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144 7. Galois Cohomology and Isomorphism Classification of Elliptic Curves

The following result is the basic property of Galois extensions, and it leads im-
mediately to the Galois correspondence between subfields of K containing F and
subgroups of Gal(K/F).

(1.2) Theorem. (1) (Dedekind) Let u 1 ,... , u n be field morphisms K → Lover
F, and suppose that there is an L-linear relation a 1 u 1 +· · · + a n u n = 0 where
the u 1 are distinct. Then all coefficients a i = 0 for i = 1,... , n.
(2) (Artin) If
⊂ Aut(K) is a finite subsemigroup, and if F is the set of all x ∈ K
with u(x) = x for all u ∈
,thenn = #
= [K : F], K is separable over F,
and
is a subgroupof Aut(K).

Proof. (1) Choose an element y ∈ K with u n−1 (y) = u n (y). Evaluate the L-linear
relation at x times u n (y) and at xy, and we obtain

0 = a 1 u 1 (x)u n (y) +· · · + a n u n (x)u n (y)
and also
0 = a 1 u 1 (xy) +· · · a n u n (xy)
= a 1 u 1 (x)u 1 (y) +· · · + a n u n (x)u n (y).
Subtracting the two resulting formulas, we eliminate one term

& '
0 = a 1 u 1 (x) [u 1 (y) − u n (y)] +· · · + a n−1 u n−1 (x) u n−1 (y) − u n (y)
where not all coefficients are zero. With this algebraic operation we reduce the num-
ber of nonzero terms in an L-linear relation by at least one, and inductively this
means that u 1 ,... , u n is L-linearly independent.
As an application of (1), the sum u 1 +· · · + u n : K → L is nonzero. If K is
separable over F, that is, for some L ⊃ K the set E of field morphisms K → L
over F has the property that x ∈ K satisfies u (x) = u (x) for all u , u ∈ E implies

x ∈ F. Then the trace

tr K/F (x) = u(x)
u∈E
is defined tr K/F : K → F, and it is a nonzero F-linear form.
(2) Show that any set α 0 ,... ,α n ∈ K is F-linearly dependent. Consider n equa-
n −1
tions in x 0 ,... , x n of the form x j u (α j ) = 0 for u ∈
. Thus there exists

j=0
elements c 0 ,... , c n ∈ K with Tr K/F (c 0 ) = 0and
n
−1
c j u α j = 0 for u ∈
.
j=0
Applying u to the relation, we have n u(c j )α j = 0 for u ∈
, and summing
j=0
n
over u ∈
,wehave Tr(c j )α j = 0 over F. Hence [K : F] ≤ n, and therefore
j=0
we have #
= n = [K : F] using Dedekind’s theorem.