## § 8.3 The Fundamental Theorem of Galois Theory: Solved problems

6. Prove that if F is a field extension of K and K = FG for a finite group G of automorphisms of F, then there are only finitely many subfields between F and K.

Solution: By Theorem 8.3.6 the given condition is equivalent to the condition that F is the splitting field over K of a separable polynomial. Since we must have G = Gal (F/K), the fundamental theorem of Galois theory implies that the subfields between F and K are in one-to-one correspondence with the subgroups of F. Because G is a finite group, it has only finitely many subgroups.

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7. Let F be the splitting field over K of a separable polynomial. Prove that if Gal (F/K) is cyclic, then for each divisor d of [F:K] there is exactly one field E with K E F and [E:K] = d.

Solution: By assumption we are in the situation of the fundamental theorem of Galois theory, so that there is a one-to-one order-reversing correspondence between subfields of F that contain K and subgroups of G = Gal (F/K). Because G is cyclic of order [F:K], there is a one-to-one correspondence between subgroups of G and divisors of [F:K]. Thus for each divisor d of [F:K] there is a unique subgroup H of index d. By the fundamental theorem, [FH: K] = [G:H], and so E = F^H is the unique subfield with [E:K] = d.

Comment: Pay careful attention to the fact that the correspondence between subfields and subgroups reverses the order.

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8. Let F be a finite, normal extension of Q for which | Gal (F/Q) | = 8 and each element of Gal (F/Q) has order 2. Find the number of subfields of F that have degree 4 over Q.

Solution: Since F has characteristic zero, the extension is automatically separable, and so the fundamental theorem of Galois theory can be applied. Any subfield E of F must contain Q, its prime subfield, and then [E:Q] = 4 iff [F:E] = 2, since [F:Q] = 8. Thus the subfields of F that have degree 4 over Q correspond to the subgroups of Gal (F/Q) that have order 2. Because each nontrivial element has order 2 there are precisely 7 such subgroups.

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9. Let F be a finite, normal, separable extension of the field K. Suppose that the Galois group Gal (F/K) is isomorphic to D7. Find the number of distinct subfields between F and K. How many of these are normal extensions of K?

Solution: The fundamental theorem of Galois theory converts this question into the question of enumerating the subgroups of D7, and determining which are normal. If we use the usual description of D7 via generators a of order 7 and b of order 2, with ba = a -1 b, then a generates a subgroup of order 7, while each element of the form ai b generates a subgroup of order 2, for 0 i < 7. Thus there are 8 proper nontrivial subgroups of D 7, and the only one that is normal is < a >, since it has |D7| / 2 elements. As you should recall from the description of the conjugacy classes of D7 conjugating one of the 2-element subgroups by a produces a different subgroup, showing that none of them are normal.

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10. Show that F = Q(, i) is normal over Q; find its Galois group over Q, and find all intermediate fields between Q and F.

Solution: It is clear that F is the splitting field over Q of the polynomial (x2+1)(x2-2), and this polynomial is certainly separable. Thus F is a normal extension of Q.

The work necessary to compute the Galois group over Q has already been done in the solution to Problem 8.1.2, which shows the existence of 3 nontrivial elements of the Galois group, each of order 2. It follows that the Galois group is isomorphic to Z2 × Z2. Since the Galois group has 3 proper nontrivial subgroups, there will be 3 intermediate subfields E with Q E F. These have been found in Problem 8.1.2, and are Q(), Q(i), and Q( i).

Note: Problem 8.1.2, which shows the existence of 3 nontrivial elements begins with the splitting field of x4+1 over Q.

Comment: Recall that Z7 is the splitting field of x7-x = x(x6-1).

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