Cyclotomic polynomials and their automorphisms

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Independent full-lesson AI review is pending. Public domain (CC0).

A supporting excerpt: definitions and the complete Theorem 12.1 proof, not the full parent lesson.

Fix the complex embedding and write $\zeta_n=e^{2\pi i/n},\qquad K_n=\mathbf Q(\zeta_n),\qquad U_n=(\mathbf Z/n\mathbf Z)^\times. \tag{1}$ We take $U_1$ to be the trivial group and $\varphi(1)=1$. Both $K_1$ and $K_2$ are $\mathbf Q$.

Cyclotomic polynomials and their automorphisms

Define $\Phi_n(X)=\prod_{\substack{1\leq a\leq n\$a,n)=1}}(X-\zeta_n^a).$ Grouping roots by their exact orders gives $X^n-1=\prod_{d\mid n}\Phi_d(X). \tag{2}$ Inductively, every \(\Phi_n$ is monic and integral. Indeed, divide $X^n-1$ by the product of the already integral monic factors for proper divisors. Monic division has integer quotient and remainder; the identity over $\mathbf C$ makes that remainder zero.

Monic factor lemma. If a monic polynomial in $\mathbf Z[X]$ is a product of two monic polynomials $f,g\in\mathbf Q[X]$, then $f,g\in\mathbf Z[X]$.

Proof. After clearing denominators and dividing the resulting coefficients by their greatest common divisor, write $f_0=af$, $g_0=bg$, where $f_0,g_0$ are primitive integer polynomials and $a,b$ are positive integers: they are the leading coefficients because $f,g$ are monic. The product of two primitive polynomials is primitive. Indeed, for each prime $p$ their reductions are nonzero polynomials over $\mathbf F_p$; the product is nonzero because that polynomial ring is a domain. Thus no prime divides every coefficient of the product. But $f_0g_0=abfg$ has coefficient greatest common divisor $ab$, since $fg$ is monic and integral. Consequently $ab=1$, so both factors were integral. $\square$

Theorem 12.1. The polynomial $\Phi_n$ is irreducible over $\mathbf Q$. Its degree is $\varphi(n)$, and the map $U_n\xrightarrow{\sim}\operatorname{Gal}(K_n/\mathbf Q), \qquad a\longmapsto\sigma_a,\quad \sigma_a(\zeta_n)=\zeta_n^a \tag{3}$ is an isomorphism.

Proof. Let $f$ be the monic minimal polynomial of $\zeta_n$. It divides the monic integral polynomial $\Phi_n$; the monic factor lemma makes its coefficients integers. Suppose $p\nmid n$ is prime and $\zeta_n^p$ is not a root of $f$. Let $g$ be its monic minimal polynomial, a distinct irreducible factor of $\Phi_n$. Since $g(\zeta_n^p)=0$, we have $f(X)\mid g(X^p)$ in $\mathbf Z[X]$.

Reduce modulo $p$. The identity $g(X^p)=g(X)^p$ there shows that any irreducible factor of $\overline f$ also divides $\overline g$. Since $fg\mid X^n-1$, the latter polynomial would have a repeated factor modulo $p$. Its derivative $nX^{n-1}$ is relatively prime to it, which is a contradiction.

The same argument works for any primitive root that is already a root of $f$. Every positive integer $a$ coprime to $n$ is a product of primes not dividing $n$. Repeated application therefore puts every $\zeta_n^a$ among the roots of $f$. Thus $f=\Phi_n$. All these roots lie in $K_n$; sending $\zeta_n$ to any one of them defines an automorphism. Composition multiplies exponents, proving (3). The cases $n=1,2$ are linear polynomials and trivial groups. $\square$

The isomorphism (3) follows from the action on roots of unity. Throughout this lesson Frobenius means the arithmetic automorphism acting on residue fields by $x\mapsto x^p$.

References

The cyclotomic irreducibility argument is compared with J. S. Milne’s freely available Fields and Galois Theory, section on cyclotomic extensions, and Algebraic Number Theory. The monic factor argument and complete Theorem 12.1 proof are written above.

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