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The $n$th cyclotomic polynomial can be expressed via the Mobius function as follows: $$\Phi_n(x) = \prod_{\substack{1\le d\le n\\d\mid n}}(x^d - 1)^{\mu(\frac{n}{d})}$$ In every reference I have encountered, the cyclotomic polynomials are defined in terms of roots of unity. It seems, though, that there are quite a few things one can say about cyclotomic polynomials that have nothing to do with roots of unity. Is it possible to sidestep this by taking the above as a definition? Can one, say, using the above, show that:

  • $\Phi_n$ is monic with integer coefficients
  • the degree of $\Phi_n$ is $\phi(n)$
  • the above expression is in fact a polynomial
  • $x^n - 1 = \displaystyle\prod_{\substack{1\le d\le n\\ d\mid n}}\Phi_d(x)$

I'm not asking for proofs of the above statements; I'm merely asking if the display works as a definition.

node196884
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1 Answers1

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Yes, this works as a definition. You do need to prove that $\Phi_n(x)$ is actually a polynomial. Since Möbius inversion makes this identity equivalent to

$$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$$

it is a little easier to see what actually needs to be proven by rewriting the definition as the recursion

$$\Phi_n(x) = \frac{x^n - 1}{\prod_{d \mid n, d < n} \Phi_d(x)}.$$

So, what needs to be proven by strong induction on $n$ is that $x^n - 1$ is divisible by the product of all $\Phi_d(x)$ for all proper divisors $d$ of $n$. Since $\Phi_d(x) \mid x^d - 1 \mid x^n - 1$ it is clear that $x^n - 1$ is divisible by each $\Phi_d(x)$ individually. So what needs to be proven is that it is divisible by all of them together. This would follow if you knew that they were relatively prime (over $\mathbb{Q}$), which is true and can be proven without referencing roots of unity, so you can try to proceed by proving that by strong induction.

Once you know that the $\Phi_n(x)$ are actually polynomials (over $\mathbb{Q}$) it follows that they are monic polynomials over $\mathbb{Z}$ since you are dividing two monic polynomials over $\mathbb{Z}$ above (by strong induction). Then writing

$$n = \deg (x^n - 1) = \sum_{d \mid n} \deg \Phi_d(x)$$

and applying Möbius inversion again shows that $\deg \Phi_n(x) = \varphi(n)$.

Qiaochu Yuan
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  • However, it seems difficult to explain why you would conjecture that this family of polynomials exists without saying something about roots of unity. The roots of $x^n - 1$ over any field $K$ are, by definition, exactly the $n^{th}$ roots of unity in $K$, and the roots of $\Phi_d(x)$ over $K$ are exactly the primitive $d^{th}$ roots of unity in $K$ (assuming the characteristic does not divide $n$). So what this factorization is saying conceptually is that every $n^{th}$ root of unity is a primitive $d^{th}$ root of unity for a unique divisor $d$ of $n$ (namely its multiplicative order). – Qiaochu Yuan Jun 24 '24 at 22:38
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    Of course we may need to pass to an algebraic extension so the roots of unity actually exist. But the point is that this discussion is not specific to $\mathbb{C}$, if that's what you were unsatisfied with. So we can run the above argument over a splitting field of $x^n - 1$ over $\mathbb{Q}$, for example. The main thing that's nice about $\mathbb{C}$ is that it's very clear there that the group of $n^{th}$ roots of unity is cyclic, and over an arbitrary field we have to prove this. – Qiaochu Yuan Jun 24 '24 at 22:45
  • (For this see: https://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic/) – Qiaochu Yuan Jun 24 '24 at 22:48