Table Of ContentBasics about linearly disjoint (cid:28)eld extensions
L L(cid:48)
Assume that and are both ((cid:28)nite, at least most of the time) extensions
K E
of the same (cid:28)eld . We shall need a bigger (cid:28)eld that contains both of
L
them so that we have e.g. a well de(cid:28)ned way of multiplying elements of
L(cid:48) E
with elements of . In case of number (cid:28)elds we can always let be the (cid:28)eld
of algebraic numbers. On other occasions we may want to use the algebraic
K L L(cid:48)
closure of for this purpose (and select isomorphic copies of and in
L/K L(cid:48)/K
there). Furthermore, if and are both separable then we can select
E/K
to be separable as well. This is not immediately obvious, but can be
done (see e.g. Jacobson’s Basic Algebra I-II).
L(cid:48)
Given this it makessense to talk about -linear combinations of elements
L
of and vice versa. Consider the following two conditions
L(cid:48) K
1. Elements of that are linearly independent over are also linearly
L
independent over .
L K
2. Elements of that are linearly independent over are also linearly
L(cid:48)
independent over .
An immediate observation is that neither condition can be true, unless
L∩L(cid:48) = K E
(the intersection obviously also taken inside ). The following
is not surprising
Lemma 1 The two conditions above are equivalent. Or one implies the
other.
Proof. It clearly su(cid:30)ces to show that (1) implies (2). So we assume that
(1) holds, and make the contrapositive assumption that we have a (cid:28)nite set
S = {(cid:96) ,...,(cid:96) } L K
1 n of elements of that is linearly independent over but is
L(cid:48)
not linearly independent over . As we shall argue by in(cid:28)nite descent we
n
can assume that is the smallest number such that a set of this type exists.
n > 1
Clearly .
S L(cid:48)
As the set is not linearly independent over we have non-trivial rela-
tion of the form
(cid:88)n
(cid:96) (cid:96)(cid:48) = 0.
i i
i=1
{(cid:96)(cid:48),...,(cid:96)(cid:48) } ⊂ L(cid:48)
This relation can also be viewed as stating that the set 1 n is
L
linearly dependent over . So by the assumption (1) it follows that this set
K
must be linearly dependent already over the smaller (cid:28)eld . Without loss of
1
(cid:96)(cid:48) K
generatlity we can assume that n can be written as a -linear combination
a ∈ K
of the others, so we can (cid:28)nd constants i such that
(cid:88)n−1
(cid:96)(cid:48) = a (cid:96)(cid:48).
n i i
i=1
Putting these two together we arrive at the relation
(cid:88)n−1
((cid:96) +a (cid:96) )(cid:96)(cid:48) = 0.
i i n i
i=1
S = {(cid:96) +a (cid:96) ,...,(cid:96) +a (cid:96) }
The set a 1 1 n n−1 n−1 n is thus linearly dependent over
L(cid:48) K S
. But it is clearly linearly independent over , because was assumed to
n
be. Therefore we have violated the minimality of , and the claim follows.
L L(cid:48) K
De(cid:28)nition 0.1 Two extension (cid:28)elds and of are said to be linearly
disjoint, if they satisfy either (and hence both) of the conditions (1) and/or
(2).
Example 1 This example shows that linear disjointness is not equivalent to
L∩L(cid:48) = K α β L = Q(α)
the condition . Let and be two cubic roots of 2. Let
L(cid:48) = Q(β) L∩L(cid:48) = Q = K ω = α/β
and , so . Then is a primitive third root
1+ω+ω2 = 0
of unity, so it satis(cid:28)es the equation . Clearing the denominators
gives us the relation
α2 +αβ +β2 = 0.
Q {1,α,α2} L(cid:48)
This shows that the -basis is linearly dependent over , and also
Q {1,β,β2} L
shows that the -basis is linearly dependent over .
[L : K] [L(cid:48) : K]
From now on assume that and are (cid:28)nite. De(cid:28)ne the set
of (cid:28)nite sums of products
(cid:88)n
LL(cid:48) = { (cid:96) (cid:96)(cid:48) | (cid:96) ∈ L,(cid:96)(cid:48) ∈ L(cid:48),n ∈ N}.
i i i i
i=1
LL(cid:48)
Lemma 2 The set is closed under multiplication, and so forms a ring.
Proof. An exercise.
K LL(cid:48)
Lemma 3 The dimension (over ) of the set is bounded from above by
[L : K][L(cid:48) : K]
.
2
{(cid:96) ,...,(cid:96) } K L {(cid:96)(cid:48),...,(cid:96)(cid:48) } K
Proof. If 1 n is a -basis for and 1 m is a -basis for
L(cid:48) nm (cid:96) (cid:96)(cid:48) LL(cid:48)
, then clearly the elements i j span all of .
LL(cid:48)
Corollary 4 The ring is a (cid:28)eld.
x ∈ LL(cid:48)
Proof. Let be non-zero. By the previous lemma there exists
t {1,x,x2,...,xt}
a smallest natural number such that the set is linearly
K t 1 xt
dependent over . The minimality of implies that both and appear
with non-zero coe(cid:30)cients in a dependency relation. Therefore we can solve
x−1 K 1,x,...,xt−1
as a -linear combination of .
LL(cid:48) E L
Lemma 5 The set is the smallest (cid:28)eld (inside ) that contains both
L(cid:48) L L(cid:48)
and (called the compositum of and ).
Proof. Obvious.
The following result is then a relatively straightforward exercise. See
proof of Lemma 3 for spanning. Also recall that any linearly independent set
can be extended to a basis.
dim LL(cid:48) = (dim L)(dim L(cid:48))
Proposition 6 Thedimensionformula K K K holds
L L(cid:48)
if and only if and are linearly disjoint.
L L(cid:48)
We then turn our attention to the case that both and are Galois
K E
extensions of . In this case it is essential that we use such a (cid:28)eld that
K LL(cid:48) E LL(cid:48)
is separable over . Because is inside , we then deduce that is
K N E
also separable over . Thus it has a normal closure inside that is the
K LL(cid:48)
smallest Galois extension of that contains the (cid:28)eld .
N = LL(cid:48) LL(cid:48)
Proposition 7 We have and thus is also a Galois extension
K L∩L(cid:48) = K G(N/K)
of . If we further assume that , then the Galois group
G(L/K)
is isomorphic to the direct product of the component Galois groups
G(L(cid:48)/K)
and .
G = G(N/K) H = G(N/L) H(cid:48) = G(N/L(cid:48)) H
Proof. Let , and , so (resp.
H(cid:48) G L L(cid:48)
) is a normal subgroup of , because (resp. ) was assumed to be a
LL(cid:48)
Galois extension. By the Galois correspondence is the (cid:28)xed (cid:28)eld of the
M = H ∩H(cid:48)
subgroup (Galois correspondence reverses inclusions). But as
G M
an intersection of two normal subgroups of the group is normal itself.
3
LL(cid:48) K
Therefore its (cid:28)xed (cid:28)eld is actually a Galois extension of . Therefore
N = LL(cid:48) M = {1 }
and G is the trivial group.
K = L∩L(cid:48)
If we also assume that then another application of the Galois
G H
correspondence tells us that is the smallest group that contains both
H(cid:48)
and . As both of these subgroups are normal, the group they generate is
HH(cid:48) = {hh(cid:48) | h ∈ H,h(cid:48) ∈ H(cid:48)} G = HH(cid:48) H ∩H(cid:48) = {1 }
, so . Together with G
G (cid:39) H × H(cid:48) H (cid:39)
this implies that . The parallelogram rule shows that
G(L/K) H(cid:48) (cid:39) G(L(cid:48)/K)
and .
L/K L(cid:48)/K L∩L(cid:48) = K
Corollary 8 If and are (cid:28)nite Galois extensions and ,
then they are linearly disjoint.
Proof. This follows because the size of the Galois group equals the degree
of the extension for Galois extensions, so
dim LL(cid:48) = |G(LL(cid:48)/K)| = [L : K][L(cid:48) : K].
K
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