**6.**
Prove that if F is a field extension of K
and K = F^{G} 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, [F^{H}: 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 D_{7}.
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 D_{7}, and determining which are normal.
If we use the usual description of D_{7} 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 a^{i} 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 |D_{7}| / 2 elements.
As you should recall from the description
of the conjugacy classes of D_{7}
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
(x^{2}+1)(x^{2}-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
**Z**_{2} × **Z**_{2}.
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 x^{4}+1 over **Q**.

*Comment:*
Recall that **Z**_{7} is the splitting field of
x^{7}-x = x(x^{6}-1).

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