4..
- Calculate the following by hand:
,
,
, and
.
- Let
be an abelian group
and let
be a positive integer.
- Prove that the
map
given by
is a group
homomorphism.
- Prove that the subset
of
of squares
of elements of
is a subgroup.
- Use Theorem 4.1.7
to show that for
prime,
- (*) Use that
is cyclic to give a
direct proof that
when
. (Hint:
There is an
of order
. Show that
.)
- (*) If
, show directly that
by the method of Exercise 4.4.
(Hint: Let
be an element of order
. Show that
, etc.)
- (*) Let
be an odd prime. In this exercise you
will prove that
if and only if
.
- Prove that
is a parameterization of the set of solutions to
,
in the sense that the solutions
are in
bijection with the
such that
.
Here
corresponds to the point
.
(Hint: if
is a solution, consider the line
through
and
, and solve for
in terms
of
.)
- Prove that the number of solutions to
is
if
and
if
.
- Consider the set
of pairs
such
that
and
. Prove that
if
and
if
. Conclude that
is odd if and only if
- The map
that swaps coordinates is a
bijection of the set
. It has exactly one fixed point if and
only if there is an
such that
and
.
Also, prove that
has a solution
with
if and only if
.
- Finish by showing that
has exactly one fixed point
if and only if
is odd, i.e., if and only if
.
Remark: The method of proof of this exercise can be generalized to give a
proof of the full quadratic reciprocity law.
- How many natural numbers
satisfy the
equation
You may assume that
is prime.
- Find the natural number
such that
. Note that
is prime.
- In this problem we will formulate an analogue of
quadratic reciprocity for a symbol like
, but without the
restriction that
be a prime. Suppose
is a positive integer,
which we factor as
. We
define the Jacobi symbol
as follows:
- Give an example to show that
need not imply that
is a perfect square modulo
.
- (*) Let
be odd and
and
be integers. Prove that the
following holds:
-
. (Thus
induces a homomorphism from
to
.)
-
.
-
if
and
otherwise.
-
- (*) Prove that for any
the integer
does not have any divisors of the form
.
William
2007-06-01