[proofplan] We prove each algebra axiom directly from the finite divisor-sum formula defining Dirichlet convolution. Pointwise addition and scalar multiplication give the usual complex [vector space](/page/Vector%20Space) structure on functions $\mathbb{N}\to\mathbb{C}$. Commutativity follows by pairing each divisor $d$ of $n$ with the complementary divisor $n/d$, associativity follows by rewriting both iterated convolutions as the same finite sum over triples $(a,b,c)\in\mathbb{N}^3$ with $abc=n$, and the identity law follows from the defining support of $\varepsilon$ at $1$. Bilinearity is checked term by term in the finite sums. [/proofplan] [step:Record the pointwise complex vector space structure] Let $\mathcal{A}=\{f:\mathbb{N}\to\mathbb{C}\}$. With pointwise addition and scalar multiplication, $\mathcal{A}$ is a complex vector space: for every vector space axiom and every $n\in\mathbb{N}$, the corresponding identity in $\mathcal{A}$ is exactly the corresponding identity in $\mathbb{C}$ after evaluation at $n$. The zero vector is the function $0_{\mathcal{A}}:\mathbb{N}\to\mathbb{C}$ defined by $0_{\mathcal{A}}(n)=0$, and the additive inverse of $f\in\mathcal{A}$ is the function $-f:\mathbb{N}\to\mathbb{C}$ defined by $(-f)(n)=-f(n)$. [/step] [step:Show that Dirichlet convolution is a well-defined binary operation] Let $f,g\in\mathcal{A}$. For each fixed $n\in\mathbb{N}$, the set \begin{align*} D(n):=\{d\in\mathbb{N}: d\mid n\} \end{align*} is finite, so the sum \begin{align*} (f*g)(n)=\sum_{d\mid n} f(d)g(n/d) \end{align*} is a finite sum of complex numbers. Thus $f*g:\mathbb{N}\to\mathbb{C}$ is a well-defined arithmetic function, so Dirichlet convolution is a binary operation on $\mathcal{A}$. [/step] [step:Prove commutativity by replacing each divisor with its complement] Let $f,g\in\mathcal{A}$ and fix $n\in\mathbb{N}$. Define the divisor-complement map \begin{align*} \theta_n:D(n)&\to D(n)\\ d&\mapsto n/d. \end{align*} This map is a bijection, and its inverse is itself. Reindexing the finite sum defining $(f*g)(n)$ by $e=\theta_n(d)=n/d$, we obtain \begin{align*} (f*g)(n) &=\sum_{d\mid n} f(d)g(n/d)\\ &=\sum_{e\mid n} f(n/e)g(e)\\ &=\sum_{e\mid n} g(e)f(n/e)\\ &=(g*f)(n), \end{align*} where the third equality uses commutativity of multiplication in $\mathbb{C}$. Since this holds for every $n\in\mathbb{N}$, $f*g=g*f$. [/step] [step:Prove associativity by summing over factor triples] Let $f,g,h\in\mathcal{A}$ and fix $n\in\mathbb{N}$. Expanding the left-associated convolution gives \begin{align*} ((f*g)*h)(n) &=\sum_{c\mid n} (f*g)(n/c)h(c)\\ &=\sum_{c\mid n}\left(\sum_{a\mid n/c} f(a)g((n/c)/a)\right)h(c). \end{align*} For each pair $(c,a)$ with $c\mid n$ and $a\mid n/c$, define \begin{align*} b:=\frac{n}{ac}. \end{align*} Then $a,b,c\in\mathbb{N}$ and $abc=n$. Conversely, if $(a,b,c)\in\mathbb{N}^3$ satisfies $abc=n$, then $c\mid n$ and $a\mid n/c$. Hence the preceding finite double sum is exactly \begin{align*} ((f*g)*h)(n) =\sum_{\substack{a,b,c\in\mathbb{N}\\ abc=n}} f(a)g(b)h(c). \end{align*} Similarly, \begin{align*} (f*(g*h))(n) &=\sum_{a\mid n} f(a)(g*h)(n/a)\\ &=\sum_{a\mid n} f(a)\left(\sum_{b\mid n/a} g(b)h((n/a)/b)\right). \end{align*} For each pair $(a,b)$ with $a\mid n$ and $b\mid n/a$, define \begin{align*} c:=\frac{n}{ab}. \end{align*} Then $a,b,c\in\mathbb{N}$ and $abc=n$, and the same converse holds. Therefore \begin{align*} (f*(g*h))(n) =\sum_{\substack{a,b,c\in\mathbb{N}\\ abc=n}} f(a)g(b)h(c). \end{align*} Thus $((f*g)*h)(n)=(f*(g*h))(n)$ for every $n\in\mathbb{N}$, so $(f*g)*h=f*(g*h)$. [guided] The key point is that every term in an iterated Dirichlet convolution corresponds to a factorisation of $n$ into three positive factors. We first expand the left-associated product. For fixed $n\in\mathbb{N}$, \begin{align*} ((f*g)*h)(n) &=\sum_{c\mid n} (f*g)(n/c)h(c)\\ &=\sum_{c\mid n}\left(\sum_{a\mid n/c} f(a)g((n/c)/a)\right)h(c). \end{align*} The outer divisor $c$ is the argument sent to $h$, and the inner divisor $a$ is the argument sent to $f$. Once $a$ and $c$ are chosen, the remaining argument sent to $g$ is forced to be \begin{align*} b:=\frac{n}{ac}. \end{align*} The divisibility condition $a\mid n/c$ ensures that $b\in\mathbb{N}$. Therefore every term in the double sum has the form $f(a)g(b)h(c)$ with $abc=n$. Conversely, suppose $(a,b,c)\in\mathbb{N}^3$ satisfies $abc=n$. Then $c\mid n$, and after dividing by $c$ we have $ab=n/c$, so $a\mid n/c$. Hence this triple appears exactly once in the expansion of $((f*g)*h)(n)$. Thus \begin{align*} ((f*g)*h)(n) =\sum_{\substack{a,b,c\in\mathbb{N}\\ abc=n}} f(a)g(b)h(c). \end{align*} Now expand the right-associated product: \begin{align*} (f*(g*h))(n) &=\sum_{a\mid n} f(a)(g*h)(n/a)\\ &=\sum_{a\mid n} f(a)\left(\sum_{b\mid n/a} g(b)h((n/a)/b)\right). \end{align*} Here the outer divisor $a$ is the argument sent to $f$, and the inner divisor $b$ is the argument sent to $g$. The remaining argument sent to $h$ is forced to be \begin{align*} c:=\frac{n}{ab}. \end{align*} Again, $b\mid n/a$ is precisely the condition that $c\in\mathbb{N}$. Conversely, every triple $(a,b,c)\in\mathbb{N}^3$ with $abc=n$ gives $a\mid n$ and $b\mid n/a$, so it appears exactly once. Therefore \begin{align*} (f*(g*h))(n) =\sum_{\substack{a,b,c\in\mathbb{N}\\ abc=n}} f(a)g(b)h(c). \end{align*} Both iterated convolutions are the same finite sum over the same indexing set, so $((f*g)*h)(n)=(f*(g*h))(n)$. Since $n$ was arbitrary, $(f*g)*h=f*(g*h)$. [/guided] [/step] [step:Verify that $\varepsilon$ is the convolution identity] Let $f\in\mathcal{A}$ and fix $n\in\mathbb{N}$. Since $\varepsilon(m)=0$ for every $m>1$ and $\varepsilon(1)=1$, the sum defining $(f*\varepsilon)(n)$ has only the term with $n/d=1$, equivalently $d=n$, contributing: \begin{align*} (f*\varepsilon)(n) &=\sum_{d\mid n} f(d)\varepsilon(n/d)\\ &=f(n)\varepsilon(1)\\ &=f(n). \end{align*} Similarly, the sum defining $(\varepsilon*f)(n)$ has only the term with $d=1$ contributing: \begin{align*} (\varepsilon*f)(n) &=\sum_{d\mid n} \varepsilon(d)f(n/d)\\ &=\varepsilon(1)f(n)\\ &=f(n). \end{align*} Thus $f*\varepsilon=f=\varepsilon*f$, so $\varepsilon$ is the multiplicative identity for Dirichlet convolution. [/step] [step:Check distributivity and scalar compatibility term by term] Let $f,g,h\in\mathcal{A}$, let $\lambda\in\mathbb{C}$, and fix $n\in\mathbb{N}$. Using finite additivity of sums and distributivity in $\mathbb{C}$, \begin{align*} (f*(g+h))(n) &=\sum_{d\mid n} f(d)(g+h)(n/d)\\ &=\sum_{d\mid n} f(d)\bigl(g(n/d)+h(n/d)\bigr)\\ &=\sum_{d\mid n} f(d)g(n/d)+\sum_{d\mid n} f(d)h(n/d)\\ &=(f*g)(n)+(f*h)(n). \end{align*} Thus $f*(g+h)=f*g+f*h$. By commutativity of convolution, this also gives $(g+h)*f=g*f+h*f$. For scalar compatibility, \begin{align*} ((\lambda f)*g)(n) &=\sum_{d\mid n} (\lambda f)(d)g(n/d)\\ &=\sum_{d\mid n} \lambda f(d)g(n/d)\\ &=\lambda\sum_{d\mid n} f(d)g(n/d)\\ &=(\lambda(f*g))(n). \end{align*} Again using commutativity, $f*(\lambda g)=\lambda(f*g)$ as well. Therefore Dirichlet convolution is bilinear over $\mathbb{C}$. [/step] [step:Conclude the algebra axioms] The set $\mathcal{A}$ is a complex vector space under pointwise addition and scalar multiplication. Dirichlet convolution is a well-defined bilinear multiplication on $\mathcal{A}$; it is associative, commutative, and has identity element $\varepsilon$. Hence $\mathcal{A}$ is a commutative associative unital $\mathbb{C}$-algebra with multiplicative identity $\varepsilon$. [/step]