4.5. Examples of polynomials with Sp as Galois group over Q.

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The next lemma gives a criterion for a subgroup of Sp to be the whole of Sp.

Lemma 4.12. Let p be a prime number. Then Sp is generated by any transposition and

any p-cycle.

Proof. After renumbering, we may assume that the transposition is τ = (12). Let the

p-cycle be σ = (i1 ・ ・ ・ ip);w e may choose to write σ so that 1 occurs in the first position,

σ = (1i2 ・ ・ ・ ip). Now some power of σ will map 1 to 2 and will still be a p-cycle (here is where

we use that p is prime). After replacing σ with the power, we may suppose σ = (12j3 . . . jp),

and after renumbering again, we may suppose σ = (123. . . p). Then we’ll have (2 3), (3 4),

(4 5), . . . in the group generated by σ and τ , and these elements generated Sp.

Proposition 4.13. Let f be an irreducible polynomial of prime degree p in Q[X]. If f

splits in C and has exactly two nonreal roots, then Gf = Sp.

Proof. Let E ⊂ C be the splitting field of f, and let α ∈ E be a root of f. Because f is

irreducible, [Q[α] : Q] = degf = p, and so p|[E : Q] = (Gf : 1). Therefore Gf contains an

element of order p (Cauchy’s theorem), but the only elements of order p in Sp are p-cycles

(here we use that p is prime again).

Let σ be complex conjugation on C. Then σ transposes the two nonreal roots of f(X)

and fixes the rest. Therefore Gf ⊂ Sp contains a transposition and a p-cycle, and so is the

whole of Sp.

32 J.S. MILNE

It remains to construct polynomials satisfying the conditions of the Proposition.

Example 4.14. Let p≥ 5 be a prime number. Choose a positive even integer m and even

integers

n1 < n2 < ・ ・ ・ < np−2.

Let f(X) = g(X) − 2, where

g(X) = (X2 + m)(X − n1)...(X − np−2).

When we write f(X) = Xp+a1Xp−1+・ ・ ・+ap, then all ai are even, and ap = −(m

_

ni)−2

is not divisible by 4. Hence Eisenstein’s criterion implies that f(X) is irreducible.

The polynomial g(X) certainly has exactly two nonreal roots. Its graph crosses the x-axis

exactly p − 2 times, and its maxima and minima all have absolute value > 2 (because its

values at odd integers have absolute value > 2). Hence the graph of f(X) = g(X) − 2 also

crosses the x-axis exactly p − 2 times.

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