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4 MARK SCHACHNER
since ε
n
d
= 0 when d < n. That ε ∗ f = f ∗ ε = f follows from Lemma 3.1.
We can now make the following important statement about A:
Theorem 3.3. The set A, along with the operations + and ∗, forms a commutative
ring.
Proof. Pointwise addition is trivially commutative and associative, and has the
identity element 0 ∈ A, where 0(n) = 0 for all n ∈ N. Furthermore, for any f ∈ A
we define −f as the function such that (−f)(n) = −(f(n)) for each n ∈ N; this
represents the additive inverse of the element f. From Lemma 3.1 and Lemma 3.2,
we have that convolution is commutative, associative, and has an identity element
ε ∈ A. Lastly, from Lemma 3.1, we have that convolution distributes over addition.
There, is, however, additional algebraic structure on A with respect to convolu-
tion as demonstrated by the following lemma:
Lemma 3.4. Suppose f ∈ A.Then there exists f
−1
∈ A such that f ∗ f
−1
= ε if
and only if f(1) 6= 0.
Proof. To show the forward direction, let f ∈ A and suppose that there exists
f
−1
∈ A such that f ∗f
−1
= ε. Observe that f(1)f
−1
(1) = (f ∗f
−1
)(1) = ε(1) = 1,
so f
−1
(1) =
1
f(1)
. Hence f(1) 6= 0.
To show the backward direction, fix f ∈ A such that f(1) 6= 0. We will show that
g(n) can be uniquely defined such that (f ∗ g)(n) = ε(n), proceeding by induction
on n :
Base Case. Let n = 1 and g(1) =
1
f(1)
. Since f(1) 6= 0, we have g(1) is well-
defined, and this definition satisfies
(f ∗ g)(1) = f(1)g(1) = 1 = ε(1).
Furthermore, any value of g(1) which satisfies the above must equal
1
f(1)
, so this
g(1) is defined uniquely.
Inductive Step. Fix n ∈ N and suppose that for each k < n we have that g(k) is
well-defined and unique. We now define g(n) as follows:
g(n) = −
1
f(1)
X
d|n,d>1
f(d)g
n
d
.
Observe that g(n) is well-defined for each n ∈ N, since f(1) 6= 0 and
n
d
∈ N,
n
d
< n,
so g
n
d
is uniquely defined by the inductive hypothesis for each d | n, d > 1. Now,
we show f ∗ g = ε. We have two cases: n = 1 and n > 1. If n = 1, then as before
(f ∗ g)(n) = f(1)g(1) = ε(n). If n > 1, then
(f ∗ g)(n) =
X
d|n
f(d)g
n
d
= f(1)g(n) +
X
d|n,d>1
f(d)g
n
d
= f(1)g(n) − f (1)g(n)
= 0
= ε(n).