Cyclotomic polynomial (${\Phi }_n(X)$)

${\Phi }_n$ is the monic polynomial whose roots are the primitive $n$th roots of unity. It belongs to $\mathbb{Z}[X]$ and is irreducible over $\mathbb{Q}$.

Partitioning the roots of $X^n-1$ by their exact orders givesInductively, the product over proper divisors is a monic polynomial in $\mathbb{Z}[X]$. Division of $X^n-1$ by this monic polynomial stays in $\mathbb{Z}[X]$ and has zero remainder, proving that ${\Phi }_n\in \mathbb{Z}[X]$.

Every cyclotomic polynomial ${\Phi }_n$ is an irreducible polynomial over $\mathbb{Q}$. If a monic irreducible factor $f\in \mathbb{Z}[X]$ contains a primitive root $\zeta$, then it also contains ${\zeta }^p$ for every prime number $p\nmid n$: otherwise, writing ${\Phi }_n=fg$, reduction modulo $p$ and the Frobenius endomorphism give $\bar{f}\mid \bar{g}(X{)}^p$. Some irreducible factor would then divide both $\bar{f}$ and $\bar{g}$, giving a repeated factor of $X^n-1$, contrary to separability of roots of unity. Iterating over the prime factors of every integer coprime to $n$ shows that $f$ contains all primitive $n$th roots, so $f={\Phi }_n$.

The thirtieth cyclotomic polynomial isIndeed,

If the prime $p$ does not divide $n$, then $X^n-1$ is square-free over ${\mathbb{F}}_p$ because its derivative $n{X}^{n-1}$ is coprime to it. Its factor ${\Phi }_n$ is therefore separable modulo $p$.

If $L$ is the splitting field of ${\Phi }_n$ over a characteristic-zero field $K$, choosing a primitive root ${\zeta }_n$ gives an injectionIt is surjective exactly when ${\Phi }_n$ is irreducible over $K$.

A cyclotomic field is generated by a root of unity. If $K\subseteq \mathbb{C}$ and ${\zeta }_n$ is primitive, every $K$-conjugate of ${\zeta }_n$ is a power ${\zeta }_n^a$ and already lies in $K({\zeta }_n)$. Thus $K({\zeta }_n)/K$ is Galois, and its Galois group embeds in $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$, so it is abelian.

Cyclotomic units include roots of unity and suitable ratios $(1-{\zeta }^a)/(1-\zeta )$ in cyclotomic fields. Compatible roots $\zeta$ make these ratios norm-compatible along cyclotomic towers. Their closed norm-limit module is a concrete submodule of global units, and its local images generate the power series in the Iwasawa main conjecture.

A cyclotomic Euler system consists of units in auxiliary cyclotomic extensions satisfying compatible norm relations. Reducing at carefully chosen auxiliary primes and descending these relations bounds class-group modules. Combined with global unit-index formulas, this supplies an alternative route to the Iwasawa main conjecture.

Every field $\mathbb{Q}({\zeta }_{59^r})$ contains $\mathbb{Q}(\sqrt{-59})$ and is totally ramified over it at the prime above $59$. The class number divisibility under total ramification therefore shows that its class number is divisible by three. Its ring of integers is not a principal ideal domain. The degrees $58\cdot 59^{r-1}$ strictly increase, giving infinitely many distinct examples.

The prime $\ell$ is totally ramified in $\mathbb{Q}({\zeta }_{{\ell }^r})$, whose degree is $\phi ({\ell }^r)$. The displayed equality is an equality of ideals in its ring of integers of a number field, and the residue field is ${\mathbb{F}}_{\ell }$.

For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ is the displayed field. A quadratic Gauss sum squares to $(-1{)}^{(\ell -1)/2}\ell$, and the cyclic Galois group has a unique subgroup of index two.

For a prime $p$, the cyclotomic polynomial of a primitive $p$th root of unity isIt is irreducible over $\mathbb{Q}$, so adjoining one root gives a field of degree $p-1$.

For an odd prime $p$ and a primitive $p$th root ${\zeta }_p$,

For a primitive fifth root ${\zeta }_5$, the element $s={\zeta }_5+{\zeta }_5^{-1}$ satisfies $s^2+s-1=0$. Thus the fixed field of complex conjugation is

The roots of unity in $\mathbb{Q}({\zeta }_n)$ form a cyclic group of order

For a primitive seventh root of unity ${\zeta }_7$,where ${\sigma }_a({\zeta }_7)={\zeta }_7^a$. Its four intermediate fields correspond to the four subgroups of the cyclic group $C_6$.

The unique quadratic subfield of $\mathbb{Q}({\zeta }_7)$ isThe conjugate period is $-1-\eta$, and

The unique cubic subfield is the maximal real subfieldandIt is the fixed field of complex conjugation.

The Galois group of $\mathbb{Q}({\zeta }_{11})/\mathbb{Q}$ isIts only subfield degrees over $\mathbb{Q}$ are $1,2,5,10$, and every subfield is Galois because this group is abelian.

The unique quadratic subfield of $\mathbb{Q}({\zeta }_{11})$ is

The fixed field of complex conjugation in $\mathbb{Q}({\zeta }_{11})$ is the degree-five field $\mathbb{Q}({\zeta }_{11}+{\zeta }_{11}^{-1})$.

For a field $K$ inside a fixed algebraic closure, the maximal cyclotomic extension is the union of the fields $K({\zeta }_n)$ generated by roots of unity.

The maximal cyclotomic extension of $\mathbb{R}$ is $\mathbb{C}$, since adjoining a primitive fourth root of unity adjoins $i$.

The maximal cyclotomic extension of $\mathbb{Q}$ is a proper subfield of $\overline{\mathbb{Q}}$. Every finite subextension inside a cyclotomic field is Galois and abelian, whereas $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal.

The maximal cyclotomic extension of ${\mathbb{F}}_p$ is ${\overline{\mathbb{F}}}_p$. A primitive $(p^m-1)$th root of unity generates ${\mathbb{F}}_{p^m}$ over ${\mathbb{F}}_p$, and these finite fields exhaust the algebraic closure.