Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

134
ADVANCED ABSTRACT ALGEBRA
field of G, then
Hence [K:L]=O(H). Also O(G(V/L)=[K:L], as K/L is seperate
.
Therefore
and
Hence Ψ is onto.
If
are subgroups of
, then the subfield left fixed by H 2
, will be left fixed by all elements
of H 1
, so this subfield is contained in the subfield left fixed by H 1
. On the other hand if
, then it is
obvious that
.
Consider the field L such that
. Suppose
is normal and
,
.
Claim:
V
Let a ∈ L , then each conjugate of a is in L.
( Let K be an extension of k, a, b K be algebraic over k, then a and b are said to be conjugate over k if
they are the roots of the same minimal polynomial over k.).
Since is a conjugate of a.
( minimal polynomial p(x) over k s.t p(a) = 0,
, a are roots of same minimal
polynomial over k).
∴
cbgh
cbgh d ibg
−1 −1 −1
σ ρ σ a = σ σ a = a σ ρσ a = a
σpd Θe
Θ∈
∃k
HK
E
L
G Ψ
σ
T
ρ
.
Now to show L k
is Galois:
If suffices to show
is normal, because we know that
is separable. Let p(x) be nonconstant
irreducible polynomial in k[x] which has one root, say a, in L. Since
and all of roots of p(x) can be expressed in the form
is normal, p(x) splits in k[x]
(a) for some
.
Let ρ ∈Ge K Lj, then
an element
such that
, for some σ ∈Ge K kj
Now ρσ = στ ρσ a = στ a = σ τ a
.
b gbgb gbg
cbg
h