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so this is a Galois extension; we will identify its Galois group. Notice that
[E : E“] = (E : E“) = | Gal(E/E“)|.
Now each element of Gal(E/E“) is also an element of Gal(E/K) and Gal(E/E“) Gal(E/K).
Notice that by definition “ Gal(E/E“), so Lagrange s Theorem implies that |“| divides
| Gal(E/E“)|. In fact we have
4.14. Proposition. For “ Gal(E/K), we have Gal(E/E“) = “ and the equations
| Gal(E/K)|
[E : E“] = | Gal(E/E“)| = |“|, [E“ : K] = .
|“|
Proof. We know that E/E“ is separable, so by the Primitive Element Theorem 3.75 it is
simple, say E = E“(u). Now let the distinct elements of “ be ³1 = id, ³2, . . . , ³h, where h = |“|.
Consider the polynomial of degree h
f(X) = (X - u)(X - ³2(u)) · · · (X - ³h(u)) " E[X].
Notice that f(X) is unchanged by applying any ³k to its coefficients since the roots ³j(u) are
permuted by ³k. Hence, f(X) " E“[X]. This shows that
[E : E“] = [E“(u) : E“] h = |“|.
Since “ Gal(E/E“), we also have
h = |“| | Gal(E/E“)| = [E : E“].
Combining these two inequalities we obtain
[E : E“] = | Gal(E/E“)| = |“| = h
and therefore “ = Gal(E/E“).
4.4. SUBFIELDS OF GALOIS EXTENSIONS AND RELATIVE GALOIS GROUPS 53
4.4. Subfields of Galois extensions and relative Galois groups
Let E/K a Galois extension and suppose that L/K E/K (i.e., K L E). Then E/L
is also a Galois extension whose Galois group Gal(E/L) is sometimes called the relative Galois
group of the pair of extensions E/K and L/K. The following is immediate.
4.15. Lemma. The relative Galois group of the pair of extensions L/K E/K is a subgroup
of Gal(E/K), i.e., Gal(E/L) Gal(E/K), and its order is | Gal(E/L)| = [E : L].
4.16. Proposition. Let L/K E/K. Then L = EGal(E/L).
Proof. Clearly L EGal(E/L). Suppose that u " E - L. By Theorem 4.8(i), there is
an automorphism ¸ " Gal(E/L) such that ¸(u) = u, hence u " EGal(E/L). This shows that
/
EGal(E/L) L and therefore EGal(E/L) = L.
We need to understand when Gal(E/L) Gal(E/K) is actually a normal subgroup. The
next result explains the connection between the two uses of the word normal which both ulti-
mately derive from their use in Galois theory.
4.17. Proposition. Let E/K be a finite Galois extension and L/K E/K.
(i) The relative Galois group Gal(E/L) of the pair of extensions L/K E/K is a normal
subgroup of Gal(E/K) if and only if L/K is a normal extension.
(ii) If L/K is normal and hence a Galois extension, then there is a group isomorphism
=
Gal(E/K)/ Gal(E/L) - Gal(L/K); ± Gal(E/L) -’! ±|L.
’!
Proof. (i) Suppose that Gal(E/L) Gal(E/K), i.e., for all ± " Gal(E/L) and ² "
Gal(E/K), we have ²±²-1 " Gal(E/L). Now if u " L, then for any ³ " Gal(E/K) and
± " Gal(E/L), ³(u) " E satisfies
±³(u) = ³(³-1±³(u)) = ³(u),
since ³-1±³ " Gal(E/L); hence ³(u) " EGal(E/L) = L. By the Monomorphism Extension
Theorem 3.49, every monomorphism L -’! K fixing K extends to a monomorphism E -’! K
which must have image E, so the above argument shows that L/K is normal.
Conversely, if L/K is normal, then for every Õ " Gal(E/K) and v " L, Õ(v) " L, so for
every ¸ " Gal(E/L), ¸(Õ(v)) = Õ(v) and therefore
Õ-1¸Õ(v) = v.
This shows that Õ-1¸Õ " Gal(E/L). Hence for every Õ " Gal(E/K),
Õ Gal(E/L)Õ-1 = Gal(E/L),
which shows that Gal(E/L) Gal(E/K).
(ii) If ± " Gal(E/K), then ±L = L since L/K is normal. Hence we can restrict ± to an
automorphism of L,
±|L : L -’! L; ±|L(u) = ±(u).
Then ±|L is the identity function on L if and only if ± " Gal(E/L). It is easy to see that the
function
Gal(E/K) -’! Gal(L/K); ± -’! ±|L
is a group homomorphism whose kernel is Gal(E/L). Thus we obtain an injective homomor-
phism
Gal(E/K)/ Gal(E/L) -’! Gal(L/K)
for which
[E : K]
| Gal(E/K)/ Gal(E/L)| = = [L : K] = | Gal(L/K)|.
[E : L]
Hence this homomorphism is an isomorphism.
54 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE
4.5. The Galois Correspondence and the Main Theorem of Galois Theory
We are now almost ready to state our central result which describes the Galois Correspon-
dence associated with a finite Galois extension. We will use the following notation. For a finite
Galois extension E/K, let
S(E/K) = the set of all subgroups of Gal(E/K);
F(E/K) = the set of all subextensions L/K of E/K.
Each of these sets is ordered by inclusion. Since every subgroup of a finite group is a finite
subset of a finite set, S(E/K) is also a finite set. Define two functions by
¦E/K : F(E/K) -’! S(E/K); ¦E/K(L) = Gal(E/L),
˜E/K : S(E/K) -’! F(E/K); ˜E/K(“) = E“.
4.18. Theorem (Main Theorem of Galois Theory). Let E/K be a finite Galois extension.
Then the functions ¦E/K and ˜E/K are mutually inverse bijections which are order-reversing.
¦E/K
F(E/K) S(E/K)
˜E/K
Under this correspondence, normal subextensions of E/K correspond to normal subgroups of
Gal(E/K) and vice versa.
Proof. We know from Proposition 4.16 that for an extension L/K in F(E/K),
˜E/K(¦E/K(L)) = ˜E/K(Gal(E/L)) = EGal(E/L) = L.
Also, by Proposition 4.14 for H " S(E/K) we have
¦E/K(˜E/K(“)) = ¦E/K(E“) = Gal(E/E“) = “.
This shows that ¦E/K and ˜E/K are mutually inverse and so are inverse bijections.
Let L1/K, L2/K " F(E/K) satisfy L1/K L2/K. Then Gal(E/L2) Gal(E/L1) since
L1 †" L2 and so if ± " Gal(E/L2) then ± fixes every element of L1. Hence ¦E/K(L2) ¦E/K(L1)
and so ¦E/K reverses order.
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