Page 283 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes
Example: 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 : K] = [G:H], and so E
H
= 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

Example: 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.

Example: Let F be a finite, normal, separable extension of the field K. Suppose that the
Galois group Gal (F/K) is isomorphic to D . Find the number of distinct subfields between F and
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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 , and determining which are normal. If we use the usual
7
description of D via generators a of order 7 and b of order 2, with ba = a b, then a generates a
-1
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subgroup of order 7, while each element of the form a b generates a subgroup of order 2, for
i
0  i < 7. Thus there are 8 proper nontrivial subgroups of D , and the only one that is normal is
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< a >, since it has |D | / 2 elements. As you should recall from the description of the conjugacy
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classes of D conjugating one of the 2-element subgroups by a produces a different subgroup,
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showing that none of them are normal.
Example: Show that F = Q ( 2,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 + 1)(x – 2), and this
2
2
polynomial is certainly separable. Thus, F is a normal extension of Q.
It follows that the Galois group is isomorphic to Z × Z . Since the Galois group has 3 proper
2
2
nontrivial subgroups, there will be 3 intermediate subfields E with Q  E  F.
The existence of 3 nontrivial elements begins with the splitting field of x +1 over Q.
4
Comment: Recall that Z is the splitting field of x – x = x(x – 1).
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6
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Self Assessment
1. If F is field and G be a subgroup of Aut(F). Then {a in F | Q(a) = a  Q in G} is called ...............
of F.

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