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For a primitive root of unity $\omega =e^{2\pi i/n}$ and labels $r,s$ modulo $n$, the displayed $n^2$ maximally entangled states form an orthonormal basis. Inner products equal ${\delta }_{s{s}^{\prime }}{n}^{-1}{\sum }_j{\omega }^{j(r^{\prime }-r)}={\delta }_{s{s}^{\prime }}{\delta }_{r{r}^{\prime }}$, by the geometric series. A nonprimitive root repeats phase labels and does not give a complete orthonormal basis.

A number-theoretic transform is a discrete Fourier transform computed in modular arithmetic. In a finite field with $p$ elements, choose a primitive root of unity $\omega$ of order $L$ dividing $p-1$. The transform above is inverted using ${\omega }^{-1}$ and $L^{-1}$. With sufficient zero padding, it computes polynomial products exactly modulo $p$, avoiding floating-point arithmetic rounding.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

An automorphism sends the primitive root of unity ${\xi }_n$ to another primitive root of unity, hence uniquely to ${\xi }_n^a$ with $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. Define $\chi (\sigma )=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore $\chi$ is an injective group homomorphism and $G$ is abelian.

The extension is finite Galois: it is the splitting field of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The Fundamental theorem of Galois theory identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives $L^H/K$ Galois, with group $G/H$.

For a non-surjective nontrivial example, take $K=\mathbb{Q}(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb{Z}/8\mathbb{Z}{)}^{\ast }$. For $K=\mathbb{Q}$ and prime $p$, ${\Phi }_p(X)=1+\cdots +X^{p-1}$ is irreducible: ${\Phi }_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the unit group's order, proving surjectivity.

For $n=7$, let $\zeta ={\xi }_7$. The cyclic Galois group of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:Here ${\sigma }_a(\zeta )={\zeta }^a$. To identify the quadratic field without guessing, put $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B{)}^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin \mathbb{Q}$, proving the quadratic identification. The real field is cubic because complex conjugation is ${\sigma }_6$, with fixed subgroup of order two.

A primitive root of unity ${\xi }_n$ satisfies ${\xi }_n^n=1$ and ${\xi }_n^j\ne 1$ for $1\le j<n$, equivalently it has multiplicative order $n$. A positive characteristic of a field must be prime: if its least positive value $p$ factored as $ab$ with $1<a,b<p$, the two nonzero field elements $a\cdot 1$ and $b\cdot 1$ would have product zero. If this prime $p$ divides $n$, put $m=n/p$. In characteristic $p$, the binomial coefficients $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ for $0<j<p$ vanish because their integer numerators contain $p$ and their denominators do not. Thus $(X-1{)}^p=X^p-1$. It follows that $({\xi }_n^m-1{)}^p={\xi }_n^n-1=0$. A field has no nonzero nilpotent element, so ${\xi }_n^m=1$, contradicting primitivity. Consequently $\boxed{\mathrm{char}K\nmid n}$.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

Let $L$ be a splitting field of $f(X)=X^m-1$ over $k$. Its formal derivative isThe assumption on the characteristic of $k$ says that $m\ne 0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f^{\prime }(\alpha )=m{\alpha }^{m-1}\ne 0$. Thus $f$ and $f^{\prime }$ have no common root, so $f$ is a separable polynomial. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the separability of roots of unity.

Fix a primitive root of unity ${\zeta }_m=e^{2\pi i/m}$. Define the $m$th cyclotomic polynomial byIt is monic and has degree given by the Euler totient function $\phi (m)$. Partitioning all $m$th roots of unity according to their exact multiplicative order gives the cyclotomic factorizationWe now prove by mathematical induction that ${\Phi }_m\in \mathbb{Z}[X]$. The base case is ${\Phi }_1=X-1$. If the result holds for every proper divisor $d$ of $m$, thenis monic and belongs to $\mathbb{Z}[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb{Z}[X]$. The factorization over $\mathbb{C}$ says that the remainder is zero and the quotient is ${\Phi }_m$, proving the claim.

It remains to prove irreducibility of cyclotomic polynomials. Let $f\in \mathbb{Z}[X]$ be a monic irreducible polynomial dividing ${\Phi }_m$, let $\epsilon$ be a root of $f$, and write ${\Phi }_m=fg$. We claim that ${\epsilon }^p$ is a root of $f$ for every prime number $p\nmid m$. Otherwise ${\epsilon }^p$ is a root of $g$, so $\epsilon$ is a root of $g(X^p)$. Since $f$ is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$, Gauss lemma for polynomials givesAfter reduction of an integer polynomial modulo a prime, the Frobenius endomorphism givesChoose an irreducible factor $h$ of the nonconstant monic polynomial $\bar{f}$. Then $h\mid \bar{g}$, so $h^2\mid \bar{f}\bar{g}={\bar{\Phi }}_m$. The cyclotomic factorization would then make $h^2$ divide $X^m-1$ over ${\mathbb{F}}_p$. This is impossible because its derivative $m{X}^{m-1}$ is coprime to it when $p\nmid m$.

Thus $f({\epsilon }^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f({\epsilon }^a)=0$ whenever $(a,m)=1$. These are all $\phi (m)$ primitive $m$th roots, soBecause $f$ divides ${\Phi }_m$, equality holds and $f={\Phi }_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb{Q}$.

Finally, direct multiplication givesAlsoApplying the cyclotomic factorization to the two factors on the right givesand thereforeSince $X^2-X+1={\Phi }_6(X)$, cancellation yieldsThe thirtieth cyclotomic polynomial is irreducible over $\mathbb{Q}$ by the theorem just proved, so $F$ is irreducible.

Let $\sigma (\alpha )={\xi }_d\alpha$ with $\alpha \ne 0$. Thenand these $d$ elements are distinct because ${\xi }_d$ is a primitive root of unity. Thus the stabilizer of $\alpha$ in the cyclic Galois group is trivial. The orbit-stabilizer form of the Fundamental theorem of Galois theory givessoMoreover,The fixed field of the whole Galois group is $K$, and thereforeThis is the basic eigenvector mechanism behind a cyclic Kummer extension.

The automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the linear independence of distinct field embeddings. Therefore the operatoris not the zero map. Choose $\beta \in L$ for which $\alpha =T(\beta )\ne 0$. Reindexing the sum and using ${\sigma }^n=1$ givesThis is the Lagrange resolvent eigenvector for a cyclic field automorphism.

Let $F=L^{\sigma }$ be the fixed field. Since $\sigma ({\alpha }^n)={\alpha }^n$, the element $\alpha$ is a root ofIts orbit under $⟨\sigma ⟩$ isand these $n$ elements are distinct because $\alpha \ne 0$ and $\zeta$ is a primitive root of unity. Hence the minimal polynomial has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so

A primitive root of unity is an $n$th root of unity of multiplicative order exactly $n$.

Let ${\zeta }_m,{\zeta }_n\in K$ be primitive and let $H$ be the finite subgroup of $K^{\times }$ that they generate. By part a, $H$ is cyclic. Its order is divisible by both $m$ and $n$, hence by $N=\mathrm{lcm}(m,n)$. On the other hand every element of $H$ has $N$th power one, so the order of the cyclic group $H$ divides $N$. Its order is therefore exactly $N$, and a generator of $H$ is a primitive $N$th root of unity in $K$.

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.

Let $n=q-1$ and let $\omega$ be a primitive root of unity in ${\mathbb{F}}_q$. The length-$n$ cyclic code with consecutive zeros of a cyclic code ${\omega }^b,\dots ,{\omega }^{b+n-k-1}$ has dimension $k$. It equals the generalized Reed-Solomon code of vectors $({\omega }^{j(1-b)}f({\omega }^j){)}_{j=0}^{n-1}$, $\mathrm{deg}f<k$. The choice $b=1$ is the unweighted Reed-Solomon code on these evaluation points; allowing other offsets makes this family closed under dual codes.

For an $n$-level qudit and primitive root of unity $\omega$, these unitary operators obey $X^n=Z^n=I$ and $ZX=\omega XZ$. The inverse of $X^s{Z}^{-r}$ is $Z^r{X}^{-s}$, with the order reversed. These operators implement the receiver corrections in qudit teleportation.

For distinct elements $a_1,\dots ,a_n$ of a finite field, a Reed-Solomon code consists of the evaluation vectors $(f(a_1),\dots ,f(a_n))$ of polynomials with degree less than $k$. It has dimension $k$ and minimum Hamming distance $n-k+1$, so it is a maximum distance separable code. A nonzero polynomial has at most its degree many distinct polynomial roots. Primitive cyclic Reed-Solomon codes use all nonzero field elements, ordered as powers of a primitive root of unity; arbitrary evaluation sets need not give a cyclic code.