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§ 8.4 Solvability by Radicals: solved problems

7. Let f(x) be irreducible over Q, and let F be its splitting field over Q. Show that if Gal (F/Q) is abelian, then F = Q(u) for all roots u of f(x).

Solution: Since F has characteristic zero, we are in the situation of the fundamental theorem of Galois theory. Because Gal (F/Q) is abelian, every intermediate extension between Q and F must be normal. Therefore if we adjoin any root u of f(x), the extension Q(u) must contain all other roots of f(x), since it is irreducible over Q. Thus Q(u) is a splitting field for f(x), so Q(u) = F.

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8. Find the Galois group of x9-1 over Q.

Solution: We can construct the splitting field F of x9-1 over Q by adjoining a primitive 9th root of unity to Q. We have the factorization

x9-1 = (x3-1)(x6+x3+1)
= (x-1)(x2+x+1)(x6+x3+1).
Substituting x+1 in the last factor yields
(x+1)6+(x+1)3+1 = x6+6x5+15x4+ 21x3+18x2+9x+3.
This polynomial satisfies Eisenstein's criterion for the prime 3, which implies that the factor x6+x3+1 is irreducible over Q. The roots of this factor are the primitive 9th roots of unity, so it follows that [F:Q] = 6. The proof of Theorem 8.4.2 (which is worth remembering) shows that Gal (F/Q) is isomorphic to a subgroup of Z9× Since Z9× is abelian of order 6, it is isomorphic to Z6. It follows that Gal (F/Q) Z6.

Comment: Section 8.5 of the text contains the full story. Theorem 8.5.4 shows that the Galois group of xn-1 over Q is isomorphic to Zn× and so the Galois group is cyclic of order (n) iff n = 2, 4, pk, or 2pk, for an odd prime p.

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9. Show that x4-x3+x2-x+1 is irreducible over Q, and use it to find the Galois group of x10-1 over Q.

Solution: We can construct the splitting field F of x10-1 over Q by adjoining a primitive 10th root of unity to Q. We have the factorization

x10-1 = (x5-1)(x5+1)
= (x-1)(x4+x3+x2+x+1) (x+1)(x4-x3+x2-x+1).
Substituting x-1 in the last factor yields
(x-1)4-(x-1)3+(x-1)2-(x-1)+1
= (x4-4x3+6x2-4x+1) - (x3-3x2+3x-1) + (x2-2x+1) - (x-1) + 1
= x4-5x3+10x2-10x+5.
This polynomial satisfies Eisenstein's criterion for the prime 5, which implies that the factor x4-x3+x2-x+1 is irreducible over Q.

The roots of this factor are the primitive 10th roots of unity, so it follows that [F:Q] = 4. The proof of Theorem 8.4.2 shows that Gal (F/Q) Z10× and so the Galois group is cyclic of order 4.

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10. Show that p(x) = x5-4x+2 is irreducible over Q, and find the number of real roots. Find the Galois group of p(x) over Q, and explain why the group is not solvable.

Solution: The polynomial p(x) is irreducible over Q since it satisfies Eisenstein's criterion for p = 2. Since p(-2) = -22, p(-1) = 5, p(0) = 2, p(1) = -1, and p(2) = 26, we see that p(x) has a real root between -2 and -1, another between 0 and 1, and a third between 1 and 2. The derivative p'(x) = 5x4-4 has two real roots, so p(x) has one relative maximum and one relative minimum, and thus it must have exactly three real roots. It follows as in the proof of Theorem 8.4.8 that the Galois group of p(x) over Q is S5, and so it is not solvable.

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