Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 3 (Constructive Galois Theory)

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.

Modulo two the polynomial becomes $h=X^5+X^2+1$. It has neither zero nor one as a root. The only irreducible quadratic over ${\mathbb{F}}_2$ is $X^2+X+1$; modulo it, $X^3=1$ and $h\equiv 1$. A reducible quintic has a factor of degree at most two, so $h$ is irreducible. It is also separable: $h^{\prime }=X^4$, which is coprime to $h$. Thus the rational polynomial is irreducible, and its group $G\le S_5$ is transitive.

Modulo three there is the exact factorizationThe quadratic has no root. The cubic takes value one at each of $0,1,2$, so it too has no root and is irreducible. The two factors are distinct and coprime. The Frobenius cycle type therefore supplies an element with disjoint cycles of lengths two and three. Its cube is a single transposition.

A transitive action of prime degree is primitive: a block size divides five, so the only sizes are one and five. To finish without classifying all degree-five groups, apply the primitive group with a transposition argument. Form the graph of all conjugate transpositions in $G$. Its connected components are blocks, so it is connected. Transpositions along the edges of a connected graph generate every transposition: along a path, conjugating adjacent swaps yields the swap of its endpoints. Hence they generate $S_5$, and

Use $R={\mathbb{F}}_2[t]$. At $t=1$ the specialization is $X^5+X^2+1$, already proved irreducible and separable. Because the original polynomial is monic, this proves irreducibility over ${\mathbb{F}}_2(t)$ and gives a transitive group on five roots.

At $t=0$ the specialization factors asNeither factor has a root in ${\mathbb{F}}_2$, so they are irreducible; the factors are coprime and the specialization is separable. For example its derivative is $(X+1{)}^4$ and its value at one is one. The unramified reduction test gives a $(2)(3)$ permutation whose cube is a transposition. The same primitive prime-degree graph argument now givesNo characteristic-zero discriminant-sign argument is being transferred to characteristic two: good specialization and the explicit cycle test are sufficient.

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.

For a finite point $p$, use the valuation $v_p$ with local parameter $u=t-p$; at infinity use $u=1/t$. Its extensions $w$ to $E$ are permuted transitively by $G$. The completed residue field is $F$, because $F$ is algebraically closed, so the local decomposition group equals its inertia group.

Write $e=e(w/v_p)$. Characteristic zero makes all local ramification tame. One can choose a uniformizer $\pi$ in the completion for whichIndeed, initially $u$ is ${\pi }^e$ times a unit; that unit has an $e$th root in $F[[\pi ]]$, by its algebraically closed residue field and the invertibility of $e$. Absorb it into $\pi$. The local automorphisms are then $\pi \mapsto \zeta \pi$, $\zeta \in {\mu }_e(F)$, and form a cyclic group of order $e$. Define $e_p=e$.

To specify an element rather than a subgroup, fix a compatible choice of primitive roots of unity ${\zeta }_e\in F$. Define ${\tau }_w$ by its uniformizer characterand let $C_p$ be its conjugacy class in $G$. Over $\mathbb{C}$, take ${\zeta }_e=e^{2\pi i/e}$, corresponding to positive local continuation around $p$. At an unramified point, $e_p=1$ and $C_p=\{1\}$.

Changing the uniformizer does not affect this character. If ${\pi }^{\prime }=a\pi$ with $a$ a unit, inertia fixes its residue, soThe character is injective: a nonidentity finite-order automorphism with derivative one would have $\tau (\pi )=\pi +b{\pi }^r+\cdots$ for $r\ge 2$, and its $N$th iterate would have coefficient $Nb$ at that first nonzero displacement. Characteristic zero contradicts finite order. Thus the chosen primitive character identifies a unique generator.

For another place $w^{\prime }=\sigma w$, conjugation gives $D_{w^{\prime }}=\sigma D_w{\sigma }^{-1}$, preserves its order, and carries ${\tau }_w$ to ${\tau }_{w^{\prime }}$, since $\sigma$ fixes the constants and hence ${\zeta }_e$. This proves that $e_p$ and the branch-cycle conjugacy class are independent of the place and uniformizer. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup conjugacy class is intrinsic: replacing ${\zeta }_e$ by another primitive root replaces the generator by a coprime power, which need not lie in the same conjugacy class.

Put $u=t-p$ and write $a(t)=u^{r}b(u)$, where $b(0)\ne 0$. The unit $b$ has a square root in $\mathbb{C}[[u]]$. Dividing the radical by that square root and by $u^{\lfloor r/2\rfloor }$ reduces the completed extension toIf $r$ is even, it is unramified and locally splits; if $r$ is odd, adjoining $u^{1/2}$ gives ramification index two. At points where $a(p)\ne 0$, $r=0$ and there is no branch. ThereforeFor infinity set $u=1/t$. If $D=\mathrm{deg}a$,The same quadratic branch parity criterion applies to the negative valuation $-D$, soThe generic irreducibility excludes $a$ being a square. Local splitting at the even-order places is compatible with that global irreducibility. The number of odd finite orders plus the possible odd infinity order is even, as also follows from the sum of orders of a rational function being zero.

Fix ordered branch positions $P=(p_1,\dots ,p_r)$ and conjugacy classes $C_i$ in $G$. Form the generating product-one setThe rigid ramification type condition is that this set is nonempty and consists of one orbit under simultaneous inner conjugation by $G$. Rationality of classes is a further arithmetic condition and is not needed for this complex uniqueness statement.

Let $U={\mathbb{P}}^1(\mathbb{C})\setminus P$. Choose a base point and positively oriented peripheral loops ${\gamma }_i$ with sole relation ${\gamma }_1\cdots {\gamma }_r=1$. A connected Galois cover over $U$ gives a surjective monodromy map ${\pi }_1(U)\to G$, and changing the labeling of one fiber conjugates it. Its peripheral images are exactly a tuple in the set above.

For two extensions of the same rigid type, rigidity conjugates their tuples to agree. The loops generate ${\pi }_1(U)$, so their entire monodromy maps then agree. This constructs an isomorphism of the unramified covers directly: label fibers at the base point identically, and transport that matching along paths. Independence of path follows from equal monodromy. The resulting map is holomorphic because the cover maps are local biholomorphisms.

Near each missing point, the cover has local form $u={\pi }^{e_i}$. The isomorphism therefore extends over the added points, equivalently by the removable-singularity theorem in these local parameters. Pullback of meromorphic functions gives an isomorphism of the function fields fixing $\mathbb{C}(t)$. Hence there is at most one extension of the prescribed rigid type. This is uniqueness for fixed branch positions and peripheral conventions, not uniqueness after arbitrarily moving the branch points.

For $n\ge 3$, take $C_1$ to be the class of $n$-cycles, $C_2$ the class of transpositions, and $C_3$ the class of an $(n-1)$-cycle with one fixed point. With rightmost permutations applied first, a witness isMultiplying an $n$-cycle by a transposition splits it at the two exchanged letters. The resulting cycles have the lengths of the two arcs between those letters in the cyclic ordering. Thus $\alpha \beta$ has type $(n-1,1)$ exactly when the exchanged letters are cyclically adjacent.

Any product-one tuple can first be conjugated so that its first entry is $\alpha$. Its second entry is then one of the $n$ adjacent transpositions. The centralizer $⟨\alpha ⟩$ acts transitively on these choices; the third entry is forced to be the inverse product. Therefore every product-one tuple lies in a single simultaneous-conjugation orbit.

The conjugates ${\alpha }^j\beta {\alpha }^{-j}$ exchange all adjacent letters around the cycle. Their connected-edge graph generates $S_n$, so the witness, and every tuple in that orbit, generates $S_n$. HenceFor $n=3$ the last two classes coincide, which causes no difficulty: the same three adjacent swaps form one orbit and generate $S_3$.

The relevant cycle types are $(5,1)$, $(2,2,2)$ and $(3,2,1)$. Conjugate the first entry to $\alpha =(12345)$, fixing six. An entry $\beta$ of type $(2,2,2)$ pairs six with one of the five other letters. Powers of $\alpha$ move that partner to one. With the pair $(16)$ fixed, only three matchings remain:Thus there is exactly one centralizer orbit of eligible second entries, consisting of five matchings before normalization. The third entry is forced to be $(\alpha \beta {)}^{-1}$, so there is at most one inner orbit of product-one tuples of these types.

For the actual displayed $g_2$, direct multiplication giveswhich belongs to the class of the displayed $g_3$. This is a product-one witness. The three displayed permutations themselves need not multiply to one; the problem concerns their conjugacy classes.

The group $⟨\alpha ,g_2⟩$ is transitive, because $\alpha$ moves the first five points transitively and $g_2$ connects six to them. Its stabilizer of six contains the five-cycle $\alpha$ and is transitive on the other points. Hence the group is two-transitive and therefore primitive. Moreover $(\alpha g_2{)}^3=(13)$ is a transposition. The primitive transposition lemma proves that the group is $S_6$. Consequently all tuples in the sole inner orbit generate $S_6$, andAs a count check, there are $6\cdot 4!=144$ first entries and five eligible second entries for each, giving $720$ tuples. A generating tuple has simultaneous stabilizer equal to the trivial center of $S_6$, so its inner orbit also has size $720$.

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.

Put $N=(q^d-1)/(q-1)$, so $N>1$ and $N\equiv 1$ in the characteristic. For a nonzero root $x$, division by $x$ givesThe projective point ${\mathbb{F}}_{q}x$ is therefore represented by $s=x^{q-1}$, satisfyingTwo nonzero roots give the same $s$ exactly when their ratio lies in ${\mathbb{F}}_q^{\times }$. Thus the $N$ projective points are precisely the roots of $P_t$. The polynomial $s^N+s+t$ is irreducible in ${\mathbb{F}}_q[s,t]$, since it is linear in $t$ and its quotient ring is ${\mathbb{F}}_q[s]$. Gauss's lemma makes it irreducible over ${\mathbb{F}}_q(t)$, so the action on projective points is transitive.

Fix one such coordinate $s$, so $t=-s^N-s$. The other coordinates satisfyWe prove its irreducibility even over ${\overline{\mathbb{F}}}_q(s)$. Make the birational change $z=Y/s$, $h=1/s$. Multiplication by $h^{N-1}$ transforms the equation intoChoose a nontrivial $N$th root of unity $\zeta$ over the algebraic closure. Since the characteristic does not divide $N$, $B$ has a simple zero at $\zeta$. Viewed as a polynomial in $h$ over ${\overline{\mathbb{F}}}_q[z]$, $h^{N-1}+B(z)$ is Eisenstein at the prime $z-\zeta$: the leading coefficient is one, every intermediate coefficient is zero, and the constant term is divisible by that prime exactly once. It is therefore irreducible.

The birational substitution preserves irreducibility after inverting $s$ or $h$. No factor supported solely on $s=0$ is lost, because $R(Y,0)=Y^{N-1}+1$ is nonzero. Gauss's lemma then proves that $R(Y,s)$ is irreducible over ${\overline{\mathbb{F}}}_q(s)$ and hence over ${\mathbb{F}}_q(s)$. Its roots are distinct: a repeated root would have $Y^{N-1}+1=0$ as well as $Y^N+Y+t=0$, forcing $t=0$, impossible generically.

The stabilizer of the chosen projective point is $\mathrm{Gal}(E/{\mathbb{F}}_q(s))$, and the irreducibility of $R$ makes it transitive on all the remaining projective points. This proves projective transitivity of a trinomial linearized polynomial, namely two-transitivity on the one-dimensional subspaces. The supplied group-theoretic result now applies to the linear subgroup from part (a), givingThe Eisenstein argument remains valid even though $N-1$ is divisible by the characteristic; it concerns irreducibility in $h$, not separability of that projection. Separability of the projective polynomial has been checked independently.

Write $d=2n$ with $n\ge 2$, and let $E$ be the splitting field. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.

The constants ${\mathbb{F}}_{q^2}$ really occur inside $E$, rather than only in the ambient algebraic closure. Choose a nonzero root $\alpha$. For every $c\in {\mathbb{F}}_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ andLet $K={\mathbb{F}}_{q^2}(t)$. Since $K/{\mathbb{F}}_q(t)$ is Galois of degree two,This index statement does not require proving that $E$ has no still larger constant field.

As an extension of $K$, $E$ is the splitting field of the same polynomial. Applying part (b) with base constant size $q^2$ and dimension $n$ givesThe special linear group is the kernel of determinant in the general linear group, so every element of $H$ normalizes it. Therefore $H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$.

In fact the full group is semilinear over ${\mathbb{F}}_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^{q}g(v)$. Field automorphisms preserve determinant one, so this also explains the normalizer behavior geometrically. This is the quadratic constant extension of a q-squared linearized splitting field, not the polynomial of part (b) with its $X^q$ term unchanged.