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  "body": "\n## Proof\n\nWe first show that $\\Phi_n$ has rational coefficients. Then we show that $\\Phi_n$ is the minimal polynomial of $\\zeta = e^{2\\pi i/n}$.\n\n1. Let $n \\ge 1$ and $\\zeta = e^{2 \\pi i / n}$, so $\\zeta^k = e^{2 \\pi i k / n}$ for every integer $k$. Since $\\zeta^n = 1$, the number $\\zeta^k$ depends only on $k$ modulo $n$.\n2. By [[the description of the $n$th roots of unity|057BH3JC9S]], each of them is $\\zeta^k$ for some $0 \\le k \u003c n$. There are exactly [[$n$ of them|0E021GF2B8]]. So $\\zeta^0, \\dots, \\zeta^{n-1}$ are pairwise distinct.\n3. Let $E$ be a field, $F \\in E[x]$, and $a_1, \\dots, a_m$ pairwise distinct roots of $F$. Then $\\prod_{i=1}^m (x - a_i)$ divides $F$. We prove this by induction on $m$; the case $m = 0$ is trivial.\n4. For $m \\ge 1$, the induction hypothesis gives $F = \\prod_{i\u003cm} (x - a_i) \\, G$. Then $0 = F(a_m) = \\prod_{i\u003cm} (a_m - a_i) \\, G(a_m)$, and each $a_m - a_i \\ne 0$. So $G(a_m) = 0$, and the [[factor theorem|0K3MWA7DMC]] gives $(x - a_m) \\mid G$.\n5. By Steps 2 to 4, $\\prod_{k=1}^{n} (x - \\zeta^k)$ divides $x^n - 1$. Both are [[monic|0BDX4MBKT4]] of degree $n$, so $x^n - 1 = \\prod_{k=1}^{n} (x - \\zeta^k)$.\n6. By the [[definition|0W0KVBM54S]], $\\Phi_d = \\prod (x - e^{2 \\pi i j / d})$, over the $j$ with $1 \\le j \\le d$ and $\\gcd(j, d) = 1$.\n7. For $1 \\le k \\le n$, let $e = \\gcd(k, n)$, $d = n / e$ and $j = k / e$. Then $d \\mid n$, $1 \\le j \\le d$, $\\gcd(j, d) = 1$ and $\\zeta^k = e^{2 \\pi i j / d}$.\n8. Conversely, for $d \\mid n$ and such a $j$, put $k = j n / d$. Then $\\gcd(k, n) = (n/d) \\gcd(j, d) = n/d$, so Step 7 gives back $(d, j)$. So Step 7 is a bijection, and Step 5 becomes $x^n - 1 = \\prod_{d \\mid n} \\Phi_d$.\n9. We show by strong induction on $n$ that $\\Phi_n \\in \\Q[x]$. First, $\\Phi_1 = x - 1$. For $n \u003e 1$, the product $h = \\prod_{d \\mid n, d \u003c n} \\Phi_d$ is monic and lies in $\\Q[x]$ by induction.\n10. [[Division by the monic $h$|0MES3KS0ZC]] in $\\Q[x]$ gives $x^n - 1 = q h + r$ with $q, r \\in \\Q[x]$ and $\\deg r \u003c \\deg h$. In $\\C[x]$ we also have $x^n - 1 = \\Phi_n h + 0$ by Step 8. Uniqueness of division in $\\C[x]$ gives $\\Phi_n = q \\in \\Q[x]$.\n11. Next, a lemma. Let $P \\in \\Z[x]$ be monic and $P = A B$ with $A, B \\in \\Q[x]$ monic. Then $A, B \\in \\Z[x]$.\n12. Proof of the lemma. Let $a \\ge 1$ be the least integer with $a A \\in \\Z[x]$. If a prime $\\ell$ divides all coefficients of $a A$, then $\\ell \\mid a$, the leading coefficient. Then $(a/\\ell) A \\in \\Z[x]$, which contradicts the choice of $a$.\n13. By Step 12 and the [[fundamental theorem of arithmetic|084HH1NDCB]], $aA$ is [[primitive|0SP14DZN0G]]. Define $b$ for $B$ in the same way, so $b B$ is primitive.\n14. The ring $\\Z$ is a [[Euclidean domain|000JFQGDXG]], hence a [[PID|0QKVE1GD9A]] and a [[UFD|055HT85T3S]]. By [[the product of primitive polynomials|08C1NWBNHA]], $(aA)(bB) = ab P$ is primitive. But $ab$ divides all its coefficients, so $ab = 1$ and $A, B \\in \\Z[x]$.\n15. Let $f$ be the [[minimal polynomial|0GK1WB9R98]] of $\\zeta$ over $\\Q$. It exists because $\\zeta$ is a root of $x^n - 1$, so $\\zeta$ is [[algebraic|0NPW1DWECM]]. By [[irreducibility|0NFEG48RDR]] and [[divisibility|06K7SARZKR]], $f$ is irreducible and $f \\mid x^n - 1$.\n16. Key claim: if $\\omega \\in \\C$, $f(\\omega) = 0$ and $p$ is a prime with $p \\nmid n$, then $f(\\omega^p) = 0$. Suppose not; we derive a contradiction in Steps 17 to 27.\n17. Since $f \\mid x^n - 1$, we get $\\omega^n = 1$. The minimal polynomial $m_\\omega$ of $\\omega$ divides $f$. Since $f$ is irreducible and both are monic of positive degree, $f = m_\\omega$.\n18. Also $(\\omega^p)^n = 1$. Let $g$ be the minimal polynomial of $\\omega^p$, so $g \\mid x^n - 1$. Write $x^n - 1 = f k$ with $k \\in \\Q[x]$ monic.\n19. Then $0 = f(\\omega^p) k(\\omega^p)$ and $f(\\omega^p) \\ne 0$, so $k(\\omega^p) = 0$. So $g \\mid k$, and $x^n - 1 = f g h$ with $h \\in \\Q[x]$ monic.\n20. By the lemma of Step 11, applied to $x^n - 1 = f \\cdot (g h)$ and then to $g h = g \\cdot h$, the polynomials $f, g, h$ lie in $\\Z[x]$.\n21. Let $G(x) = g(x^p)$, a monic polynomial in $\\Z[x]$. Then $G(\\omega) = g(\\omega^p) = 0$. By Step 17 and divisibility, $G = f k_2$ with $k_2 \\in \\Q[x]$ monic. By the lemma, $k_2 \\in \\Z[x]$.\n22. Let $\\F_p$ be the [[field $\\Z/p\\Z$|0NV6E43P4A]], and let $u \\mapsto \\bar{u}$ reduce coefficients modulo $p$. It is a ring homomorphism $\\Z[x] \\to \\F_p[x]$, by the [[universal property|0DQMDSVSXJ]]. So $x^n - 1 = \\bar{f} \\bar{g} \\bar{h}$ and $\\bar{G} = \\bar{f} \\bar{k}_2$.\n23. The ring $\\F_p[x]$ has [[characteristic|0P7CH1CND5]] $p$. So the [[Frobenius map|0GHDDVAH8P]] $u \\mapsto u^p$ is a ring homomorphism. With $\\bar{g} = \\sum_i b_i x^i$ and [[Fermat's little theorem|0BMXWXDVJZ]] $b_i^p = b_i$, we get $\\bar{g}^p = \\sum_i b_i x^{ip} = \\bar{G}$.\n24. The polynomial $\\bar{f}$ is monic with $\\deg \\bar{f} = \\deg f \\ge 1$. Let $q$ be a monic divisor of $\\bar{f}$ of least degree among those of degree at least $1$.\n25. Then $q$ is irreducible. Suppose $q = a b$ with $\\deg a, \\deg b \\ge 1$. Then $a$, made monic, is a smaller such divisor, by the [[degree of a product|0T233AZ1GE]].\n26. The ring $\\F_p[x]$ is a [[Euclidean domain|0DQG39K2NH]], hence a UFD, so $q$ is [[prime|0QRCFZKXH8]]. Since $q \\mid \\bar{f} \\mid \\bar{G} = \\bar{g}^p$, the definition of a [[prime element|0MNHXBSX3M]], used $p - 1$ times, gives $q \\mid \\bar{g}$.\n27. So $q^2 \\mid \\bar{f} \\bar{g}$, and $x^n - 1 = q^2 s$ for some $s \\in \\F_p[x]$. The [[formal derivative|0AD4VS02H3]] and its product rule give $n x^{n-1} = q (2 q' s + q s')$. Since $p \\nmid n$, $n$ is a unit of $\\F_p$, so $q \\mid x^{n-1}$. Then $q$ divides $x \\cdot x^{n-1} - (x^n - 1) = 1$, which is impossible for $\\deg q \\ge 1$.\n28. This proves the key claim. Now let $k \\ge 1$ with $\\gcd(k, n) = 1$. We show $f(\\zeta^k) = 0$ by induction on the number of prime factors of $k$. For $k = 1$ this holds by the choice of $f$.\n29. Otherwise $k = p m$ with $p$ prime. Then $p \\nmid n$ and $\\gcd(m, n) = 1$, since both $p$ and $m$ divide $k$. By induction $f(\\zeta^m) = 0$. The key claim with $\\omega = \\zeta^m$ gives $f(\\zeta^k) = 0$.\n30. By Steps 2, 6 and 29, the roots $\\zeta^k$ of $\\Phi_n$ are $\\deg \\Phi_n$ distinct roots of $f$. A [[polynomial has at most $\\deg f$ roots|0KJVQV3E0F]], so $\\deg f \\ge \\deg \\Phi_n$.\n31. By Step 10, $\\Phi_n \\in \\Q[x]$, and $\\Phi_n(\\zeta) = 0$. So $f \\mid \\Phi_n$, say $\\Phi_n = f u$. 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