450
9. FIELD EXTENSIONS – SECOND LOOK
Proposition 9.7.2.
Let
f
2
KOExŁ
be an irreducible separable polynomial.
Let
E
be a splitting field for
f
, and let
˛
1
; : : : ; ˛
n
be the roots of
f .x/
in
E
. The Galois group
Aut
K
.E/
, regarded as a group of permutations of
the set of roots, is contained in the alternating group
A
n
if, and only if, the
discriminant
ı
2
.f /
of
f
has a square root in
K
.
Proof.
The discriminant has a square root in
K
if, and only if,
ı.f /
2
K
.
But
ı.f /
2
K
if, and only if, it is fixed by the Galois group, if, and only
if, the Galois group consists of even permutations.
n
We do not need to know the roots of
f
in order to compute the dis
criminant. Because the discriminant of
f
D
P
i
a
i
x
i
is the discriminant
of the monic polynomial
.1=a
n
/f
multiplied by
a
2n
1
n
, it suffices to give
a method for computing the discriminant of a monic polynomial. Suppose,
then, that
f
is monic.
The discriminant is a symmetric polynomial in the roots and, there
fore, a polynomial in the coefficients of
f
, by Theorem
9.6.6
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 Fall '08
 EVERAGE
 Algebra, Group Theory, Permutations, Polynomials, Galois group

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