Suppose is a subgroup of a finite group and
is a -module. For each , there is a natural map
called restriction. Elements of can be viewed as classes
of -cocycles, which are certain maps
, and the restriction maps restricts these cocycles to
.
If is a normal subgroup of , there is also an inflation map
given by taking a cocycle
and precomposing with the quotient map to obtain a cocycle for .
Proof.
Our proof follows [
Ser79, pg. 117] closely.
We see that
by looking at cochains. It remains
to prove that
is injective and that the image of
is the kernel of .
- That
is injective: Suppose
is a
cocycle whose image in is equivalent to 0 modulo
coboundaries. Then there is an such that
, where we identify with the map that is
constant on the costs of . But depends only on the costs of
modulo , so
for all
, i.e.,
(as we see by adding to both
sides and multiplying by
).Thus , so is
equivalent to 0 in
.
- The image of
contains the kernel of :
Suppose is a cocycle whose
restriction to is a coboundary, i.e., there is such
that
for all
.
Subtracting the coboundary
for
from , we may assume
for all
.
Examing the equation
with shows that is constant on the cosets of .
Again using this formula, but with
and , we see
that
so the image of is contained in . Thus defines a cocycle
, i.e., is in the image of
.
This proposition will be useful when proving
the weak Mordell-Weil theorem.
Example 11.3.2
The sequence of Proposition
11.3.1 need not be
surjective on the right. For example, suppose
,
and let
act trivially on the cyclic group
.
Using the
interpretation of
, we see
that
, but
has order
.
Remark 11.3.3
On generalization of Proposition
11.3.1 is to
a more complicated exact sequence involving the ``transgression map''
tr:
Another generalization of Proposition
11.3.1
is that if
for
, then
there is an exact sequence
Remark 11.3.4
If
is a not-necessarily-normal subgroup of
, there are also
maps
for each
. For
this is the trace map
, but the definition for
is more involved.
One has
.
Taking
we see
that for each
the group
is annihilated by
.
William Stein
2012-09-24