Lemma 2

Let f, g and h be monic polynomials in $\mathbf{Q}[X]$ such that $f = gh$. If f has integer coefficients, the same is true of g and h.

Let $a$ (resp. $b$) be the least of the integers $\alpha \geq 3$ (resp. $\beta \geq 1$) such that $\alpha g$ (resp. $\beta h$) has integer coefficients; we put $g' = ag$ and $h' = bh$ and show by reductio ad absurdum that $a = b = 1$. If this were not so, there would be a prime divisor $p$ of $ab$. If $u \in \mathbf{Z}[X]$, let us write $\bar{u}$ for the polynomial with coefficients in the field $\mathbf{F}_p$ obtained by reduction mod $p$ of the coefficients of $u$. We have $g'h' = abf$, whence $\bar{g}'\bar{h}' = 0$; since the ring $\mathbf{F}_p[X]$ is an integral domain (IV, p. 9, Prop. 8) we thus have $\bar{g}' = 0$ or $\bar{h}' = 0$. In other words, $p$ divides all the coefficients of $g'$ or all those if $h'$ and this contradicts the hypothesis.

We have the relation

For we have $X^n - 1 = \prod_{\zeta \in \mu_n(\mathbf{Q})} (X - \zeta)$ and the sets $S_d$ for $d$ dividing $n$ form a partition of $\mu_n(\mathbf{Q})$.

The formula (6) determines $\Phi_n(X)$ when we know $\Phi_d(X)$ for all the divisors $d < n$ of $n$; since $\Phi_1(X) = X - 1$, we thus have a recursive procedure for calculating $\Phi_n$. For example for prime $p$ we have

whence

and

Let us list the values of the polynomials $\Phi_n(X)$ for $1 \leq n \leq 12$:

The values of $\Phi_1, \Phi_2, \Phi_3, \Phi_4, \Phi_5, \Phi_7, \Phi_8, \Phi_9$ and $\Phi_{11}$ follow directly from (7); now we have $\Phi_1 \Phi_2 \Phi_3 \Phi_6 = X^6 - 1$ and $\Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12} = X^{12} - 1$, whence $\Phi_4 \Phi_{12} = \frac{X^{12} - 1}{X^6 - 1} - X^6 + 1$ and finally $\Phi_{12} = \frac{X^6 + 1}{X^2 + 1} = X^4 - X^2 + 1$. The cases $n = 6$ and $n = 10$ may be treated similarly.