Number Theory: An Introduction to Mathematics

3 Multiplicative Functions 153

For any two functionsf,g:N→C, we define theirsum f+g:N→Cin the
natural way:

(f+g)(n)=f(n)+g(n).

It is obvious that addition is commutative and associative, and that the distributive law
holds:

f∗(g+h)=f∗g+f∗h.

The functionδ:N→C,definedby

δ(n)=1 or 0 according asn=1orn> 1 ,

acts as an identity element for Dirichlet multiplication:

δ∗f=f for everyf:N→C,

since

δ∗f(n)=

∑

d|n

δ(d)f(n/d)=f(n).

Thus the setAof all arithmetical functions is indeed a commutative ring.
For any function f :N→Cwhich is not identically zero, put|f|=v(f)−^1 ,
wherev(f)is the least positive integernsuch thatf(n)=0, and put|O|=0. Then

|f∗g|=|f||g|,|f+g|≤max(|f|,|g|) for allf,g∈A.

Hence the ringA of all arithmetical functions is actually an integral domain. It is
readily shown that the set of allf∈Asuch that|f|<1 is an ideal, but not a prin-
cipal ideal. (AlthoughA is not a principal ideal domain, it may be shown that it is a
factorialdomain.)
The next result shows that the functionsf∈Asuch that|f|=1aretheunitsin
the ringA:

Lemma 25For any function f:N→C, there is a function f−^1 :N→Csuch that
f−^1 ∗ f =δif and only if f( 1 )= 0. The inverse f−^1 is uniquely determined and
f−^1 ( 1 )f( 1 )= 1.

Proof Supposeg:N→Chas the property thatg∗f=δ.Theng( 1 )f( 1 )=1. Thus
g( 1 )is non-zero and uniquely determined. Ifn>1, then
∑

d|n

g(d)f(n/d)= 0.

Hence

g(n)f( 1 )=−

∑

d|n,d<n

g(d)f(n/d).

It follows by induction thatg(n)is uniquely determined for everyn∈N.Conversely,
ifgis defined inductively in this way, theng∗f=δ.