Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 3 / 1

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.