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For distinct or equal prime numbers $\ell ,\lambda$, the polynomial $X^{\ell }-\lambda$ is Eisenstein at $\lambda$. Its splitting field over $\mathbb{Q}({\zeta }_{\ell })$ has Galois group a subgroup of $C_{\ell }$, which is nontrivial because a degree-$\ell$ radical cannot belong to a degree-$(\ell -1)$ cyclotomic field. Thus the full Galois group has order $\ell (\ell -1)$. On roots indexed by $j\in {\mathbb{F}}_{\ell }$, all automorphisms act as $j\mapsto uj+b$. Equality of orders identifies the group with the full affine group over a finite field.

// Target: algebra.bigb

For a polynomial $f\in K[x]$ of degree $n$, its splitting field $L$ has $[L:K]\mid n!$, including in positive characteristic when the polynomial is inseparable. Prove this by induction on the degree. If $f$ is irreducible, adjoining a root gives degree $n$ and leaves a polynomial of degree $n-1$ to split, so the tower law for field extensions gives the divisibility. If $f=gh$ with positive degrees $r,s<n$, split $g$ first and then $h$ over the resulting field. The degree divides $r!s!$, which divides $(r+s)!$. This separate reducible case avoids the invalid step of assuming that the degree of an arbitrarily chosen irreducible factor must divide $n$.

A cubic polynomial in one variable has polynomial degree three. Over a splitting field, its factorization of polynomials contains three polynomial factors of polynomial degree one when multiplicity of a root is counted.

If $a^3=3$ and ${\zeta }_3$ is a primitive cube root of unity, then ${\pi }^3=1+2{\zeta }_3$ and ${\pi }^6=-3$. The Eisenstein polynomial $X^6+3$ therefore presents the entire degree-six splitting field, and $\pi$ is a uniformizer. Its normalised valuation gives $v_L(3)=6$, $v_L(a)=2$ and $v_L({\zeta }_3-1)=3$. The order-three automorphisms displace $\pi$ with valuation four, whereas transpositions displace it with valuation one.

A nonzero polynomial is factorwise separable if every irreducible factor is a separable polynomial. Unlike the convention that all roots must be distinct, this condition allows repeated irreducible factors. Its splitting field is a separable field extension precisely when the polynomial is factorwise separable: in one direction every root has a separable minimal polynomial; in the other, adjoining finitely many roots with separable minimal polynomials gives a separable extension. Thus the implication from a separable splitting extension to a polynomial with distinct roots needs the extra squarefree hypothesis. The example $(X-1{)}^2$ already distinguishes the conventions.

In a splitting field of $x^{p^n}-x$ over a prime field, the polynomial roots form a field. Frobenius identities close the polynomial roots under addition and multiplication, and nonzero polynomial roots have inverses which are again polynomial roots. The derivative is minus one, so the polynomial has exactly $p^n$ distinct polynomial roots. This constructs a field of every prime-power order.

Let a monic integer polynomial have distinct roots in its splitting field. Outside the finitely many primes dividing its polynomial discriminant or ramifying in that field, its roots reduce to distinct residue roots. The Frobenius automorphism permutes them by $a\mapsto a^p$; a reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed. Distinctness of reduction identifies this with a fixed root of the Galois permutation itself. Combining this observation with a derangement in a transitive group action and the Chebotarev density theorem proves that an irreducible polynomial of degree at least two has rootless reduction at a positive Dirichlet density of primes. The degree bound is essential: every monic linear integer polynomial has a root modulo every prime.

The splitting field of $T^3+T-1$ has discriminant square class $-31$ and Galois group $S_3$. It is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-31})$: the relative extension is unramified cyclic cubic, and the ideal class group of Q of square root minus thirty-one has order three. Locally at $31$, the simple residue root $28$ lifts and the remaining quadratic splits over ${\mathbb{Q}}_{31}(\sqrt{-31})$, so the prime of $K$ above $31$ splits completely in its Hilbert class field. The inverse generator satisfies $T^3-T^2-1=0$.

The three reduced primitive binary quadratic forms of discriminant $-23$ are $(1,1,6)$ and $(2,\pm 1,3)$, so the class number is three. The displayed cubic has discriminant $-23$ and is irreducible. Its splitting field has Galois group $S_3$ and is cyclic cubic over $\mathbb{Q}(\sqrt{-23})$. Modulo $23$ it factors as $(X-10{)}^2(X-3)$. Lift the simple factor; the remaining quadratic has discriminant valuation one, so its local inertia group has order two. Its intersection with $A_3$ is trivial. Thus the relative cyclic cubic extension is everywhere unramified and equals the Hilbert class field.

François Viète studied the connection between positive roots of a polynomial and the coefficients of the powers of the unknown. His 1591 presentation of algebraic notation used letters for both known and unknown quantities, with consonants for known quantities and vowels for unknown quantities. That move makes it possible to express a relation for whole families of polynomials, rather than only a numerical instance. His historical setting concerned positive roots and expressions satisfying dimensional homogeneity; the modern splitting field formulation above is a broader statement.^[ref]

Over a field containing ${\mathbb{F}}_q$, a sum of $q$-power monomials is ${\mathbb{F}}_q$-linear. If its leading coefficient and linear coefficient are nonzero, its root space has exactly $q^d$ elements and dimension $d$ over ${\mathbb{F}}_q$. Automorphisms of its splitting field act faithfully and linearly on that root space, embedding the Galois group in $GL(d,q)$.

Irreducibility of the cyclotomic polynomial gives $[K:\mathbb{Q}]=\phi (15)=8$. The extension is Galois, since $K$ is the splitting field of $T^{15}-1$ in characteristic zero. The injection in part (a) is consequently an isomorphism, and the Chinese remainder theorem givesThe Fundamental theorem of Galois theory identifies proper intermediate fields strictly above $\mathbb{Q}$ with proper nontrivial subgroups. Such subgroups have order two or four. The two-torsion subgroup is $C_2\times C_2$, with three nonidentity elements and hence three order-two subgroups. There are also three order-four subgroups: they are the kernels of the three nonzero group homomorphisms $C_2\times C_4\to C_2$. Distinct such maps have distinct kernels because the target has only one nonidentity element. Therefore

A nonconstant polynomial is separable when it has distinct roots in a splitting field, equivalently when it is coprime to its formal derivative. For such $f$, Bézout's identity for polynomials gives $a,b\in F[T]$ with $af+b{f}^{\prime }=1$. This identity remains valid over any extension $E/F$. A repeated root $\gamma \in E$ would satisfy $f(\gamma )=f^{\prime }(\gamma )=0$, giving the contradiction $1=0$. Thus there are no repeated roots in any field extension. The argument in fact applies to every separable polynomial, whether irreducible or not.

We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following symmetric Hermite integral obstruction will do both jobs. Supposewhere $k\ne 0$ and $b_j$ are integers, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in \mathbb{Z}[X]$ vanishing at those exponents, with $P(0)\ne 0$, and a positive integer $c$ for which every $c{\beta }_j$ is an algebraic integer. For a large prime number $p$, letAt zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^{D}P(0{)}^p$. All higher derivatives at zero are integers divisible by $p$. At each ${\beta }_j$, derivatives below $p$ vanish and all higher derivatives belong to $p{\mathcal{O}}_L$ in a Galois splitting field $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D{\beta }_j^u$ is integral for $u\le D$. ConsequentlyGalois invariance makes the sum rational, and integrality makes it an integer. For prime numbers not dividing $kcP(0)$ it is nonzero.

But $F_p-F_p^{\prime }=f_p$, so integration along the straight segment to $\beta$ givesMultiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all polynomial and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to 0$, contradicting its being a nonzero integer. The obstruction is proved.

If $e$ were algebraic, an integral polynomial relation would give $a_0+{\sum }_{j=1}^s{a}_j{e}^j=0$ with $a_0\ne 0$. Integer exponents and weights satisfy the obstruction, so $e$ is transcendental.

If $\pi$ were algebraic, put $\alpha =i\pi$ and let ${\alpha }_1,\dots ,{\alpha }_d$ be its algebraic conjugates. Since $1+e^{\alpha }=0$, the product ${\prod }_j(1+e^{{\alpha }_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive integer $k\ge 1$; the nonzero sums give a Galois-stable list ${\beta }_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence $\pi$ is transcendental.

The general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin \beta =0$ would give a forbidden relation between the exponentials of the distinct algebraic numbers $i\beta ,-i\beta$. The quotient identity would giveThe four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts linear independence. Therefore

Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.

A Hilbertian field is a field over which generic irreducibility can be retained by specializing a parameter. More precisely, for polynomials $f_i(T,X)$ irreducible over $K(T)$ and separable in $X$, and a nonzero exclusion polynomial $h(T)$, the Hilbert subsetis infinite. Denominators, leading coefficients and discriminants may be incorporated into $h$. The infinite condition permits simultaneous avoidance of any finitely many bad places. Separability is part of the definition, particularly in positive characteristic.

The field-theoretic interpretation is useful in constructive Galois theory. If $L/K(T)$ is finite separable and generated by a root of $f$, specialization at a good $a$ is the residue extension at a place over $T-a$. Irreducibility of $f(a,X)$ means its residue degree is the entire degree $[L:K(T)]$. Thus a Hilbertian field has infinitely many unramified rational places which are completely inert in each prescribed finite separable extension. For a finite family, one can work in a primitive-element presentation of their separable compositum; preserving its full degree preserves the degrees of all the subextensions.

Hilbert's irreducibility theorem says that $\mathbb{Q}$ is Hilbertian, and gives the corresponding assertion for number fields, including infinitely many integral specializations in its classical form. Rational function fields $k(U)$ over arbitrary fields $k$ are also Hilbertian. In particular, global function fields are Hilbertian just as number fields are. These standard field classes and the specialization viewpoint are discussed in Haran and Jarden's introduction to Hilbertian fields. The theorem does not say every parameter is good: $X^2-T$ is generically irreducible, but a square parameter makes it split. For simultaneous irreducibility, one excludes more exceptional parameters, not the entire field.

Hilbertianity survives finite extensions. For the separable case, let $K$ be Hilbertian, $M/K$ finite separable and $f\in M(T)[X]$ irreducible and separable. A primitive element of $M(T)(\alpha )$ over $K(T)$ has degree $[M:K]\mathrm{deg}f$. Apply Hilbertianity to its minimal polynomial and exclude the presentation's denominators. The resulting residue field contains the constant copy of $M$ and has the same total degree over $K$, so its degree over $M$ is $\mathrm{deg}f$. Therefore $f(a,X)$ remains irreducible over $M$, even for infinitely many $a$ chosen in $K$. This argument also handles finitely many polynomials by a compositum. Purely inseparable finite extensions are treated by separating the separable part and transporting its rational places through the unique purely inseparable extension; the finite-extension stability statement remains valid.

The property is not inherited by arbitrary algebraic extensions. An algebraic closure is not Hilbertian: in characteristic zero, $X^2-T$ is irreducible over $K(T)$ but every specialization has a root in $K$. In characteristic $p$, use $X^p-X-T$ instead; a hypothetical rational-function root would give its right side a pole of order divisible by $p$, contradicting its simple pole at infinity. All constant specializations split over the algebraically closed field.

Finite fields also fail. For $K={\mathbb{F}}_q$ of characteristic $p$, considerIts right-hand additive datum has a pole of order $q+1$, not divisible by $p$, and cannot be $b^p-b$ for a rational function $b$. The polynomial is therefore irreducible over $K(T)$: its roots are $\alpha +c$ for $c\in {\mathbb{F}}_p$, so its splitting-field group has prime order unless a root already belongs to the base field. At every $a\in K$, however, $a^q-a=0$ and the specialization is $X^p-X$, which splits. This gives a concrete failure of the required specialization property.

The main constructive application is Galois-preserving Hilbert specialization. Suppose $E/K(T)$ is finite Galois with group $G$, and use a primitive element of the entire splitting field, not merely a generator of one root field. A good specialization retaining degree $|G|$ has residue group embedded in $G$ by unramified reduction. Since its order is still $|G|$, it is exactly $G$. Preserving only the irreducibility of the original degree-$n$ polynomial would prove transitivity, which is weaker than preserving its full group. Resolvents or the primitive generator eliminate that ambiguity.

Consequently a regular realization of $G$ over $K(T)$, with constant field $K$, yields realizations over the Hilbertian field $K$. This is the bridge from branch-cycle constructions to the Inverse Galois problem over number fields. A realization over $\mathbb{C}(T)$ alone cannot be specialized to $\mathbb{Q}$; descent of its coefficients and constants is an additional step. Rational rigidity can supply that descent under its appropriate hypotheses, while ordinary complex rigidity answers only the geometric uniqueness question.

For example, the family $X^n-X-T$ has full symmetric geometric monodromy in characteristic zero. It is irreducible because it is linear in $T$. Its derivative has $n-1$ distinct critical roots satisfying $n{X}^{n-1}=1$, and their distinct critical values are $-(n-1)X/n$. Each simple critical point gives a transposition. These finite branch cycles generate the monodromy, since the infinity loop is the inverse product of the finite loops. Their transitive connected-edge action gives $S_n$. The arithmetic group is also $S_n$, so the constant field is $\mathbb{Q}$. Hilbert specialization then produces symmetric Galois groups over $\mathbb{Q}$. For $n=2$ the same conclusion follows directly from the square-root family.

Finally, the good parameters form a large arithmetic set without being every point or generally a Zariski-open set. Reducibility can occur at infinitely many exceptional points, as the square example shows. The language of thin sets packages those exceptions as images of generically finite separable covers of degree greater than one, together with finitely many points. Hilbertianity asserts that such finitely many exceptional constructions cannot exhaust the rational line. It is an existence principle with powerful constructive consequences, not by itself an algorithm that certifies a particular chosen parameter.

The PDF has $q$-power exponents; the polynomial is a linearized polynomial over a finite field, not the degree-$d$ polynomial produced by the converted TeX. Let $V$ be its root set in an algebraic closure. Because $(x+y{)}^{q^i}=x^{q^i}+y^{q^i}$ and $c^{q^i}=c$ for $c\in {\mathbb{F}}_q$,Thus $V$ is an ${\mathbb{F}}_q$-vector space. Its derivative is $a_0\ne 0$, and its degree is $q^d$, so it has exactly $q^d$ distinct roots. A finite vector space of that size has dimension $d$.

Every splitting-field automorphism fixes ${\mathbb{F}}_q$ and carries roots to roots. Hence it acts linearly on $V$. The action is faithful because the splitting field is generated by the roots. Choosing a basis identifies this injection withDifferent bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.