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The nonzero cubes modulo thirteen are $\{1,5,8,12\}$, which excludes three. Thus this cubic has no root in its base field and is irreducible. For a root $\alpha$, the extension ${\mathbb{F}}_{13}(\alpha )$ has degree three. The base field already contains the primitive cube root of unity $3$, since $3^3=1$ and $3\ne 1$. Hence the other two roots are $3\alpha$ and $9\alpha$, already in this extension, making it the full splitting field.

The Frobenius endomorphism satisfies ${\alpha }^{13}=\alpha ({\alpha }^3{)}^4=3\alpha$, so it cycles all three roots. Therefore

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.

The cyclotomic polynomial isBy the allowed irreducibility, this is the minimal polynomial of $\xi$ and its degree is $\phi (n)$. The field $\mathbb{Q}(\xi )$ contains every $n$th root of unity and is the splitting field of $X^n-1$. This polynomial has no repeated roots in characteristic zero, so the extension is Galois.

An automorphism sends $\xi$ to another primitive root ${\xi }^a$, where $a$ is a unit modulo $n$. Conversely each such conjugate defines an embedding, and it is an automorphism of $\mathbb{Q}(\xi )$ because $a$ has an inverse modulo $n$. Composition corresponds to multiplication of exponents. Therefore

A finite solvable group is a group admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chainhas quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every subgroup of a solvable group is solvable, because intersecting the chain with the subgroup gives factors isomorphic to subgroups of the original abelian factors. A Galois group of a polynomial of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the polynomial is reducible or has repeated roots.

For the substitution, the characteristic-zero solvability by radicals theorem says that a polynomial has solvable Galois group if and only if its splitting field is contained in a finite tower formed by adjoining successive roots of elements of the preceding field. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th root of unity and, for every nonzero $\alpha$, an $m$th root ${\beta }_{\alpha }$. These are radical adjunctions, and all roots of $x^m-\alpha$ are ${\beta }_{\alpha }{\zeta }_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the splitting field of $f(x^m)$. The theorem implies

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

A primitive eighth root of unity is ${\zeta }_8=(1+i)/\sqrt{2}$, and $\mathbb{Q}({\zeta }_8)=\mathbb{Q}(i,\sqrt{2})$. Indeed $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$, while the displayed expression gives the reverse containment. Since $\mathbb{Q}(\sqrt{2})\subset \mathbb{R}$, adjoining $i$ has degree two; adjoining $\sqrt{2}$ to $\mathbb{Q}$ also has degree two. By the tower law,

A primitive seventh root of unity ${\zeta }_7$ generates all roots. Its minimal polynomial of an algebraic element is the permitted irreducible cyclotomic polynomial $1+X+\cdots +X^6$. Hence the splitting field has

If $f$ is integrable and $2\pi$-periodic and $n$ is a positive integer, then ${\int }_0^{2\pi }f(nx)e^{imx}dx=0$ for every integer $m$ with $0<|m|<n$. Split the integral into $n$ equal intervals and substitute $y=nx-2\pi j$ to obtainThe root of unity sum vanishes by the finite geometric series formula. The zero mode is $\int f$, so a mean-zero dilated function is orthogonal to all trigonometric polynomials of degree at most $n-1$.

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

For $d\ge 2$, nonzero roots define projective-point coordinates $s=X^{q-1}$, satisfying $s^N+s+T=0$, where $N=(q^d-1)/(q-1)$. This polynomial is irreducible over ${\mathbb{F}}_q(T)$. After fixing one root $s$, the remaining coordinates satisfy $R(Y,s)=(Y^N-s^N)/(Y-s)+1$. Substitute $z=Y/s$ and $h=1/s$: the equation becomes $h^{N-1}+(z^N-1)/(z-1)=0$. At a nontrivial $N$th root of unity the second term has a simple zero, so the polynomial is Eisenstein over the algebraic closure's rational function field. This proves irreducibility of $R$ and hence two-transitivity on projective points.

The symmetric quantum integer is the displayed Laurent polynomial, with $[0{]}_q=0$. It obeys $[n+1{]}_q=q^{-1}[n{]}_q+q^n$ and $[r{]}_q[s+1{]}_q-[s{]}_q[r+1{]}_q=[r-s{]}_q$. At a primitive odd $\ell$th root of unity, $[n{]}_q=0$ exactly when $\ell$ divides $n$.

If a finite extension of the P-adic numbers contains a primitive $p$th root of unity $\zeta$, the identity $p={\prod }_{j=1}^{p-1}(1-{\zeta }^j)$ gives $e=(p-1)v_L(1-\zeta )\ge p-1$. The ratios of these factors reduce to the nonzero integers $j$ in the residue field. Thus an extension with $e<p-1$ has only roots of unity of order prime to $p$.

For finite matrices satisfying $YX=qXY$, with $q$ not a root of unity, $Y$ sends the generalized eigenspace of $X$ at $\lambda$ into the one at $q^{-1}\lambda$. Finiteness of the spectrum makes $Y$ nilpotent on the sum of the nonzero $X$ generalized eigenspaces. Exchanging $X,Y$ gives the corresponding statement for $X$. Splitting the zero generalized eigenspace of $X$ by that of $Y$ gives the displayed invariant direct sum; both operators are nilpotent on $U$.

Let $S$ be a finite multiset of nonzero complex numbers such that taking the $m^2$th powers permutes it for every positive integer $m$. Taking $m=2$ and iterating the resulting permutation shows that each $z\in S$ obeys $z^{4^r}=z$ for some $r\ge 1$, hence is a root of unity. Let $d$ be a common multiple of their orders. The $d^2$th powers are all one, and the assumed permutation then forces every original element to be one. For a unitary matrix, the spectral theorem for normal operators further forces the matrix itself to be the identity.

An integer matrix with determinant $\pm 1$ induces an invertible transformation of the real torus, preserving normalized Haar measure. It is a mixing measure-preserving transformation exactly when none of its eigenvalues is a root of unity. Indeed Fourier frequencies evolve by $A^T$: a nonzero integer frequency cannot revisit a fixed frequency unless some power has eigenvalue one. Conversely a root of unity gives a nonzero integer vector fixed by a power of $A^T$, and its nonconstant character obstructs mixing.

In a mixed-characteristic discretely valued field, a primitive $p^a$th root of unity reduces to one. For every $j$ prime to $p$, the ratio $({\zeta }^j-1)/(\zeta -1)$ is a unit, so all factors of ${\Phi }_{p^a}(1)=p$ have the same valuation. Taking valuations gives the formula. Because the left side is a positive integer for a normalized valuation, the existence of such a root forces $p^{a-1}(p-1)\mid v(p)$, bounding the possible $p$-power torsion.