Prove that , where is the th cyclotomic polynomial and is the Möbius function on defined by , if is squarefree with distinct prime factors, and if is divisible by a square.

[I consulted this document from the Berkeley Math Circle when preparing this solution.]

Before we approach this (seemingly magical) identity, let’s build up some machinery.

Given an abelian group , let denote the set of all functions . Recall that is itself an abelian group under pointwise addition.

Now let be a commutative ring with 1 and let be a left -module. Define an operator by . Note that since is positive, this sum is finite, and that and are nonzero. We will usually drop the subscripts on . This operator is called the *Dirichlet convolution* of and .

Lemma 1: Let and . Then . Proof: We have

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= | ||

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= |

As desired.

Lemma 2: Let and . Then (using pointwise addition) we have and . Proof: We have

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= | ||

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= |

as desired. The other equality is analogous.

Corollary 1: is a ring with pointwise addition and Dirichlet convolution, and is a left module via Dirichlet convolution. Proof: Follows from Lemmas 1 and 2.

Let denote the Kronecker delta (whose value is 1 if and 0 otherwise) and let be the Möbius function.

Lemma 3: For all , . Proof: We have . Now the delta function is 0 unless , in which case , and so this sum is precisely as desired.

Now let denote the constant function whose value is 1.

Lemma 4: . Proof: We have . If , then we have . Now suppose , with the prime. Now , where . Note that if any is greater than 1, then (by definition) . So the only summands which contribute to those with . Moreover, in this case, the value of depends only on whether the number of nonzero is even or odd. There are subsets of , half of which have an even number of elements and half odd. So . Hence .

Now given , define – so . Then .

Now let be the abelian group . (The nonzero rational functions over in one variable). This is an abelian group, hence a -module, where we write our operator multiplicatively and our -action by exponentiation. Let be given by , and let . Now , which expanded out gives the identity as desired.