Multiplicative functions form abelian group under Dirichlet product
Contents
Statement
Suppose is the set of natural numbers and
is a commutative unital ring. Let
be the set of all multiplicative functions from
to
, i.e., all the functions
satisfying the following:
-
,
-
whenever
are relatively prime.
Consider the Dirichlet product, a binary operation defined on functions from to
:
.
Then, forms an abelian group under the Dirichlet product, with multiplicative identity given by the function
that takes the value
at
and
elsewhere.
Proof
Closure under Dirichlet product
We first need to show that the Dirichlet product is a well-defined binary operation on multiplicative functions; in other words, that a Dirichlet product of multiplicative functions is multiplicative.
Given: Multiplicative functions .
To prove: is multiplicative.
Proof: Suppose and
are relatively prime natural numbers. Consider:
.
Given any divisor of
,
can be expressed uniquely as
, where
and
. Conversely, given
and
,
. Thus, we have:
.
Next, since and
are relatively prime and
and
are also relatively prime, and
and
are multiplicative:
.
Next, we rearrange:
.
Thus:
.
Associativity and commutativity
These follow from the fact that the Dirichlet product is associative and commutative for all functions:
Identity element
This follows from the fact that the function is an identity element for the Dirichlet product for all functions: Identity element for Dirichlet product is indicator function for one.
Inverses
Given a multiplicative function , the inverse of
can be defined inductively as follows:
.
Since , this clearly satisfies:
.
Thus, . Since the multiplication is commutative,
is a two-sided inverse for
.
Next, we need to verify that is multiplicative. Fill this in later