformizer criterion for lower ramification groups gives $\sigma \in G_m\setminus G_{m+1}$ exactly when $v_L((\sigma (\pi )-\pi )/\pi )=m$. The Newton polygon root valuation theorem therefore identifies a slope $-m$ in the coefficient-index convention with a lower ramification jump at $m$. Its horizontal length is the number of automorphisms in that difference. Under the reflected Newton polygon convention the same jump has slope $+m$.

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

For $L={\mathbb{Q}}_3({\zeta }_3,\alpha )$ with ${\alpha }^3=2$, the shifted cubic for $\alpha +1$ and quadratic for ${\zeta }_3-1$ are Eisenstein. Their compositum is totally ramified of degree six. The element $\pi =({\zeta }_3-1)/(\alpha +1)$ has valuation one. Order-three automorphisms move it with valuation two and transpositions with valuation one. The lower breaks are zero and one, the upper breaks zero and one-half, and the different exponent from ramification groups is seven.

For odd $p$, the splitting field $K={\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. Withthe lower ramification groups are

To verify the filtration, write ${\alpha }^p=p$ and $\lambda ={\zeta }_p-1$. The Eisenstein polynomials for $\alpha$ and $\lambda$ give coprime degrees $p$ and $p-1$, so their compositum has degree $p(p-1)$. Its element $\varpi =\lambda /\alpha$ has $p$-adic valuation $1/[p(p-1)]$, forcing total ramification. For an automorphism ${\sigma }_{a,b}$ with $\alpha \mapsto {\zeta }_p^a\alpha$ and ${\zeta }_p\mapsto {\zeta }_p^b$,Its residue is $b$, so the normalized displacement valuation is one if $b\ne 1$. If $b=1$ and $a\ne 0$, that valuation is $1+v_K({\zeta }_p^{-a}-1)=p+1$. The uniformizer criterion for lower ramification groups now gives every group displayed above.

In $L={\mathbb{Q}}_2(\sqrt{2})$, set $\pi =\sqrt{2}$ and normalize $v_L(\pi )=1$, so $v_L(2)=2$. The two terms of $a+b\pi$ have valuations of opposite parity, giving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$. The nonidentity automorphism sends $\pi$ to $-\pi$, and $v_L(-2\pi )=3$. The lower ramification numbering therefore gives the displayed groups. The different exponent is three.

The ramification polynomial records the normalised differences of conjugates of a uniformizer, including the zero difference from the identity automorphism. Its quotient by $X$ is the reduced ramification polynomial. In the usual coefficient-index convention, the finite slopes of its lower Newton polygon are negatives of the valuations of its nonzero roots; the coefficient at the zero root is treated as having infinite valuation.

NFU has a unary predicate $S$ for sets, atoms have no members, extensionality applies to sets, and comprehension applies to every stratified formula. A stratification assigns integer types to variables so that equality uses equal types and $x\in y$ requires $\mathrm{type}(y)=\mathrm{type}(x)+1$. A unary set predicate adds no type difference.

Start with the actual set structure $B=(V_{\omega +\omega },\in )$. This structure satisfies extensionality and every pure-membership separation instance: the subset defined in a set $a$ of rank below $\omega +\omega$ still has rank below $\omega +\omega$. We do not assert that $B$ satisfies replacement or all of ZFC. Expand it by Skolem functions. In it use the sequence of rank objects $V_{\omega +n+6}$, $n<\omega$. For two such objects $d<e$ in sequence order, we havewith quantifiers in $B$. The second statement holds because every subset of $V_{\alpha }$ belongs to $V_{\alpha +1}$, and the later rank is at least $\alpha +1$.

Introduce constants $d_i$, $i\in \mathbb{Z}$, and require them to be order indiscernibles in the Skolem language, satisfying these two assertions whenever $i<j$. Also require $V_{\omega +5}\subseteq d_i$ for every $i$; this fixed rank object is first-order definable in $B$. Every finite part of these requirements together with $\mathrm{Th}(B^{\ast })$ is satisfiable: color finite increasing tuples of the original rank sequence by the truth values of the finitely many formulas mentioned, and use Ramsey's theorem to get a homogeneous subsequence. The rank assertions hold for every increasing tuple there. The compactness theorem gives a model of the full theory. In its Skolem hull $N$ generated by the $d_i$, the integer shift extends to an automorphism $j$ with $j(d_i)=d_{i+1}$, by the term-transport proof of the Ehrenfeucht-Mostowski theorem. Thus $N$ still satisfies extensionality and all pure-membership separation instances, andThis is a set-sized first-order model; no satisfaction predicate for a proper-class universe is being assumed.

Externally let $D_i=\{x\in N:N\models x\in d_i\}$. The automorphism bijects $D_i$ onto $D_{i+1}$. Interpret a typed structure with sort $i$ equal to $D_i$. At sort $i+1$, declare $y$ a set exactly when $N\models y\subseteq d_i$; otherwise it is an atom. Define adjacent-sort membership byThe guard is necessary: a typed atom might have some ordinary $N$-members in $D_i$, and those must not become its typed members. Set-extensionality follows from extensionality in $N$, because the ordinary members of a typed set all lie in $D_i$. For any typed formula on sort $i$, its quantifiers can be restricted to the corresponding $d_l$, its set predicates replaced by subset assertions, and its memberships replaced by the guarded relation. This is an ordinary pure-membership formula of $N$ with finitely many parameters. Separation gives its extension $X\subseteq d_i$, and the displayed closure assertion gives $X\in d_{i+1}$. It is therefore a typed set witnessing comprehension. The automorphism shifts these sorted domains and preserves the set flags and guarded memberships.

Collapse the types onto $D_0$. DefineObjects not satisfying $S$ have no $E$-members. If two sets have the same $E$-members, their images under $j$ are subsets of $d_0$ with the same ordinary members, so are equal by extensionality of $N$; injectivity of $j$ gives equality of the original objects.

To verify every stratified comprehension instance, assign a variable $v$ its stratification type $t(v)$ and translate its value $a\in D_0$ to $j^{t(v)}(a)\in D_{t(v)}$. Equality is preserved. An atomic $E(x,y)$ translates to the guarded adjacent-sort membership because $t(y)=t(x)+1$; the unary predicate $S(y)$ translates to $j^{t(y)}(y)\subseteq d_{t(y)-1}$. Induction on formulas, using the domain bijections for quantifiers, preserves truth. If the free variable $x$ has type $t$, typed comprehension produces its extension $X\subseteq d_t$, with $X\in d_{t+1}$. Put $b=j^{-(t+1)}(X)\in D_0$. Then $S(b)$ holds, and for every $a\in D_0$,This is exactly the required NFU set. No formula containing the external automorphism $j$ has been used in a separation or comprehension scheme; $j$ only transports values after the pure formula is formed.

The construction also permits the usual infinity requirement. The ordinary $\omega$ and its successor graph $f$ are definable in $N$, hence fixed by $j$, and lie below $V_{\omega +5}$, so belong to all the required domains. Under $E$, the members of $\omega$ are still its ordinary $N$-members. The new Kuratowski pair satisfiesSince $j(f)=f$, membership of this pair in $f$ under $E$ is equivalent to $⟨x,y{⟩}_N\in f$. Thus $f$ is an internal injection of $\omega$ into itself omitting zero, giving a Dedekind-infinite set.

We have built a model of NFU, with infinity if it is included in the formulation, in ordinary ZFC metatheory:In particular $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{NFU})$. This proof does not assume $\mathrm{Con}(\mathrm{ZFC})$ as an additional internal axiom: its starting structure is the set $V_{\omega +\omega }$, not a model of ZFC.

A rational structure over ${\mathbb{F}}_q$ is a model over that finite field together with its identification after extension to ${\overline{\mathbb{F}}}_q$. It determines a Frobenius endomorphism of an algebraic group, whose fixed points are the rational points. Conjugating a standard Frobenius by an algebraic group automorphism transports the rational structure, even if the automorphism is not itself rational for the original structure.

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.

A transitive group action is regular if every point stabilizer is trivial. Equivalently each ordered pair of points is connected by exactly one group element. Identifying one point with the identity identifies the action with left multiplication on the group. A regular normal subgroup inside a larger primitive group can supply translations or nonabelian left multiplication; the whole primitive group usually has a nontrivial stabilizer acting by automorphisms of this regular subgroup.