Let
℘B
=
P
e
1
1
P
e
2
2
· · ·
P
e
g
g
, where
P
1
,P
2
,...,P
g
are distinct prime ideals of
B
and
e
1
,e
2
,...,e
g
are their ramification indices. As noted in the proof of the Theorem in
§
2.1, we have
℘B
∩
A
=
℘
=
P
e
i
i
∩
A
, and we have an isomorphism of
A/℘
–vector spaces
B/℘B
≃
g
circleplusdisplay
i
=1
B/P
e
i
i
.
Let us set
¯
A
=
A/℘
and
¯
B
=
B/℘B
. For
x
∈
B
, let ¯
x
denote the image of
x
in
¯
B
. Note that we
clearly have
Tr
¯
B/
¯
A
(¯
x
) =
Tr
L/K
(
x
)
for all
x
∈
B.
Now if
{
α
1
,...,α
n
}
is any
K
–basis of
L
contained in
B
such that
{
¯
α
1
,...,
¯
α
n
}
is an
¯
A
–basis of
¯
B
,
then using the above identity for traces, we see that
D
¯
B/
¯
A
(¯
α
1
,...,
¯
α
n
) =
D
L/K
(
α
1
,...,α
n
)
.
(1)
Next, we show that if
¯
B
≃
¯
B
1
⊕
...
⊕
¯
B
g
, where the isomorphism is of
¯
A
–vector spaces, then we
have
D
¯
B/
¯
A
=
g
productdisplay
i
=1
D
¯
B
i
/
¯
A
.
(2)
6
This assumption would always be satisfied in number theoretic applications since
A/℘
would usually be a finite
field.
26
To see the above identity, it suffices to consider the case when
g
= 2 since the general case would
follow by induction on
g
. For convenience of notation, let us denote the element of
B
corresponding
to (
u,
0)
∈
¯
B
1
⊕
¯
B
2
by
u
itself and, similarly, the element of
B
corresponding to (0
,v
)
∈
¯
B
1
⊕
¯
B
2
by
v
itself.
It is clear that we can choose
¯
A
–bases
{
u
1
,...,u
r
}
and
{
v
1
,...,v
s
}
of
¯
B
1
and
¯
B
2
respectively such that
{
u
1
,...,u
r
,v
1
,...,v
s
}
is an
¯
A
–basis of
¯
B
. In view of the above convention,
we see that
u
i
v
j
= 0. Thus Tr
¯
B/
¯
A
(
u
i
v
j
) = 0, and so
D
¯
B/
¯
A
(
u
1
,...,u
r
,v
1
,...,v
s
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
Tr(
u
i
u
i
′
)

0
......

......
0

Tr(
v
j
v
j
′
)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
D
¯
B
1
/
¯
A
(
u
1
,...,u
r
)
D
¯
B
2
/
¯
A
(
v
1
,...,v
s
)
.
Since
¯
A
is a field and the nonvanishing of any of the above discriminants is independent of the
choice of the corresponding
¯
A
–bases, the desired equality of discriminant ideals follows. Thus we
have proved (2).
Now suppose
℘
is a ramified prime. Then
e
i
>
1 for some
i
and thus the ring
B/P
e
i
i
contains
a nonzero nilpotent element (which may be taken to be any element of
P
e
i
−
1
i
\
P
e
i
i
), and hence
so does
¯
B
.
Let
β
∈
B
be such that
¯
β
∈
¯
B
is a nonzero nilpotent element.
Extend
{
¯
β
}
to an
¯
A
–basis
{
¯
β
1
,...,
¯
β
n
}
of
¯
B
with
β
1
=
β
.
Since
¯
β
1
is nilpotent, so is
¯
β
1
¯
β
j
for 1
≤
j
≤
n
.
Hence
Tr(
¯
β
1
¯
β
j
) = 0 for 1
≤
j
≤
n
[because if
u
∈
¯
B
is nilpotent, then 0 is clearly the only eigenvalue
of the linear transformation
x
mapsto→
ux
of
¯
B
→
¯
B
and Tr(
u
) equals the sum of all eigenvalues of
this linear transformation].
Consequently,
D
¯
B/
¯
A
(
¯
β
1
,...,
¯
β
n
) = 0, and so
D
¯
B/
¯
A
is the zero ideal.
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 Spring '11
 sdd
 Algebra, Number Theory, The Land, Algebraic number theory, Galois group, L/K