Remark

Tag 06OS, permanent. Cite it as [Bourbaki, Tag 06OS].

Algebra, chapter V, § 11. Abelian extensions

\* For every integer $n > 0$ a function $\mu(n)$ is defined as follows: if $n$ is divisible by the square of a prime number, we put $\mu(n) = 0$ , otherwise $\mu(n) = (-1)^h$ if $n$ is the product of $h$ distinct prime numbers (« Mobius function »). It may be shown that

\Phi_n(X) = \prod_{d|n} (X^{n/d} - 1)^{\mu(d)},

or more explicitly

\Phi_n(X) = \prod_{p_1 < \ldots < p_h} (X^{n/p_1 \cdots p_h} - 1)^{(-1)^h}

where $(p_1, \ldots, p_h)$ runs over the set of all strictly increasing sequences of prime divisors of $n$ (cf. Lie, II, p. 207, Ex. 1).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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