Proof. If \(x\in\Gamma_q^*\) and \(z=\sigma x/x\), then \[0=\frac{x+\sigma x+\sigma^2x}{x}=1+z+z\sigma z.\] Since \(\sigma z=z^q\), this gives \(z^{q+1}+z+1=0\), so \(z\in\Lambda_q\). Also \(z\ne0\), and \(z\ne-1\) because substituting \(z=-1\) in \(1+z+z\sigma z\) gives \(1\).

Conversely, let \(z\in\Lambda_q\). Then \(z^q=-1-z^{-1}=\tau(z)\). From \(1+z+z\sigma z=0\) one checks that \[z\sigma z\sigma^2z=1,\] so \(\mathop{\mathrm{Nm}}_{L/K}(z)=1\). By multiplicative Hilbert 90 there is \(x\in L^*\) with \(z=\sigma x/x\). Then \[x+\sigma x+\sigma^2x=x(1+z+z\sigma z)=0,\] so \(x\in\Gamma_q^*\). The ambiguity in \(x\) is exactly multiplication by elements of \(K^*\), which proves the bijection. Since \(L/K\) is a finite separable extension, the trace map \(\mathop{\mathrm{Tr}}_{L/K}:L\to K\) is a nonzero \(K\)-linear functional and hence is surjective. Therefore \(\Gamma_q\) has \(K\)-dimension \(2\), so \(|\Gamma_q|=q^2\), and the cardinality follows. ◻

Proof. The zero element is fixed by \(F_{r,B}\). The nonzero points are controlled by the quotient map of 1. For \(c\in K^*\) and \(x\in\Gamma_q^*\), \[F_{r,B}(cx)=c^rF_{r,B}(x), \qquad (cx)^{q-1}=x^{q-1}.\] Thus, on a nonzero quotient fiber on which \(B\) does not vanish, the map along the \(K^*\)-fiber is multiplication by \(c^r\); it is bijective exactly when \(\gcd(r,q-1)=1\).

Assume first that \(F_{r,B}\) permutes \(\Gamma_q\). If \(B(z)=0\) for some \(z\in\Lambda_q\), choose \(x\in\Gamma_q^*\) with \(x^{q-1}=z\) by 1; then \(F_{r,B}(x)=0=F_{r,B}(0)\), contradicting injectivity. Hence condition (ii) holds, and the fiber calculation above gives condition (i). For \(z=x^{q-1}\in\Lambda_q\) we have \(F_{r,B}(x)\in\Gamma_q^*\), so \[\frac{\sigma(F_{r,B}(x))}{F_{r,B}(x)} =z^r\frac{B^{(q)}(\sigma z)}{B(z)} =z^r\frac{B^{(q)}(\tau(z))}{B(z)} =R_{r,B}(z)\] lies in \(\Lambda_q\). The quotient map induced by a bijection of \(\Gamma_q^*\) is bijective on \(\Gamma_q^*/K^*\simeq\Lambda_q\), so condition (iii) follows.

Conversely, assume (i)–(iii). If \(x\in\Gamma_q^*\) and \(z=x^{q-1}\), then \(B(z)\ne0\), so \(F_{r,B}(x)\ne0\). The same computation gives \[\sigma(F_{r,B}(x))/F_{r,B}(x)=R_{r,B}(z)\in\Lambda_q.\] By the defining equation of \(\Lambda_q\), this implies \[F_{r,B}(x)+\sigma(F_{r,B}(x))+\sigma^2(F_{r,B}(x))=0,\] so \(F_{r,B}(x)\in\Gamma_q^*\). Condition (iii) gives bijectivity on the quotient \(\Lambda_q\), and condition (i) gives bijectivity on every \(K^*\)-fiber. Together with \(F_{r,B}(0)=0\), this proves that \(F_{r,B}\) permutes \(\Gamma_q\). ◻

Proof. There is nothing to prove in characteristic zero. Assume \(p>0\) and \(p\nmid e\). Write \[N_e=(-1)^e(z+1)^e-z^e, \qquad D_e=z^e-1, \qquad C_e=\gcd(N_e,D_e),\] and \(h_e=n_e/d_e\) with \(N_e=C_en_e\) and \(D_e=C_ed_e\). If \(e=1\), then \[H_1^{\rm raw}(z)=\frac{-2z-1}{z-1};\] in characteristic \(3\) this quotient is constant, while in every other characteristic it has degree one and is separable. Thus assume \(e\ge2\).

A direct calculation gives \[N_e'D_e-N_eD_e'=eF_e(z), \qquad F_e(z)=z^{e-1}-(-1)^e(z+1)^{e-1}(z^{e-1}+1),\] and, after cancellation, \[N_e'D_e-N_eD_e'=C_e^2(n_e'd_e-n_ed_e').\] The polynomial \(F_e\) is not identically zero, because its leading term is \(-(-1)^e z^{2e-2}\). Since \(e\ne0\) in \(k\), the Wronskian \(n_e'd_e-n_ed_e'\) is not identically zero. A nonconstant rational function over a perfect field of characteristic \(p\) is inseparable exactly when its Wronskian is identically zero. Hence \(h_e\) is separable. ◻

Proof. Put \[a=z,\qquad b=\tau(z),\qquad c=\tau^2(z).\] Then \(abc=1\). Let \(A=a^d\), \(B=b^d\), and \(C=c^d\), so \(ABC=1\). From the definition, \[H_d^{\rm raw}(a)=A\frac{B-1}{A-1}.\] Therefore \[\begin{align} \tau(H_d^{\rm raw}(a)) &=-1-\frac{1}{H_d^{\rm raw}(a)} \\ &=-\frac{H_d^{\rm raw}(a)+1}{H_d^{\rm raw}(a)} \\ &=-\frac{AB-1}{A(B-1)} \\ &=\frac{1-AB}{A(B-1)}. \end{align}\] Since \(C=(AB)^{-1}\), \[H_d^{\rm raw}(b)=B\frac{C-1}{B-1} =B\frac{(AB)^{-1}-1}{B-1} =\frac{1-AB}{A(B-1)}.\] Thus \(H_d^{\rm raw}(\tau z)=\tau(H_d^{\rm raw}(z))\) on the dense open set where the displayed expressions are defined, and hence as rational functions on \(\mathbb{P}^1\). Passing to the equivalent reduced rational function gives the statement for \(h_d\). ◻

Proof. Assume first that \(\operatorname{char}k\ne3\). The automorphism \(\tau\) has two fixed points, the primitive cube roots \(\omega,\omega^2\), and is diagonalizable in \(\operatorname{PGL}_2(k)\). The displayed coordinate sends these fixed points to \(0\) and \(\infty\), so \(\tau\) becomes multiplication by a primitive cube root, which we denote again by \(\omega\). The equivariance relation is \[\Phi(\omega u)=\omega\Phi(u).\] Thus \(\Phi(u)/u\) is invariant under \(u\mapsto\omega u\), and the invariant field is \(k(u^3)\). Hence \(\Phi(u)=uR(u^3)\).

In characteristic \(3\), every nontrivial order-three element of \(\operatorname{PGL}_2(k)\) is unipotent and hence conjugate to \(u\mapsto u+1\). The equivariance relation becomes \[\Phi(u+1)=\Phi(u)+1.\] Therefore \(\Phi(u)-u\) is invariant under translation by \(1\). The invariant field for this Artin–Schreier action is \(k(u^3-u)\), so \(\Phi(u)=u+R(u^3-u)\). ◻

Proof. The denominator \(D_d=z^d-1\) is squarefree. Its roots are the elements \(\zeta\in\mu_d\). Such a root is canceled by the numerator precisely when \[0=N_d(\zeta)=(-1)^d(\zeta+1)^d-\zeta^d.\] Since \(\zeta^d=1\), this is equivalent to \[(-1)^d(\zeta+1)^d=1.\] But \[\tau(\zeta)^d=\left(-\frac{\zeta+1}{\zeta}\right)^d =(-1)^d(\zeta+1)^d,\] again because \(\zeta^d=1\). Hence the common roots of \(N_d\) and \(D_d\) are exactly the \(\zeta\in\mu_d\) with \(\tau(\zeta)\in\mu_d\).

The map \[\zeta\longmapsto (\zeta,\zeta\tau(\zeta))=(\zeta,-1-\zeta)\] identifies this set with \[L_0\cap\mu_d^2, \qquad L_0:\;1+X+Y=0.\] Because \(D_d\) is squarefree, each cancellation is simple, so the cancellation degree is \(\delta_d\).

Let \(C_d=\gcd(N_d,D_d)\). Since \(D_d\) is squarefree, \(C_d\) is a product of exactly the \(\delta_d\) canceled linear factors. After cancellation, \[d_d=D_d/C_d, \qquad \deg d_d=d-\delta_d.\] The numerator satisfies \[n_d=N_d/C_d, \qquad \deg n_d\le d-\delta_d,\] because \(\deg N_d\le d\) and the same common factor \(C_d\) of degree \(\delta_d\) has been removed. Hence \[\deg h_d=\max(\deg n_d,\deg d_d)=d-\delta_d.\] This also covers the constant case: if \(d-\delta_d=0\), both reduced numerator and denominator are constant and the convention gives morphism degree \(0\). ◻

Proof. This is the denominator computation in the proof of 6. The denominator \(D_d=z^d-1\) is squarefree, so exactly the \(\delta_d\) canceled linear factors are removed from \(D_d\) and no other denominator factor is lost. ◻

Proof. This is the second formulation of 6. The connection with vanishing sums of roots of unity is classical; see [4]–[6] and [7] for broader background. ◻

Proof. In characteristic \(p\), \[z^d-1=(z^{d_0}-1)^{p^s}\] and \[(-1)^d(z+1)^d-z^d =\left((-1)^{d_0}(z+1)^{d_0}-z^{d_0}\right)^{p^s}.\] This proves \(H_d^{\rm raw}=(H_{d_0}^{\rm raw})^{p^s}\), and therefore \(h_d=h_{d_0}^{p^s}\) after canceling common factors.

On the toric line \(L_0\) we use the coordinate \(X\), with \(Y=-1-X\), and localize away from \(X=0\) and \(1+X=0\). The intersection with \(\boldsymbol{\mu}_d^2\) is defined by \[X^d-1=0, \qquad (-1-X)^d-1=0.\] These two equations are the \(p^s\)-th powers of the corresponding equations for \(d_0\). At every closed point of the reduced intersection for \(d_0\), the functions \(X^{d_0}-1\) and \((-1-X)^{d_0}-1\) vanish simply because \(p\nmid d_0\) and the point lies in the torus. Thus, in the completed local ring \(k[[u]]\), both tame local equations are unit multiples of \(u\), while the wild local equations generate \((u^{p^s})\). Each reduced point therefore contributes length \(p^s\). The degree identity follows from 6 applied to \(d_0\) and the equality \(\deg h_d=p^s\deg h_{d_0}\).

For the finite-field assertion, fix \(q=p^k\), \(L=\mathbb{F}_{q^3}\), and \(K=\mathbb{F}_q\). Then \[X^{dq}-X^d=(X^{d_0q}-X^{d_0})^{p^s}.\] The \(p^s\)-power Frobenius is a bijection of \(L\) and preserves \(\Gamma_q=\ker\mathop{\mathrm{Tr}}_{L/K}\), so the permutation property is unchanged. ◻

Proof. Write \[\binom d{j+1}=\rho\binom dj \qquad(1\le j\le d-2)\] for some \(\rho\in k^*\). The identity \[(j+1)\binom d{j+1}=(d-j)\binom dj\] gives \[(j+1)\rho=d-j \qquad(1\le j\le d-2).\] Taking \(j=1\) and \(j=2\) and subtracting gives \(\rho=-1\), without dividing by \(2\) or \(3\). Then \(2\rho=d-1\) gives \(d+1=0\) in \(k\).

Since \(\binom d1=d=-1\) in \(k\) and the ratio is \(-1\), the interior coefficients satisfy \[\binom dj=(-1)^j\qquad(1\le j\le d-1).\] For \(j=d-1\) this says \(d=(-1)^{d-1}\); as \(d=-1\) in \(k\), we get \((-1)^d=1\), so the endpoint \(\binom dd=1\) also equals \((-1)^d\). Hence \[(1+z)^d=\sum_{j=0}^d(-1)^jz^j.\] Multiplying by \(1+z\) gives \[(1+z)^{d+1}=1+z^{d+1}.\] By Lucas’ theorem [8], equivalently by writing \(d+1\) in base \(p\), all intermediate binomial coefficients of \((1+z)^{d+1}\) vanish in characteristic \(p\) if and only if \(d+1\) is a power of \(p\). Thus \(d+1=p^a\). ◻

Proof. Write \(N_d=(-1)^d(z+1)^d-z^d\) and \(D_d=z^d-1\). Since the reduced form of \(N_d/D_d\) has degree at most one, there are constants \(A,B,C,E\), not all zero, such that \[\label{eq:linear-relation} (Cz+E)N_d=(Az+B)D_d.\tag{4}\] A constant map would force \(N_d\) to be a scalar multiple of \(D_d\), which is impossible for \(d\ge4\) because the coefficient of \(z\) in \(N_d\) is \((-1)^d d\ne0\) whereas \(D_d\) has no \(z\)-term. Thus 4 represents a genuine degree-at-most-one reduced map.

Let \(n_j\) be the coefficient of \(z^j\) in \(N_d\) for \(1\le j\le d-1\). Then \(n_j=(-1)^d\binom dj\) and \(n_1=n_{d-1}=(-1)^d d\ne0\). We first show \(C,E\ne0\). If \(E=0\), then comparing coefficients of \(z^j\) for \(2\le j\le d-1\) in 4 gives \(Cn_{j-1}=0\), hence \(n_1=0\), contradiction. If \(C=0\), the same comparison gives \(En_j=0\) for \(2\le j\le d-1\), hence \(n_{d-1}=0\), again a contradiction.

Now for \(2\le j\le d-1\), the coefficient of \(z^j\) on the right side of 4 is zero, because \((Az+B)(z^d-1)\) has only the terms \(Az^{d+1}\), \(Bz^d\), \(-Az\), and \(-B\). Hence \[Cn_{j-1}+En_j=0 \qquad(2\le j\le d-1).\] Since \(C,E\ne0\) and \(n_1\ne0\), the sequence \(n_1,\ldots,n_{d-1}\) is nonzero and geometric. Therefore the same is true of the interior binomial coefficients. ◻

Proof. The small cases are explicit. For \(d=1\), \[H_1^{\rm raw}(z)=\frac{-(z+1)-z}{z-1}=\frac{-2z-1}{z-1}.\] For \(d=2\), \[H_2^{\rm raw}(z)=\frac{(z+1)^2-z^2}{z^2-1}=\frac{2z+1}{z^2-1},\] which has degree \(2\) unless characteristic \(3\), in which case \(d=2=3^1-1\). For \(d=3\) and \(p\ne3\), \[H_3^{\rm raw}(z)=\frac{-(z+1)^3-z^3}{z^3-1} =-\frac{2z+1}{z-1}.\] In the small overlaps with the branch \(d=p^a-1\), these formulas also give the asserted identity \(h_d=\tau^2\): in characteristic \(2\) for \(d=1\) and \(d=3\), and in characteristic \(3\) for \(d=2\), the displayed expressions reduce to \(-1/(z+1)=\tau^2(z)\).

Now assume \(d\ge4\) and \(p=0\). If \(\deg h_d\le1\), the same coefficient comparison as in 3 implies that the integer binomial coefficients \(\binom d1,\ldots,\binom d{d-1}\) form a geometric progression in \(\mathbb{Q}^*\). This is impossible for \(d\ge4\) since the ratios \(\binom d2/\binom d1=(d-1)/2\) and \(\binom d3/\binom d2=(d-2)/3\) are unequal over \(\mathbb{Q}\).

Finally assume \(p>0\) and \(d\ge4\). By 3 2, degree at most one forces \(d=p^a-1\). Conversely, if \(d=p^a-1\), then in characteristic \(p\), \[(z+1)^{p^a}=z^{p^a}+1.\] Using \(d=p^a-1\) gives \((-1)^d=1\) in the ground field. Therefore \[(z+1)^d=\frac{z^{p^a}+1}{z+1}, \qquad z^d=\frac{z^{p^a}}{z}.\] Thus \[\begin{align} H_d^{\rm raw}(z) &=\frac{(z^{p^a}+1)/(z+1)-z^{p^a}/z}{z^{p^a}/z-1} \\ &=\frac{z-z^{p^a}}{z(z+1)}\cdot\frac{z}{z^{p^a}-z} =-\frac{1}{z+1}=\tau^2(z) \end{align}\] as an identity in \(k(z)\), so the reduced quotient is \(h_d=\tau^2\). This completes the proof. ◻

Proof. By 6, the inequality is equivalent to \(\deg h_d\le1\). Apply 8. ◻

Proof. The recurrence \(R_j=0\) is exactly the coefficient comparison in the degrees \(3\le j\le d-1\), because the right side \(P(z)(z^d-1)\) has no terms in those degrees. The boundary equations are obtained by comparing the remaining coefficients in degrees \(0,1,2,d,d+1,d+2\) and using \(b_2=d(d-1)/2\).

For \(\varepsilon=1\), substituting \(S=d(T+1)\) into the first and third boundary equations gives two linear equations in \(T\): \[(1-d^2)T+\left(b_2-d^2\right)=0, \qquad \left(b_2-d^2\right)T+(1-d^2)=0.\] Their resultant is \[(1-d^2)^2-\left(b_2-d^2\right)^2 =\frac{(d-2)(d+1)^2(3d-2)}{4}.\] Since \(d+1\ne0\), this gives \((d-2)(3d-2)=0\). If \(d=2\), the boundary equations give \(S=0\) and \(T=-1\), hence \(R_3=b_1-b_3=d-b_3\). On the branch \(3d-2=0\), the first two boundary equations give \[S=\frac{d(d-2)}{2d-2}, \qquad T=-\frac{d}{2d-2},\] and substitution into \(R_3\) and \(R_4\) gives the two displayed formulas.

For \(\varepsilon=-1\), if \(d=3\) in the ground field, the displayed reductions are immediate from the three boundary equations and the formula for \(R_3=b_1-Sb_2+Tb_3\). Now assume \(d-3\ne0\). If \(d+3=0\), the first two boundary equations are inconsistent: they reduce to \(S+T+2=0\) and \(S+T+1=0\). Hence \(d+3\ne0\). Solving the first two boundary equations gives \[S=\frac{d(d+2)}{2d+6}, \qquad T=\frac{d}{2d+6}.\] Substitution into the third boundary equation gives \[-\frac{(d-3)(d+2)(d+6)}{4(d+3)}=0,\] so, under \(d-3\ne0\), one obtains \((d+2)(d+6)=0\). Substitution into \(R_3\) gives \[R_3+\frac{d^3+3d^2+2d-12}{6} =\frac{(d-1)(d+2)(d+6)}{2(d+3)},\] which proves the displayed residual modulo \((d+2)(d+6)\). Similarly, after clearing the factor \(120\) and the already-invertible factor \(d+3\), one has \[\begin{align} &(d+3)\left(120R_5+(d-2)(d-1)(3d^3+3d^2-32d-72)\right) \\ &\qquad=(d-3)(d-2)(d-1)(d+2)(d+6)^2 . \end{align}\] Since \(d+3\ne0\), the cleared \(R_5\) identity holds on the branch \(d=-6\). Finally, the \(j=3\) residual on \(d=-6\) is \(22\), so its vanishing forces characteristic \(11\); only after this point do we divide by \(120\). ◻

Proof. The case \(d=2\) follows from \[H_2^{\rm raw}(z)=\frac{2z+1}{z^2-1}.\] It has degree \(2\) in characteristic zero and in characteristic \(p\ne2,3\); in characteristic \(3\) it is the degree-one case \(d=2=3^1-1\). A direct calculation in characteristic \(11\) gives \[\gcd\bigl(z^5-1,-(z+1)^5-z^5\bigr)=z^3-4z^2+4z-1,\] (the Sage verification in 23 records this computation reproducibly), and hence \[h_5(z)=\frac{-2z^2-2z+1}{z^2+4z+1}.\] Thus the listed cases have degree \(2\).

Conversely, suppose \(\deg h_d=2\) and \(d\ge4\). By 8, we may assume that we are not in a degree-one case. By 1, the reduced denominator has degree \(2\), equivalently exactly two denominator factors of \(D_d\) remain uncanceled. Let \[Q(z)=z^2-Sz+T, \qquad T\ne0,\] be the denominator of the reduced form of \(h_d\). Since the unreduced denominator is \(z^d-1\) and is squarefree, \(Q\) is the product of the two uncanceled denominator factors. There is a polynomial \(P\) of degree at most \(2\) such that \[\label{eq:quadratic-identity} Q(z)N_d(z)=P(z)D_d(z).\tag{5}\] Put \(b_j=\binom dj\) and \(\varepsilon=(-1)^d\). For \(1\le j\le d-1\) the coefficient of \(z^j\) in \(N_d\) is \(\varepsilon b_j\). In the coefficient-comparison argument below, equations such as \(d=-1\), \(3d-2=0\), or \(d\equiv5\pmod {11}\) are equations for the image of the integer \(d\) in the ground field unless the text explicitly says “as an integer.” Comparing coefficients of \(z^j\) in 5 for \(3\le j\le d-1\) gives \[\label{eq:quadratic-recurrence} b_{j-2}-S b_{j-1}+T b_j=0 \qquad(3\le j\le d-1).\tag{6}\] The boundary coefficients in degrees \(0,1,2,d,d+1,d+2\) will be used repeatedly below.

First assume \(p=2\). Then \(d\) is odd and \(\varepsilon=1\). The boundary coefficients specialize to \[b_2-Sd+T=0, \qquad S=d(T+1), \qquad b_2T-Sd+1=0.\] Since \(d=1\) in the ground field, the first two equations give \[b_2=1, \qquad T=S+1,\] and the third equation is then redundant. Thus 6 becomes \[b_{j-2}+S b_{j-1}+(S+1)b_j=0.\] Here every \(b_j\) lies in \(\mathbb{F}_2\), and \(b_1=b_2=1\). If no consecutive pair \(b_{j-1},b_j\) differs, then all interior coefficients are already equal to \(1\). Otherwise, for some \(j\) one has \(b_{j-1}\ne b_j\); the displayed recurrence then determines \[S=b_{j-2}+b_j\in\mathbb{F}_2.\] If \(S=0\), the recurrence is \(b_j=b_{j-2}\), so the initial values force all \(b_j=1\). If \(S=1\), the recurrence is \(b_{j-2}=b_{j-1}\) for every \(j\) in the range; shifting the index gives \(b_3=b_2=1\), \(b_4=b_3=1\), and so on, while \(b_{d-1}=d=1\). Hence again all interior binomial coefficients are \(1\). They are therefore nonzero and geometric, so 2 gives \(d=2^a-1\), a degree-one case. This contradiction excludes characteristic \(2\).

Now assume \(p=3\). If \(\varepsilon=-1\), the boundary coefficients specialize to \[b_2-Sd=0, \qquad d(T+1)=0, \qquad b_2T-Sd=0.\] Since \(d\ne0\) in the ground field, the second equation gives \(T=-1\); comparing the first and third then gives \(b_2=0\), and hence \(S=0\). Thus \[T=-1, \qquad b_2=0, \qquad S=0.\] Then 6 gives \(b_j=b_{j-2}\) for all \(3\le j\le d-1\). Since \(d\) is odd in this branch, \(d-1\) is even, so \(b_{d-1}=b_2=0\), contradicting \(b_{d-1}=d\ne0\). If \(\varepsilon=1\), the boundary coefficients are \[b_2-Sd+T=0, \qquad S=d(T+1), \qquad b_2T-Sd+1=0.\] If \(d=1\) in the ground field, then \(b_2=0\), and the first two equations give \(T=S\) and \(S=T+1\), a contradiction. Hence \(d=-1\) in the ground field; then \(b_2=d(d-1)/2=1\), and the second equation gives \(S+T=-1\) while the third gives the same relation. Thus \[d=-1, \qquad b_2=1, \qquad S+T=-1.\] The recurrence then forces \(b_j=(-1)^j\) for every \(1\le j\le d-1\): if \(T\ne0\) this follows by induction from \(b_1=-1\) and \(b_2=1\), and if \(T=0\) the shifted recurrence gives \(b_j=-b_{j-1}\) up to \(j=d-2\), while \(b_{d-1}=d=-1\) supplies the last term. Hence the interior binomial coefficients are nonzero and geometric, so 2 gives \(d=3^a-1\), again a degree-one case. Thus characteristic \(3\) gives no degree-two cases.

It remains to consider characteristic zero or characteristic \(p\ge5\). The eliminations used below are those of 4. We split according to \(\varepsilon\).

Suppose first that \(\varepsilon=1\). Comparing the six boundary coefficients gives \[\label{eq:eps-plus} b_2-Sd+T=0, \qquad S=d(T+1), \qquad b_2T-Sd+1=0.\tag{7}\] If \(d=-1\) in the ground field, then 7 gives \(S+T=-1\), and the recurrence, with initial values \(b_1=-1\) and \(b_2=1\), forces the interior coefficients to be geometric as above. This is a degree-one case, contrary to our assumption. Hence \(d+1\ne0\).

Eliminating \(S\) and \(T\) from 7 , using \(b_2=d(d-1)/2\), gives \[(d-2)(3d-2)=0\] in the ground field. If \(d=2\) in the ground field, then 7 gives \(S=0\) and \(T=-1\). The recurrence at \(j=3\) would then give \(b_3=b_1=d=2\), whereas the binomial identity gives \[b_3=\frac{d(d-1)(d-2)}{6}=0,\] a contradiction. Therefore \(3d-2=0\).

On the branch \(3d-2=0\), the recurrence coefficient at \(j=3\) reduces to \[-\frac{d(d-3)(d+1)}{3}.\] Since \(d\ne0\) and \(d+1\ne0\), this can vanish only in characteristic \(7\). In characteristic \(7\) the same branch gives \(d=3\) in the ground field. Then \(d\) cannot be the integer \(4\), so \(j=4\) lies in the recurrence range. The recurrence coefficient at \(j=4\) reduces on this branch to \[-\frac{d(d-4)(d+1)(5d-6)}{48},\] which is nonzero at \(d=3\) in characteristic \(7\). This contradiction excludes the branch \(\varepsilon=1\).

Finally suppose that \(\varepsilon=-1\). The boundary equations are \[\label{eq:eps-minus} b_2-Sd+3T=0, \qquad 3S=d(T+1), \qquad b_2T-Sd+3=0.\tag{8}\] If \(d=3\) in the ground field and \(d>3\) as an integer, then 8 gives \(T=S-1\), and applying 6 at \(j=3\) gives \(S=1\), hence \(T=0\), impossible. Thus, apart from the already excluded integer \(d=3\), we may assume \(d-3\ne0\).

For \(d-3\ne0\), equations 8 imply \[(d+2)(d+6)=0.\] After substituting the corresponding value of \(T\), the recurrence coefficient at \(j=3\) is \[-\frac{d^3+3d^2+2d-12}{6}.\] The branch \(d=-2\) gives the nonzero value \(2\). On the branch \(d=-6\), this coefficient is \(22\), so the characteristic must be \(11\). Thus \(d\equiv5\pmod{11}\). If \(d=5\), this is the exceptional case already verified. If \(d>5\), then \(j=5\) lies in the range of 6 . At this point the previous residual has already forced characteristic \(11\), so \(120\) is invertible. By the cleared \(R_5\) identity of 4, the corresponding recurrence coefficient is \[-\frac{(d-2)(d-1)(3d^3+3d^2-32d-72)}{120},\] which becomes \(9\in\mathbb{F}_{11}\) after substituting \(d\equiv5\pmod{11}\). This contradiction proves that \(d=5\) is the only remaining case. ◻

Proof. The set \(\mu_d\) is \(H\). For each \(x\in H\), the equation \(1+x+y=0\) determines \(y=-1-x\). Therefore the intersection count in 6 is exactly \(N_e(Q)\). ◻

Proof. The subgroup indicator is \[1_H(x)=\frac{1}{e}\sum_{\chi\in\mathcal{X}_e}\chi(x)\] for every \(x\in\mathbb{F}_Q\), because all characters in the sum vanish at \(0\). Hence \[N_e(Q)=\sum_{x\in\mathbb{F}_Q}1_H(x)1_H(-1-x) =\frac{1}{e^2}\sum_{\chi,\psi\in\mathcal{X}_e}J_-(\chi,\psi).\] The main term is \(J_-(\varepsilon,\varepsilon)=Q-2\). If exactly one of \(\chi, \psi\) is trivial, the corresponding sum has absolute value at most \(1\). If both are nontrivial and \(\chi\psi=\varepsilon\), the corresponding Jacobi sum has absolute value \(1\). If \(\chi\), \(\psi\), and \(\chi\psi\) are all nontrivial, the standard Jacobi-sum bound gives absolute value \(\sqrt Q\); see [9], [10], or Storer’s classical cyclotomy reference [11]. Since \(1\le\sqrt Q\), every non-main term has absolute value at most \(\sqrt Q\), and there are \(e^2-1\) such terms. This proves the estimate. ◻

Proof. The case \(e=1\) is immediate: \(H=\mathbb{F}_Q^*\), and the only excluded \(x\) are \(0\) and \(-1\).

For \(e=2\), the subgroup \(H\) is the square subgroup and \[1_H(x)=\frac{\varepsilon(x)+\chi(x)}{2},\] where \(\varepsilon\) is the trivial multiplicative character extended by \(\varepsilon(0)=0\). Hence \[N_2(Q)=\frac{1}{4}\sum_{x\in\mathbb{F}_Q} (\varepsilon(x)+\chi(x))(\varepsilon(-1-x)+\chi(-1-x)).\] The four sums are \[\sum_x\varepsilon(x)\varepsilon(-1-x)=Q-2,\] \[\sum_x\varepsilon(x)\chi(-1-x)=\sum_x\chi(-1-x)-\chi(-1)= -\chi(-1),\] \[\sum_x\chi(x)\varepsilon(-1-x)=\sum_x\chi(x)-\chi(-1)= -\chi(-1),\] and \[\sum_x\chi(x)\chi(-1-x)=\chi(-1)\sum_x\chi(x(1+x))=-\chi(-1),\] because \(x(1+x)\) is a nonconstant quadratic polynomial with distinct roots. Adding gives the formula for \(N_2(Q)\), and the degree formula follows from 5. ◻

Proof. This is immediate from 4. If \(Q\equiv1\pmod4\), then \(\deg h_d=(Q+3)/4\), which equals \(2\) only for \(Q=5\) and never equals \(0\) or \(1\) for an odd prime power. If \(Q\equiv3\pmod4\), then \(\deg h_d=(Q-3)/4\), which equals \(0,1,2\) for \(Q=3,7,11\), respectively. ◻

Proof. The indicator of the subgroup of cubes is \[I(x)=\frac{\varepsilon(x)+\chi(x)+\chi^2(x)}{3},\] with all characters extended by zero at \(0\). Thus \[N_3(Q)=\frac{1}{9}\sum_{i,j=0}^2\sum_x\chi^i(x)\chi^j(-1-x).\] Substitute \(x=-t\). Since \(\chi(-1)=1\) for cubic characters, the inner sums are Jacobi sums \(J(\chi^i,\chi^j)\). The \((i,j)=(0,0)\) term is \(Q-2\). The four terms with exactly one of \(i,j\) equal to \(0\) are all \(-1\). The two mixed nontrivial terms \(J(\chi,\chi^2)\) and \(J(\chi^2,\chi)\) are also \(-1\). The remaining terms are \(J(\chi,\chi)\) and \(J(\chi^2,\chi^2)\). This gives the formula. The bound follows from \(|J(\alpha,\beta)|=\sqrt Q\) when \(\alpha\), \(\beta\), and \(\alpha\beta\) are nontrivial. ◻

Proof. By 5 10, \[\begin{align} \deg h_d &=\frac{Q-1}{e}-N_e(Q) \\ &\ge \frac{Q-1}{e}-\frac{Q-2}{e^2}-\frac{e^2-1}{e^2}\sqrt Q \\ &=\frac{(e-1)Q+(2-e)-(e^2-1)\sqrt Q}{e^2}. \end{align}\] If \(\deg h_d\le M\), then \[(e-1)T^2-(e^2-1)T+(2-e-e^2M)\le0.\] Solving this quadratic inequality for \(T\ge0\) gives the displayed bound. ◻

Proof. For (i), \(P_1=\sigma-1\). Its kernel on \(\Gamma_q\) is \(K\cap\Gamma_q\), which is the kernel of multiplication by \(3\) on \(K\). Hence the kernel is zero if and only if \(p\ne3\).

For (ii), if \(q\equiv0\pmod3\), then \(1\in\Lambda_q\cap\mu_3\). If \(q\equiv1\pmod3\), a primitive cube root \(\omega\in K\) satisfies \(\omega^{q+1}+\omega+1=\omega^2+\omega+1=0\), so \(\omega\in\Lambda_q\cap\mu_3\). If \(q\equiv2\pmod3\) and \(\lambda\in\Lambda_q\cap\mu_3\), then \(\lambda^{q+1}=1\), so \(1+\lambda+1=0\), hence \(\lambda=-2\). This is incompatible with \(\lambda^3=1\) unless \(p=3\), which is excluded by \(q\equiv2\pmod3\).

For (iii), \(\mu_d=\mathbb{F}_{2^a}^*\). Let \(A:z\mapsto z^{2^k}\) on \(\mathbb{F}_{2^a}\) and let \[\tau(z)=1+z^{-1}\] in characteristic two. The maps \(A\) and \(\tau\) commute, and \(\tau^3=1\) on \(\mathbb{P}^1\). The automorphism \(A\) has order \[n=\frac{a}{\gcd(a,k)}=\frac{a}{g}\] on \(\mathbb{F}_{2^a}\). A denominator point is exactly a solution of \[A(z)=\tau(z),\qquad z\in\mathbb{F}_{2^a}^*.\]

Assume first that \(3\nmid n\). If \(A(z)=\tau(z)\), then \[z=A^n(z)=\tau^n(z),\] so \(z\) is fixed by \(\tau\) or by \(\tau^2\). In characteristic two the two fixed-point equations are the same: \[z^2+z+1=0.\] Thus any solution lies in \(\mathbb{F}_4\setminus\mathbb{F}_2\). Such points lie in \(\mathbb{F}_{2^a}\) if and only if \(a\) is even, and on them \(\tau\) acts trivially. Hence the equation \(A(z)=\tau(z)\) is solvable in this case if and only if \(A\) also acts trivially on \(\mathbb{F}_4\), i.e. if and only if \(k\) is even. Therefore, when \(3\nmid n\), denominator points exist exactly when \(a\) and \(k\) are both even, equivalently exactly when \(g\) is even.

Now assume \(3\mid n\). Put \(Q=2^g\) and write \(a=gn\), \(k=gk'\) with \(\gcd(k',n)=1\). Then \(3\nmid k'\) and \(\mathbb{F}_{Q^3}\subseteq\mathbb{F}_{2^a}\). On \(\mathbb{F}_{Q^3}\) the map \(A\) is the \(k'\)-th power of the \(Q\)-Frobenius. If \(k'\equiv1\pmod3\), choose a root of \[z^Q=\tau(z), \qquad\text{equivalently}\qquad z^{Q+1}+z+1=0.\] This polynomial has degree \(Q+1\), is squarefree, and every root lies in \(\mathbb{F}_{Q^3}\) by iterating the equation three times. Thus a nonzero solution exists in \(\mathbb{F}_{Q^3}\subseteq\mathbb{F}_{2^a}\), and it satisfies \(A(z)=\tau(z)\). If \(k'\equiv2\pmod3\), choose instead a root of \[z^Q=\tau^2(z), \qquad\text{equivalently}\qquad z^{Q+1}+z^Q+1=0.\] The same squarefreeness and iteration argument gives \(Q+1\) roots in \(\mathbb{F}_{Q^3}\), and for such a root \(A(z)=z^{Q^2}=\tau(z)\). Hence denominator points exist whenever \(3\mid n\).

Combining the two cases, \(\mu_d\cap\Lambda_q\) is empty exactly when \(3\nmid n\) and \(g\) is odd, as claimed. ◻

Proof. By 7, the \(p\)-power part of \(d\) does not affect the permutation property. Thus assume \(p\nmid d\). By 8, the only degree-one quotient cases are \(d=1\), \(d=3\), and \(d=p^a-1\). The descent criterion 2 requires the fiber condition \(\gcd(d,q-1)=1\), the denominator condition \(\mu_d\cap\Lambda_q=\varnothing\), and quotient bijectivity.

For \(d=1\), this is exactly 5(i). For \(d=3\), the standing hypothesis gives \(p\ne3\). If \(P_3\) permutes \(\Gamma_q\), then the denominator condition in 2 holds, and 5(ii) gives \(q\equiv2\pmod3\). Conversely, if \(q\equiv2\pmod3\), then \(\gcd(3,q-1)=1\), and 5(ii) gives \(\mu_3\cap\Lambda_q=\varnothing\). The reduced quotient is the degree-one map \[h_3(z)=-\frac{2z+1}{z-1}\] from the proof of 8; it is defined over \(\mathbb{F}_p\) and commutes with \(\tau\) by 3. Therefore, for every \(z\in\Lambda_q\), \[h_3(z)^q=h_3(z^q)=h_3(\tau z)=\tau(h_3(z)),\] so \(h_3\) preserves \(\Lambda_q\). Being an automorphism of \(\mathbb{P}^1\), it restricts to a bijection of this finite set. Thus all three conditions of 2 hold exactly when \(q\equiv2\pmod3\).

For \(d=p^a-1\), the fiber condition \(\gcd(d,q-1)=1\) is restrictive. If \(p\) is odd, then \(p^a-1\) and \(p^k-1\) are both even, so it fails. Hence \(p=2\). In that case \[\gcd(2^a-1,2^k-1)=2^{\gcd(a,k)}-1,\] so the fiber condition is \(\gcd(a,k)=1\). Under this condition, 5(iii) gives the denominator condition \(3\nmid a\). The reduced quotient map is \(h_d=\tau^2\), an automorphism of the twisted quotient set, so the quotient condition is automatic. This proves the classification. ◻

Proof. Under the Mersenne hypotheses, 8 gives \(h_d=\tau^2\), and \(\tau\) has order three. Hence the inverse quotient map is \(\tau\). Thus if \(w_y=y^{q-1}\) and \(z=\tau(w_y)\), a lift over the desired quotient point is obtained by solving the \(K\)-linear equation \(\sigma(u)=zu\). Such a nonzero solution exists by multiplicative Hilbert 90, equivalently because \(z\in\Lambda_q\) has norm one; the solution space is a one-dimensional \(K\)-vector space. For any nonzero solution, \[u+\sigma(u)+\sigma^2(u)=u(1+z+zz^q)=0,\] so \(u\in\Gamma_q^*\) and \(u^{q-1}=z\). Since \(\lambda(P_d(u))=h_d(z)=w_y=\lambda(y)\), the ratio \(y/P_d(u)\) lies in \(K^*=\mathbb{F}_q^*\). If \(q=2\), then \(K^*\) is trivial, so this ratio is \(1\) and \(c=1\). If \(q>2\), multiplication by \(c\in K^*\) on the \(K^*\)-fiber above \(z\) changes \(P_d(u)\) by the factor \(c^d\). Because \(\gcd(d,q-1)=1\), exponentiation by \(d\) is invertible on \(K^*\), with any inverse exponent \(e\) satisfying \(ed\equiv1\pmod{q-1}\). This gives the stated formula.

It remains only to check independence of the Hilbert–90 lift. If \(u'=tu\) with \(t\in K^*\), then \(P_d(u')=t^dP_d(u)\). For \(q>2\) the corresponding scalar is \[c'=\left(\frac{y}{P_d(u')}\right)^e =t^{-de}c=t^{-1}c,\] because \(ed\equiv1\pmod{q-1}\) on \(K^*\). Hence \(c'u'=cu\). For \(q=2\) there is no nontrivial choice of \(t\in K^*\), so the same conclusion is automatic. ◻

Proof. Put \[A=(1-u)^d, B=(1-\omega u)^d, C=(1-\omega^2u)^d.\] Since \[M(u)=\omega\frac{1-\omega u}{1-u}, \qquad M(\omega u)=\omega\frac{1-\omega^2u}{1-\omega u},\] we have \[H_d^{\rm raw}(M(u)) =\omega^d\frac{\omega^dC-B}{\omega^dB-A}.\] Applying \(M^{-1}(z)=(z-\omega)/(z-\omega^2)\) gives \[\Phi_d(u) =\frac{\omega A+\omega^{d+2}B+\omega^{2d}C}{\omega^2A+\omega^{d+1}B+\omega^{2d}C}.\] The coefficient of \(u^j\) in the numerator is \[(-1)^j\binom dj \bigl(\omega+\omega^{d+2+j}+\omega^{2d+2j}\bigr).\] The root-of-unity factor in parentheses is nonzero precisely when \[j\equiv 2-d\pmod3,\] and in that case it equals \(3\omega\). It is zero for the other two residue classes. Thus the numerator is the corresponding root-of-unity projection of the binomial expansion; in positive characteristic some projected terms may still vanish because the binomial coefficient itself vanishes in \(k\). Similarly, the root-of-unity factor in the denominator is nonzero precisely when \[j\equiv1-d\pmod3,\] and in that case it equals \(3\omega^2\), with the same possible further vanishing of binomial coefficients. Dividing the two filtered sums gives the displayed formula. ◻

Proof. Part (i) is the final computation in 8.

For (ii), since \(d=r+1\), the Frobenius identity gives \[H_{r+1}^{\rm raw}(z)=\frac{z^r+z+1}{z^{r+1}-1}.\] If \(r\equiv1\pmod3\), then \(M(u)^r=M(u^r)\); if \(r\equiv2\pmod3\), then \(M(u)^r=M(u^{-r})\). For arbitrary \(u,v\) a direct substitution gives \[M^{-1}\left( \frac{M(v)+M(u)+1}{M(u)M(v)-1} \right)=\frac{\omega^2}{uv}.\] Taking \(v=u^r\) in the first case and \(v=u^{-r}\) in the second gives the two formulas.

For (iii), in characteristic \(3\) the coordinate \(u=1/(z-1)\) satisfies \[u(\tau z)=u(z)+1.\] Again \[H_{r+1}^{\rm raw}(z)=\frac{z^r+z+1}{z^{r+1}-1}.\] Writing \(z=1+u^{-1}\) and using \(r=3^a\), we get \[z^r=1+u^{-r},\] so \[H_{r+1}^{\rm raw}(z) =\frac{u^{-r}+u^{-1}}{u^{-1}+u^{-r}+u^{-r-1}} =\frac{u^r+u}{u^r+u+1}.\] Therefore \[\Phi_{r+1}(u)=\frac{1}{H_{r+1}^{\rm raw}(z)-1}=-(u^r+u+1).\] ◻

Proof. For \(p\ne3\), 13 conjugates the map to \(u\mapsto \omega^2u^m\). Since \(a\ge1\), the integer \(m\) is prime to \(p\), so this is a separable Kummer cover of degree \(|m|\) and is Galois with cyclic group generated by multiplication of \(u\) by an \(|m|\)-th root of unity.

For \(p=3\), the equation \(t=-(u^{3^a}+u+1)\) is equivalent to \[u^{3^a}+u=-t-1.\] The additive polynomial \(X^{3^a}+X\) is separable and has \(3^a\) roots. Translation by any root preserves the cover, and these translations act simply transitively on a geometric generic fiber. Hence the geometric monodromy group is the elementary abelian group \(\ker(X^{3^a}+X)\). ◻

Proof. For \(p\ne3\), if \(q\equiv1\pmod3\), then Frobenius fixes \(\omega\) and \[M(u)^q=M(u^q).\] The equation \(z^q=\tau(z)\) becomes \(M(u^q)=M(\omega u)\), hence \(u^q=\omega u\) on \(\mathbb{P}^1\). This gives the first line. If \(q\equiv2\pmod3\), then Frobenius swaps \(\omega\) and \(\omega^2\) and \[M(u)^q=M(u^{-q}).\] Thus \(M(u^{-q})=M(\omega u)\), equivalently \(u^{q+1}=\omega^2\); neither \(0\) nor \(\infty\) satisfies this equation.

For \(p=3\), the coordinate conjugates \(\tau\) to \(u\mapsto u+1\) and commutes with \(q\)-Frobenius. Hence \(z^q=\tau(z)\) becomes \(u^q=u+1\) for finite \(u\). The point \(u=\infty\) is fixed by both Frobenius and translation, and corresponds to the fixed point \(z=1\) of \(\tau\). ◻

Proof. Assume first that \(p\ne3\). By 13, the map is conjugate to \(u\mapsto \omega^2u^m\), with \[m=-(p^a+1)\quad\text{if }p^a\equiv1\pmod3, \qquad m=p^a-1\quad\text{if }p^a\equiv2\pmod3.\] The congruence \(m\equiv1\pmod3\) holds in both cases. Therefore the monomial preserves the cosets described in 6. On the nonzero coset it is bijective exactly when \(m\) is invertible modulo the size of the underlying cyclic group. Thus, if \(q\equiv1\pmod3\), the condition is \(\gcd(|m|,q-1)=1\), while if \(q\equiv2\pmod3\), the condition is \(\gcd(|m|,q+1)=1\).

For \(p=2\) this is precisely the stated criterion, since \(|m|=2^a+1\) when \(a\) is even and \(|m|=2^a-1\) when \(a\) is odd, while \(q=2^k\) satisfies \(q\equiv1\pmod3\) for \(k\) even and \(q\equiv2\pmod3\) for \(k\) odd. If \(k\) is odd and \(a\) is even, a common prime divisor of \(2^a+1\) and \(2^k+1\) would have the order of \(2\) dividing both \(2a\) and \(2k\) but neither \(a\) nor \(k\); this is impossible because \(a\) is even and \(k\) is odd. If \(k\) is odd and \(a\) is odd, a common divisor of \(2^a-1\) and \(2^k+1\) would have odd order dividing \(a\) and also dividing \(2k\) but not \(k\), again impossible. Hence every odd \(k\) satisfies the criterion.

If \(p\) is odd and \(p\ne3\), then \(|m|\) is even. But \(q-1\) is even when \(q\equiv1\pmod3\), and \(q+1\) is even when \(q\equiv2\pmod3\). The required gcd is therefore never \(1\).

It remains to treat \(p=3\). By 13 6, the finite part of \(\Lambda_{3^k}\) is an affine coset \(u_0+\mathbb{F}_{3^k}\) with \(u_0^{3^k}-u_0=1\), and the point \(\infty\) is fixed. Since \(h_{3^a+1}\) is defined over \(\mathbb{F}_3\) and commutes with \(\tau\), it preserves \(\Lambda_{3^k}\); equivalently, the following affine formula is a self-map of the coset \(u_0+\mathbb{F}_{3^k}\). On this coset, \[u_0+x\longmapsto -(u_0^{3^a}+u_0+1)-(x^{3^a}+x), \qquad x\in\mathbb{F}_{3^k}.\] Thus the induced map is bijective if and only if the \(\mathbb{F}_3\)-linear part \[x\longmapsto x^{3^a}+x\] has zero kernel on \(\mathbb{F}_{3^k}\). A nonzero kernel element satisfies \(x^{3^a-1}=-1\). Let \(g=\gcd(a,k)\). Since \[\gcd(3^a-1,3^k-1)=3^g-1,\] the equation \(x^{3^a-1}=-1\) is solvable in \(\mathbb{F}_{3^k}^*\) if and only if \(3^g-1\) divides \((3^k-1)/2\). This happens if and only if \[\frac{3^k-1}{3^g-1}=1+3^g+\cdots+3^{k-g}\] is even, equivalently if and only if \(k/g\) is even. Therefore bijectivity holds exactly when \(k/g\) is odd. ◻

Proof. The fiber condition in 2 is exactly \(\gcd(d,q-1)=1\), so it is necessary. Assume it holds.

If \(a\) is even, then 13 gives a monomial of degree \(d\), so the torsion defect is zero and hence \(\mu_d\cap\Lambda_q=\varnothing\). The quotient condition follows from 14: for \(k\) even it is the same gcd condition, and for \(k\) odd the gcd \(\gcd(2^a+1,2^k+1)\) is automatically \(1\) by the order argument used in the proof of 14.

If \(a\) is odd, then \(d\) is divisible by \(3\). The fiber condition forces \(k\) to be odd, since \(2^k-1\) is divisible by \(3\) for \(k\) even. For odd \(k\), 14 gives the quotient condition. The normal form in 13 has degree \(2^a-1=d-2\), so the cancellation degree is \(2\). The primitive cube roots, equivalently the two roots of \(z^2+z+1=0\), lie in \(\mu_d\) and are fixed by \(\tau\); hence they are canceled denominator points. Conversely, if \(z\in\mu_d\cap\Lambda_q\), then \(z^q=\tau(z)\) and therefore \(\tau(z)\in\mu_d\), so \(z\) is a canceled denominator point. Since there are exactly two cancellations, these are all the possible denominator points. In characteristic two they are the two nontrivial elements of \(\mathbb{F}_4^*\). For odd \(k\) Frobenius interchanges them, so they do not satisfy \(z^q=\tau(z)=z\). Thus \(\mu_d\cap\Lambda_q=\varnothing\). The descent criterion now proves sufficiency. ◻

Proof. By the torsion-defect formula, the only issue is to count pairs \((x,y)\) of \(d\)-th roots of unity with \(1+x+y=0\). These equations involve only \(d\)-th roots of unity and hence are defined over a cyclotomic field; choosing an embedding of that cyclotomic field into \(\mathbb{C}\) preserves algebraic existence and nonexistence. We may therefore count complex roots of unity. Since \(|x|=|y|=1\) and \(y=-1-x\), the equality \(|1+x|=1\) gives \[2+x+\bar x=1, \qquad\text{hence}\qquad \operatorname{Re}(x)=-\frac{1}{2}.\] Thus \(x\) is one of the two primitive third roots of unity, and then \(y\) is the other one. Hence the defect is \(2\) if \(3\mid d\) and is \(0\) otherwise. Applying 6 gives the degree formula. The fixed-degree classification follows immediately by solving \(m=d\) with \(3\nmid d\) and \(m=d-2\) with \(3\mid d\). ◻

Proof. First \(x=0\) and \(x=1\) are not critical points in the stated range. The raw denominator is nonzero at both points, so the following local expansions are expansions of the reduced map. At \(x=0\), \[K_d^{\rm raw}(x)=-\varepsilon-\varepsilon d\,x+O(x^2),\] whose linear coefficient is nonzero because \(p\nmid d\). At \(x=1\), put \(u=1-x\). If \(\varepsilon=1\), then \[K_d^{\rm raw}(x)=d\,u+O(u^2),\] and if \(\varepsilon=-1\), then \[K_d^{\rm raw}(x)=-2-d\,u+O(u^2),\] where in the second case \(2\ne0\) under the hypothesis \(p\nmid d(d-1)\). In both cases the first nonconstant coefficient is nonzero. Thus any critical point satisfying the hypotheses has \(x\notin\{0,1\}\).

At an uncanceled non-pole point the equation \(k_d(x)=T\) is equivalently the raw equation \(K_d^{\rm raw}(x)=T\), namely \[(T+1)x^d-T(1-x)^d=\varepsilon.\] At a critical point over this value, differentiating with respect to \(x\) and using \(p\nmid d\) gives \[(T+1)x^{d-1}+T(1-x)^{d-1}=0.\] Write \(A=x^m\) and \(B=(1-x)^m\). Since \(x\notin\{0,1\}\), both \(A\) and \(B\) are nonzero. The derivative equation gives \[T(A+B)+A=0.\] If \(A+B=0\), then this equation gives \(A=0\), a contradiction. Hence \(A+B\ne0\), so \(T=-A/(A+B)\) and \(T+1=B/(A+B)\). Substituting into the equation for \(K_d^{\rm raw}(x)=T\) gives \[\frac{B}{A+B}\,xA+\frac{A}{A+B}\,(1-x)B=\frac{AB}{A+B}=\varepsilon.\] Thus \(AB=\varepsilon(A+B)\), which is ?? . The formulas ?? follow by replacing \(A+B\) with \(AB/\varepsilon\) in \(T=-A/(A+B)\) and \(T+1=B/(A+B)\). ◻

Proof. By 8, equality of critical values gives \[(1-y)^m=(1-x)^m.\] Hence \(1-y=\zeta(1-x)\) for some \(\zeta\in\mu_m\). Put \(s=1-x\), so \(y=1-\zeta s\). Since \(x\) and \(y\) are both critical, we have \[s^m(1-s)^m=\varepsilon\bigl(s^m+(1-s)^m\bigr)\] and \[s^m(1-\zeta s)^m=\varepsilon\bigl(s^m+(1-\zeta s)^m\bigr).\] Subtracting gives \[\bigl((1-s)^m-(1-\zeta s)^m\bigr)(s^m-\varepsilon)=0.\] The alternative \(s^m=\varepsilon\) is impossible, because substituting it in the critical equation gives \(0=1\). Therefore \[(1-\zeta s)^m=(1-s)^m,\] so \(1-\zeta s=\eta(1-s)\) for some \(\eta\in\mu_m\). The equality \(x\ne y\) gives \(\zeta\ne1\). If \(\eta=1\), then \(1-\zeta s=1-s\), so \(s=0\), contradicting the critical equation. If \(\eta=\zeta\), then \(1-\zeta s=\zeta(1-s)\), so \(\zeta=1\), again a contradiction. Hence \(\eta\ne1\) and \(\eta\ne\zeta\). Solving gives \[s=\frac{\eta-1}{\eta-\zeta}.\] Now \[1-s=\frac{1-\zeta}{\eta-\zeta}.\] Dividing the critical equation \[s^m(1-s)^m=\varepsilon\bigl(s^m+(1-s)^m\bigr)\] by \(s^m(1-s)^m\) gives \[\frac{1}{s^m}+\frac{1}{(1-s)^m}=\varepsilon,\] which is exactly 10 . For the converse, the cross-ratio equation is exactly the critical equation after substituting the displayed value of \(s\), and the same root-of-unity relations give equality of the two critical values. Reversing the argument is valid on the open locus where the reduced map is defined and the raw denominator has not become an uncanceled pole. The stated proviso excludes precisely the pole and canceled-denominator cases that are outside the critical-point locus of the theorem. ◻

Proof. The cross-ratio equation and the nondegeneracy conditions involve only \(m\)-th roots of unity and rational operations, so they are defined over a cyclotomic field. Choosing an embedding of that cyclotomic field into \(\mathbb{C}\) preserves the existence of such a solution; hence it suffices to work over \(\mathbb{C}\). Suppose, to the contrary, that a nondegenerate solution exists. The equation is symmetric in \(\zeta\) and \(\eta\), so write \[\zeta=e^{2\pi i r/m},\qquad \eta=e^{2\pi i s/m}, \qquad 0<r<s<m.\] Put \[i=s-r, \qquad j=r, \qquad k=m-s, \qquad \theta=\frac{\pi}{m}, \qquad S_a=\sin(a\theta).\] Then \(i,j,k\) are positive and \(i+j+k=m\). The elementary identity \[e^{2iu}-e^{2iv}=2ie^{i(u+v)}\sin(u-v)\] gives \[\left(\frac{\eta-\zeta}{\eta-1}\right)^m =(-1)^j\left(\frac{S_i}{S_k}\right)^m, \qquad \left(\frac{\eta-\zeta}{1-\zeta}\right)^m =(-1)^k\left(\frac{S_i}{S_j}\right)^m.\] Since \(\varepsilon=(-1)^d=(-1)^{m+1}\), the cross-ratio equation is equivalent, after multiplying by \(S_j^mS_k^m\), to \[\label{eq:charzero-trig-collision} (-1)^j(S_iS_j)^m +(-1)^k(S_iS_k)^m +(-1)^m(S_jS_k)^m=0.\tag{11}\] If \(i,j,k\) have the same parity, then all three signs in 11 are equal, which is impossible. Otherwise exactly one of \(i,j,k\) has parity different from the other two. Let \(a\) be this parity outlier and let \(b,c\) be the remaining two indices. Then 11 is equivalent to \[\label{eq:outlier-sine-equation} 1=\left(\frac{S_a}{S_b}\right)^m +\left(\frac{S_a}{S_c}\right)^m .\tag{12}\] Indeed, the term not involving the outlier has the sign opposite to the other two terms.

Equation 12 forces \(S_a<S_b\) and \(S_a<S_c\). Since \(a+b<m\) and \(a+c<m\), the sign of \(S_u-S_v\) agrees with the sign of \(u-v\) for each pair \((u,v)=(a,b),(a,c)\); this follows from \[\sin(u\theta)-\sin(v\theta) =2\cos\left(\frac{(u+v)\theta}{2}\right) \sin\left(\frac{(u-v)\theta}{2}\right),\] with the cosine positive. Hence \(a<b\) and \(a<c\). Thus \(b,c\ge a+1\) and \[m=a+b+c\ge 3a+2, \qquad\text{so}\qquad a\le \frac{m-2}{3}.\] Moreover \(S_b,S_c\ge S_{a+1}\), because on the interval \(a+1\le u\le m-a-1\) the sine \(\sin(u\theta)\) is minimized at the endpoints, where both endpoint values equal \(S_{a+1}\). Therefore \[\label{eq:sine-ratio-bound} 1 \le 2\left(\frac{S_a}{S_{a+1}}\right)^m .\tag{13}\] For \(m\ge5\) we have \[(a+1)\theta\le \frac{m+1}{3m}\pi\le \frac{2\pi}{5}.\] Since \(\cot t\) is decreasing and positive on \((0,2\pi/5]\), \[\begin{align} m\log\frac{S_{a+1}}{S_a} &=m\int_{a\theta}^{(a+1)\theta}\cot t\,dt \\ &\ge \pi\cot((a+1)\theta) \ge \pi\cot\frac{2\pi}{5} >\log2 . \end{align}\] The final strict inequality is explicit: \(\cot(2\pi/5)=1/\sqrt{5+2\sqrt5}\), so \(\pi\cot(2\pi/5)>1>\log2\). Thus \(2(S_a/S_{a+1})^m<1\), contradicting 13 . For \(m<5\), there is nothing to check for \(m=1,2\); for \(m=3\) the only triple is \((1,1,1)\), where all signs agree; and for \(m=4\) the only triples are permutations of \((2,1,1)\), where the parity outlier has the larger sine, already contradicting 12 . Hence no nondegenerate solution exists.

The final assertion follows immediately from 16: any equality of critical values for two distinct finite critical points in the stated open locus would produce such a nondegenerate solution. ◻

Proof. The first identity is the direct calculation \[\begin{align} \frac{1}{d}(N_d'D_d-N_dD_d') &=\bigl((-1)^d(z+1)^{d-1}-z^{d-1}\bigr)(z^d-1) \\ &\quad-\bigl((-1)^d(z+1)^d-z^d\bigr)z^{d-1} \\ &=z^{d-1}-(-1)^d(z+1)^{d-1}(z^{d-1}+1). \end{align}\] The second identity follows by writing \(N_d=C_dn_d\) and \(D_d=C_dd_d\) and expanding; the terms involving \(C_d'\) cancel. ◻

Proof. If \(d=1\), the nonconstancy hypothesis excludes the characteristic-\(3\) constant quotient; the reduced quotient is then a degree-one map, so there is no ramification to check. Hence assume \(d\ge2\). Under the positive-characteristic tame hypothesis \(p\nmid d(d-1)\), this also forces \(p\ne2\), since one of \(d\) and \(d-1\) is even. Thus the later canceled-point and infinity calculations never require a hidden division by \(2\) in characteristic \(2\). Let \(m=d-1\) and put \(\epsilon=(-1)^d\). Suppose first that \(\alpha\) is a multiple zero of the unreduced skeleton \[F_d(z)=z^m-\epsilon(z+1)^m(z^m+1)\] which is not a canceled denominator point. Since \(F_d(0)\) and \(F_d(-1)\) are nonzero, we have \(\alpha(\alpha+1)\ne0\). The equations \(F_d(\alpha)=0\) and \(F_d'(\alpha)=0\) give \[\alpha^m=\epsilon(\alpha+1)^m(\alpha^m+1)\] and, because \(p\nmid m\), \[\alpha^{m-1} =\epsilon(\alpha+1)^{m-1}\bigl(2\alpha^m+\alpha^{m-1}+1\bigr).\] Dividing the second equation by the first gives \[\frac{1}{\alpha} =\frac{2\alpha^m+\alpha^{m-1}+1}{(\alpha+1)(\alpha^m+1)}.\] After clearing denominators, \[(\alpha+1)(\alpha^m+1)=\alpha(2\alpha^m+\alpha^{m-1}+1),\] hence \[\alpha^{m+1}=1,\] that is \(\alpha^d=1\). Since \(D_d'(\alpha)=d\alpha^{d-1}\ne0\), the equality \(N_d'D_d-N_dD_d'=0\) at \(\alpha\) forces \(N_d(\alpha)=0\). Thus \(\alpha\) is a common zero of \(N_d\) and \(D_d\), contrary to the assumption. Hence, away from the cancellation locus, the reduced Wronskian has no multiple finite zeros.

It remains to check that the cancellation points themselves do not become critical after reduction. Let \(\alpha\) be a common zero of \(N_d\) and \(D_d\). Then \(\alpha^d=1\) and \(\tau(\alpha)^d=1\). Since the cancellation is simple, write \(u=z-\alpha\) and expand \[N_d=a_1u+a_2u^2+O(u^3), \qquad D_d=b_1u+b_2u^2+O(u^3),\] with \(a_1=N_d'(\alpha)\), \(b_1=D_d'(\alpha)\), \(a_2=N_d''(\alpha)/2\), and \(b_2=D_d''(\alpha)/2\). The derivative at \(\alpha\) of the quotient after canceling \(u\) is \[\frac{a_2b_1-a_1b_2}{b_1^2} =\frac{N_d''(\alpha)D_d'(\alpha)-N_d'(\alpha)D_d''(\alpha)}{2D_d'(\alpha)^2}.\] Since \(2\) is invertible in the present cases, the derivative of the reduced quotient at \(\alpha\) is nonzero if and only if \[N_d''(\alpha)D_d'(\alpha)-N_d'(\alpha)D_d''(\alpha)\ne0.\] Using \(\alpha^d=1\) and \((-1)^d(\alpha+1)^d=1\), one computes \[N_d'(\alpha)=d\left(\frac{1}{\alpha+1}-\frac{1}{\alpha}\right), \qquad D_d'(\alpha)=\frac{d}{\alpha},\] \[N_d''(\alpha)=d(d-1)\left(\frac{1}{(\alpha+1)^2}-\frac{1}{\alpha^2}\right), \qquad D_d''(\alpha)=\frac{d(d-1)}{\alpha^2}.\] Therefore \[N_d''(\alpha)D_d'(\alpha)-N_d'(\alpha)D_d''(\alpha) =-\frac{d^2(d-1)}{\alpha^2(\alpha+1)^2},\] which is nonzero under the tame hypothesis. Thus no canceled point is ramified in the reduced map.

Every finite pole of \(h_d\) is a simple pole, since \(D_d=z^d-1\) is squarefree and cancellation removes only simple factors. Hence finite poles are unramified over the branch value \(\infty\). Finally, in the local coordinate \(w=1/z\) at infinity, if \(d\) is even then \(H_d^{\rm raw}=dw+O(w^2)\), and if \(d\) is odd and the characteristic is not \(2\) then \(H_d^{\rm raw}=-2-dw+O(w^2)\). Thus infinity is also unramified. This proves the theorem. ◻

Proof. By 18, every finite critical point is simple, and no pole or point over infinity is ramified. By the Morse hypothesis, the finite critical values are pairwise distinct. Thus every finite branch cycle is a transposition. The cover \(\mathbb{P}^1\to\mathbb{P}^1\) is connected, so its geometric monodromy group is transitive. A transitive subgroup of \(S_n\) generated by transpositions is \(S_n\): the graph with vertices the \(n\) sheets and edges the transpositions is connected, and the group generated by edge transpositions of a connected graph is the full symmetric group. Hence \(G_{h_d}=S_n\). ◻

Proof. By 15, the displayed degree formula is \(n=d\) when \(3\nmid d\) and \(n=d-2\) when \(3\mid d\). The degree-one cases are exactly \(d=1,3\), so assume \(n>1\).

Work in the coordinate \(x=z/(z+1)\). The point \(x=\infty\), which corresponds to \(z=-1\), is unramified: if \(d\) is even, then it is a simple pole of the raw expression \(K_d^{\rm raw}\) and hence of the reduced map; if \(d\) is odd, writing \(t=1/x\) gives \[K_d^{\rm raw}=\frac{-t^d-1}{1+(1-t)^d} =-\frac{1}{2}-\frac{d}{4}t+O(t^2),\] with the evident harmless modification in the excluded case \(d=1\). The point \(x=1\), which corresponds to \(z=\infty\), is unramified by the infinity calculation in 18. All remaining finite points of the \(x\)-line correspond to finite \(z\)-points, so 18 gives simple finite critical points after removing cancellations and shows that finite poles and canceled denominator points are unramified. By 17, no two finite non-pole critical points in this affine locus have the same critical value.

Consequently all ramification of \(h_d\) occurs at simple finite critical points, these ramified points have pairwise distinct branch values, and there is no ramification at poles, canceled points, \(x=1\), or \(x=\infty\). Equivalently, \(h_d\) is Morse in the sense of 4; each finite branch cycle is a single transposition, and no additional branch cycle arises over \(\infty\) or over a canceled point.

Now apply 19. Equivalently, the transposition branch cycles generate a transitive subgroup of \(S_n\) because the cover is connected; hence they generate the full symmetric group. ◻

Proof. Choose primitive coprime representatives \(n_d,d_d\in\mathbb{Z}[z]\) for the reduced characteristic-zero quotient, and put \(n=\deg h_d\). By 1, \(\deg d_d=n\). Put \[W_d=n_d'd_d-n_dd_d', \qquad B_d(T)=\operatorname{Res}_z(n_d(z)-Td_d(z),W_d(z))\in\mathbb{Z}[T].\] To record the local condition at infinity explicitly, set \[\widehat n_d(w)=w^n n_d(1/w),\qquad \widehat d_d(w)=w^n d_d(1/w),\] and define \[\iota_d=\widehat n_d'(0)\widehat d_d(0)-\widehat n_d(0)\widehat d_d'(0)\in\mathbb{Z}.\] The characteristic-zero calculation in 18 gives \(\iota_d\ne0\); it is the numerator of the local derivative of \(h_d\) at infinity in the source coordinate \(w=1/z\).

By 18 17, the characteristic-zero quotient is Morse, so the relevant resultants and discriminants below are nonzero. It is enough to take \(M_d\) divisible by \[\begin{align} &d(d-1)\,\iota_d\,\operatorname{Res}(n_d,d_d)\, \operatorname{disc}(d_d)\, \operatorname{Res}(W_d,d_d)\, \operatorname{disc}(W_d)\,\operatorname{disc}(B_d) \\ &\qquad\text{and by the nonzero leading coefficients of }n_d,d_d,W_d,B_d. \end{align}\] Factors corresponding to constant polynomials are omitted. For primes outside this finite set, reduction preserves the degree of the quotient, the numerator-denominator coprimality, the pole divisor, the simple finite critical divisor, the unramified local behavior at infinity, and the separation of finite critical values. More explicitly, the leading-coefficient factors preserve the degrees of \(n_d\), \(d_d\), \(W_d\), and \(B_d\); \(\operatorname{Res}(n_d,d_d)\) prevents a new numerator-denominator gcd and hence prevents new cancellation after reduction; \(\operatorname{disc}(d_d)\) and \(\operatorname{Res}(W_d,d_d)\) preserve simple unramified finite poles; \(\iota_d\) preserves the nonzero local derivative at infinity; \(\operatorname{disc}(W_d)\) preserves the reduced finite critical divisor; and \(\operatorname{disc}(B_d)\) preserves distinct finite critical values. The tame hypothesis \(p\nmid d(d-1)\) is also included in the definition of \(M_d\). Thus the reduced quotient modulo \(p\) remains Morse in the sense of 4, and 19 applies. ◻

Proof. The automorphism \(\alpha\) commutes with Frobenius because it is defined over \(\mathbb{F}_p\). Hence \(\Theta^r=\mathop{\mathrm{Frob}}_{q^r}\). If \(\lambda\) is an eigenvalue of \(\Theta\), then \(\lambda^r\) is an eigenvalue of \(\Theta^r\), hence of \(\mathop{\mathrm{Frob}}_{q^r}\). Deligne’s weight bound [12] gives \(|\lambda^r|\le q^{ri/2}\); for \(i=1\) this gives \(|\lambda|\le q^{1/2}\). ◻

Proof. Since \(f\) is defined over \(\mathbb{F}_p\) and satisfies \(f\circ\tau=\tau\circ f\), it also satisfies \(f\circ\Theta_q=\Theta_q\circ f\) for every \(q=p^k\); hence \(f\) preserves the twisted fixed sets, and \(U\), \(V\), and \(C_f\) are stable under \(\Theta_q\). The fixed points of \(\Theta_q\) on \(C_f\) are precisely the ordered off-diagonal pairs \((x,y)\) with \(x,y\in\Lambda_q\cap V\) and \(f(x)=f(y)\). We use the standard equivariant Grothendieck–Lefschetz trace formula for a finite-order automorphism composed with Frobenius. Here \(\tau\) is defined over \(\mathbb{F}_p\) and has finite order, so \(\Theta_q=\tau^{-1}\mathop{\mathrm{Frob}}_q\) is exactly such an endomorphism of the separated curve \(C_f\); see, for example, [13] or [14] for the trace formula in etale cohomology. This gives the first formula.

If \(\deg f=1\), then \(B=\varnothing\), \(U=V=\mathbb{P}^1\), and \(C_f=(\mathbb{P}^1\times_{\mathbb{P}^1}\mathbb{P}^1)\setminus\Delta=\varnothing\). The displayed formulas hold with \(N_f(q)=0\) and \(a_f(q)=0\). Hence assume \(\deg f>1\). By Riemann–Hurwitz the branch locus is nonempty, so \(U\) and \(V\) are nonempty affine curves. Since \(V\to U\) is finite etale, the off-diagonal fiber square \(C_f\) is a smooth affine curve over \(\overline{\mathbb{F}}_p\), possibly disconnected, with no proper zero-dimensional components. Hence \(H_c^i(C_f,\mathbb{Q}_\ell)=0\) for \(i\notin\{0,1,2\}\) and \(H_c^0=0\). The group \(H_c^2(C_f,\mathbb{Q}_\ell)\) is a direct sum of one copy of \(\mathbb{Q}_\ell(-1)\) for each geometric irreducible component of \(C_f\). Thus, after the Tate twist, the induced action is the permutation representation on \(\pi_0(C_{f,\overline{\mathbb{F}}_p})\). A component fixed by \(\Theta_q\) contributes \(q\) to the untwisted top-cohomology trace, and a component in a longer cycle contributes zero to the trace. By the definition of \(a_f(q)\), the top term is \(a_f(q)q\).

The remaining term is \(H_c^1\). Applying 7 with \(\alpha=\tau\) and \(r=3\) shows that every eigenvalue of \(\Theta_q\) on \(H_c^1\) has absolute value at most \(q^{1/2}\). The dimension of \(H_c^1\) depends only on \(f\). This gives the estimate. Choose \(Q_f\) so that the error term has absolute value less than \(q\) for all \(q>Q_f\). If \(a_f(q)>0\) and \(q>Q_f\), then \(N_f(q)>0\), so \(f\) has an off-diagonal collision on \(\Lambda_q\). If this happens for all sufficiently large \(q=p^k\), then only finitely many \(k\) can give bijectivity on \(\Lambda_q\), which is exactly the negation of \(\tau\)-twisted exceptionality. ◻

Proof. In the present rational-map setting, \(V_{\overline{\mathbb{F}}_p}\) is a dense open subset of \(\mathbb{P}^1_{\overline{\mathbb{F}}_p}\) and is therefore connected. For a connected finite etale cover \(V\to U\), the connected components of the fiber product \(V\times_UV\) over \(\overline{\mathbb{F}}_p\) correspond to the orbits of the geometric fundamental group of \(U\) on \(\Omega\times\Omega\). The diagonal component corresponds to the diagonal orbit. Removing it leaves the orbits on ordered distinct pairs. The image of the geometric fundamental group in \(\operatorname{Sym}(\Omega)\) is precisely \(G_f\), giving the indexing set. The assertion about \(H_c^2(1)\) follows because the top compactly supported cohomology of each smooth affine irreducible component is one-dimensional after Tate twist. ◻

Proof. The connected-cover hypothesis in 10 applies because \(V\) is a dense open subset of \(\mathbb{P}^1\). If \(G_f\) is \(2\)-transitive, then it has one orbit on \(\Omega^{(2)}\). The connected-cover hypotheses used in 10 hold here because \(V\) is a nonempty open subset of \(\mathbb{P}^1\). Thus \(C_f\) is geometrically irreducible. Since there is only one geometric component, every arithmetic automorphism preserving the cover, in particular every \(\Theta_q\), fixes it. Therefore \(a_f(q)=1\) for every \(q\), and Theorem 21 gives \(N_f(q)=q+O_f(q^{1/2})\). This is positive for all sufficiently large \(q\). ◻

Proof. By 19, the geometric monodromy group is the full symmetric group, hence is \(2\)-transitive. Apply 11. ◻

Proof. A separable rational map of degree two has geometric monodromy \(S_2\), which is \(2\)-transitive on a two-point fiber. Apply Corollary 11. The degree-two cases in 9 are separable because their degree is smaller than the characteristic in the sporadic case and because the excluded characteristics are precisely the inseparable or collapsed cases for \(d=2\). ◻

Proof. After completing at \(y\), the points of \(X\) above \(y\) correspond to the extensions of the completed local field of \(Y\) occurring in the tensor product with the function field of \(X\). In the Galois closure, the decomposition group acts transitively on the places above a fixed point \(x\), and the quotient by inertia records the residue extension. Since the base field is algebraically closed, the residue extensions are trivial, so the inertia orbit over \(x\) has size equal to the ramification index \(e_x(\phi)\). This is the standard local description of branch cycles via completed discrete valuation rings; compare [15].

In the final situation, all unramified sheets are singleton inertia orbits and hence are fixed by every inertia element. On the unique nontrivial orbit of length \(m\), the inertia image is a transitive permutation group. If \(m=2\), any nonidentity element on that orbit is a transposition. If \(m\) is prime, the order of the transitive inertia image is divisible by \(m\); by Cauchy’s theorem it contains an element of order \(m\), and such an element is an \(m\)-cycle on the length-\(m\) orbit. Since the remaining orbits are singletons, this \(m\)-cycle fixes the unramified sheets. ◻

Proof. The displayed identity is a direct calculation: \[\begin{align} \frac{1}{d}(N_d'D_d-N_dD_d') &=\bigl((-1)^d(z+1)^{d-1}-z^{d-1}\bigr)(z^d-1) \\ &\quad-\bigl((-1)^d(z+1)^d-z^d\bigr)z^{d-1} \\ &=z^{d-1}-(-1)^d(z+1)^{d-1}(z^{d-1}+1). \end{align}\] The separability assumption ensures that the geometric monodromy action is defined on \(n\) separable sheets; the simple-branch-point condition is then a statement about this separable cover. After canceling common factors, the simple branch point hypothesis says that over the corresponding branch value there is one point of ramification index \(2\) and all other points are unramified. By 8, including in wild characteristic, the inertia image on a generic fiber contains a transposition. Indecomposability of a rational function over an algebraically closed field is equivalent to primitivity of its geometric monodromy action; see Fried–MacRae [16] for the separated-variables framework underlying this equivalence. A primitive permutation group containing a transposition is the full symmetric group \(S_n\). Thus \(G_{h_d}=S_n\), and Corollary 11 proves non-exceptionality. ◻

Proof. By 7, \(h_d=\mathop{\mathrm{Frob}}_{p^s}\circ h_{d_0}\) on values. This Frobenius factor does not affect quotient-level twisted exceptionality: for every \(q=p^k\), the map \(\mathop{\mathrm{Frob}}_{p^s}\) commutes with \(\Theta_q=\tau^{-1}\mathop{\mathrm{Frob}}_q\) because \(\tau\) is defined over \(\mathbb{F}_p\), hence \(\mathop{\mathrm{Frob}}_{p^s}\) restricts to a bijection of the twisted fixed set \(\mathbb{P}^1(\overline{\mathbb{F}}_p)^{\Theta_q}\). It does, however, multiply the full morphism degree by \(p^s\). Thus the separable cover relevant for monodromy and quotient-level twisted exceptionality is \(h_{d_0}\). If \(\deg h_{d_0}\le1\), then 8 gives the low-degree branch. The constant subcase is excluded by 2, which only applies to nonconstant rational functions; every nonconstant degree-one quotient is an automorphism commuting with \(\tau\) and hence is bijective on each twisted fixed set. No trace-zero fiber or denominator condition is part of this quotient-level statement. If \(d_0=p^a+1\) with \(a\ge1\), 14 gives exactly the sparse quotient-permutation criteria. If \((p,d_0)=(19,6)\), this theorem makes no further claim beyond retaining it as one of the stated alternatives. All remaining quotients of degree greater than one have \(2\)-transitive monodromy by hypothesis, and Corollary 11 rules out twisted exceptionality. Therefore every twisted exceptional quotient in the stated range belongs to one of the three listed alternatives. ◻

Proof. Reducing \[K_6^{\rm raw}(x)=\frac{1-x^6}{x^6-(1-x)^6}\] modulo \(19\), the numerator and denominator have common factor \(x^2-x+1\). Division gives the displayed formula for \(K_6\). Since \(\phi(t)=t/(t+1)\), the conjugated self-map in the \(x\)-coordinate is \[\widetilde{h}_6(x)=\phi(K_6(x))=\frac{K_6(x)}{K_6(x)+1},\] and simplifying gives the displayed formula for \(\widetilde{h}_6\).

The computer-assisted certificate in 23 verifies the required polynomial identities and matrix products. In particular, it verifies \[K_6\circ\gamma_i=K_6\qquad (i=0,1,2,3),\] checks that the four projective classes are pairwise distinct, checks that each nonidentity class has square equal to the identity, and checks closure under multiplication. Hence these four transformations form a Klein four-group in \(\operatorname{PGL}_2(\mathbb{F}_{19})\). Because \(\widetilde{h}_6=\phi\circ K_6\) differs from \(K_6\) only by a target automorphism, the same transformations are deck transformations of \(\widetilde{h}_6\). Since \(|V_4|=\deg K_6=\deg \widetilde{h}_6=4\), the fixed field \(k(x)^{V_4}\) has degree four in \(k(x)\) and equals both \(k(K_6)\) and \(k(\widetilde{h}_6)\). Hence the cover is the quotient by this group, its geometric monodromy group is \(V_4\), and away from ramification its fibers are exactly \(V_4\)-orbits.

The displayed conjugation relations are likewise verified in 23 by multiplication in \(\operatorname{PGL}_2(\mathbb{F}_{19})\), using \(\tau_x(x)=1/(1-x)\). The same certificate also verifies \[\widetilde{h}_6\circ\tau_x=\tau_x\circ\widetilde{h}_6,\] which is also the coordinate conjugate of the original relation \(h_6\circ\tau=\tau\circ h_6\). It remains to check that the cyclic permutation of the off-diagonal components gives actual injectivity on the twisted fixed sets, including possible branch-point intersections of the graphs.

Because \(19\nmid |V_4|\), the quotient is tame and Galois. Thus a source point lies over a branch value if and only if its stabilizer in \(V_4\) is nontrivial; equivalently, the ramification points are exactly the fixed points of nontrivial deck transformations. The certificate verifies that the fixed-point equations \(x=\gamma_i(x)\) for the three nonidentity deck transformations are, up to nonzero scalar factors, the quadratics \[x^2-x-3, \qquad x^2+4x+1, \qquad x^2-6x+6,\] respectively. It also checks that each nontrivial \(\gamma_i\) has nonzero lower-left matrix entry, so none fixes \(\infty\); hence the three quadratics account for all ramification points. These quadratics are irreducible over \(\mathbb{F}_{19}\). The branch-point exclusions used below are the following explicit gcd checks in \(\mathbb{F}_{19}[x]\): \[\begin{array}{c|c|c|c} \text{deck involution} & f_i(x) & S_i & \text{odd-}k\text{ test polynomial} \\ \hline \gamma_1 & x^2-x-3 & 1 & (1-x)(1-x)-1 \\ \gamma_2 & x^2+4x+1 & -4 & (-4-x)(1-x)-1 \\ \gamma_3 & x^2-6x+6 & 6 & (6-x)(1-x)-1 \end{array}\] Here \(S_i\) is the sum of the two roots of \(f_i\). In each row, \[\gcd(f_i,x^2-x+1)=1, \qquad \gcd\bigl(f_i,(S_i-x)(1-x)-1\bigr)=1.\] If \(k\) is even, then Frobenius acts trivially on their roots, so the twisted fixed-point equation would force \[x=\tau_x(x), \qquad\text{equivalently}\qquad x^2-x+1=0.\] This quadratic is coprime to each of the three fixed-point quadratics above. If \(k\) is odd, Frobenius acts on a root of \(x^2-Sx+P\) as \(x\mapsto S-x\). The twisted fixed-point equation is then \[S-x=\frac{1}{1-x}.\] For the three values \(S=1,-4,6\), the gcd calculation recorded in 23 with the corresponding fixed-point quadratic again gives no common root. Thus no branch point of the \(V_4\)-cover lies in any twisted fixed set.

Now suppose \(x\) and \(y\) are two distinct \(\tau_x^{-1}\mathop{\mathrm{Frob}}_{19^k}\)-fixed points with \(\widetilde{h}_6(x)=\widetilde{h}_6(y)\). Since \(\phi\) is a target automorphism, this is equivalent to \(K_6(x)=K_6(y)\). By the fixed-field description above, the fiber away from the checked branch locus is a \(V_4\)-orbit, so there is a nontrivial \(\gamma_i\) with \(y=\gamma_i x\). The graph of \(\gamma_i\) is sent by \(\tau_x^{-1}\mathop{\mathrm{Frob}}_{19^k}\) to the graph of \(\tau_x^{-1}\gamma_i\tau_x\), which is the next nontrivial graph in the displayed three-cycle. Because the pair \((x,y)\) is itself fixed, it lies in the intersection of two distinct nontrivial graphs. Hence \(x\) is fixed by a nontrivial deck transformation, contradicting the branch-point check. Therefore no off-diagonal collision occurs on the twisted fixed set.

Since \(\widetilde{h}_6\) is \(\tau_x\)-equivariant, it maps the finite \(\tau_x^{-1}\mathop{\mathrm{Frob}}_{19^k}\)-fixed set to itself. The injectivity just proved is therefore bijectivity. Conjugating by \(\phi\) gives the equivalent \(\tau\)-twisted exceptionality statement for the original quotient \(h_6\) in the \(z\)-coordinate. ◻

Proof. The sporadic assertion is 23 under its certificate-execution hypothesis. The exclusion of all quotients outside the three alternatives is exactly the conclusion of 22. ◻

Proof. The reproducible Sage certificate in 23, summarized in Remark 7, gives \(\deg h_5=5\) and shows that the branch polynomial has two simple roots, namely \(T=2\) and \(T=-3\). By 18, the corresponding branch points are simple critical values, so the geometric monodromy contains a transposition. Since the degree is the prime number \(5\), the monodromy action is primitive. A primitive subgroup of \(S_5\) containing a transposition is the full symmetric group. The final assertion follows from 11. ◻

Proof. Since \(d-1=p^s e\), the Frobenius identity gives \[z^{d-1}=(z^e)^{p^s}, \qquad (z+1)^{d-1}=((z+1)^e)^{p^s}, \qquad z^{d-1}+1=(z^e+1)^{p^s}.\] For odd \(p\), \(p^s\) is odd and \((-1)^d=(-1)^{e+1}\); for \(p=2\) all signs are \(1\). Hence \[F_d(z)=\left(z^e-(-1)^{e+1}(z+1)^e(z^e+1)\right)^{p^s}=F_{e+1}(z)^{p^s}.\] The final assertion is an affine statement on \(\mathbb{A}^1_z\) and follows from 9: before cancellation the skeleton is \(F_d\), while after cancellation one has \(N_d'D_d-N_dD_d'=C_d^2W(h_d)\). Thus finite affine divisorial multiplicities in the reduced Wronskian are those of \(F_d\) with the \(2\,\operatorname{div}(C_d)\) cancellation contribution removed. The projective point \(z=\infty\) is not encoded by this affine polynomial identity and must be handled by a separate local expansion. In wild characteristic, the order of vanishing of the Wronskian is a different exponent, not automatically the tame ramification index minus one. The local ramification index must be read from the first nonzero term in the local expansion of \(h_d\). ◻

Proof. If \(m=2\), the primitive-transposition theorem gives \(G_{h^{\mathrm{sep}}}=S_n\). If \(m\ge3\), then the hypotheses give a primitive group containing an \(m\)-cycle with \(n-m\ge3\) fixed points, so 9 applies and gives \(A_n\le G_{h^{\mathrm{sep}}}\). Thus in all cases \(A_n\le G_{h^{\mathrm{sep}}}\le S_n\). The non-exceptionality of \(h^{\mathrm{sep}}\) follows from 11. Finally, \(h_d=\mathop{\mathrm{Frob}}_{p^s}\circ h^{\mathrm{sep}}\) on values by 1. Since \(\tau\) is defined over \(\mathbb{F}_p\), one has \(\mathop{\mathrm{Frob}}_{p^s}\Theta_q=\Theta_q\mathop{\mathrm{Frob}}_{p^s}\) for every \(q=p^k\); hence Frobenius is a bijection on every twisted fixed set. Thus composing with this Frobenius factor cannot remove or create off-diagonal collisions, so the full quotient \(h_d\) is not \(\tau\)-twisted exceptional either. ◻

Proof. The identities \[x^{cr+1}=x(x^r)^c, \qquad (1-x)^{cr+1}=(1-x)(1-x^r)^c\] are immediate from \(r=p^a\). Substituting them into the definition of \(K_d^{\rm raw}\) gives the formula. Expanding \((1-x^r)^c\) gives at most \(c+1\) terms, and multiplying by \((1-x)\) gives the union of two adjacent translates, hence at most \(2c+2\) ordinary monomial terms. The numerator has at most two terms before cancellation. ◻

Proof. This is 12 with \(d-1=cr\). Explicitly, \[z^{d-1}=(z^c)^r, \qquad (z+1)^{d-1}=((z+1)^c)^r, \qquad z^{d-1}+1=(z^c+1)^r,\] and the sign agrees with the sign in \(F_{c+1}\) because \(r\) is odd for odd \(p\), while in characteristic two all signs are trivial. ◻

Proof. Use the filtered normal form of 7. Since \[(1-u)^{2r+1}=(1-u)(1-u^r)^2 =1-u-2u^r+2u^{r+1}+u^{2r}-u^{2r+1},\] only the six displayed exponents can occur.

If \(r\equiv1\pmod3\), then \(d\equiv0\pmod3\), so the numerator filter has residues congruent to \(2\) and the denominator filter has residues congruent to \(1\). Writing \(T=u^3\), this gives \[A_{d,2}(T)=2T^{(r-1)/3}+T^{2(r-1)/3}, \qquad A_{d,1}(T)=-1-2T^{(r-1)/3}.\] Substitution in 7 gives \[\Phi_{2r+1}(u) =-\omega^2u^r\frac{u^{r-1}+2}{2u^{r-1}+1}.\] There is no common factor between numerator and denominator, and the degree is \(2r-1\).

If \(r\equiv2\pmod3\), then \(d\equiv2\pmod3\). The numerator filter has residue \(0\) and the denominator filter has residue \(2\), giving \[A_{d,0}(T)=1+2T^{(r+1)/3},\] and \[A_{d,2}(T)=-2T^{(r-2)/3}-T^{(2r-1)/3}.\] Again 7 gives the stated expression \[\Phi_{2r+1}(u) =-\omega^2u^{-r}\frac{1+2u^{r+1}}{2+u^{r+1}}.\] Its degree is \(2r+1\).

It remains to check ramification. In the first case, ignoring the nonzero scalar \(-\omega^2\), write \[f(u)=u^r\frac{u^{r-1}+2}{2u^{r-1}+1}.\] Since \(r=0\) and \(r-1=-1\) in the ground field, differentiation on the affine line gives \[f'(u)=\frac{3u^{2r-2}}{(2u^{r-1}+1)^2}.\] Thus the only finite affine critical point away from poles is \(u=0\). Locally \(f(u)=2u^r+O(u^{2r-1})\), so its ramification index at \(0\) is \(r\). At infinity, in the coordinate \(v=1/u\), one has \(f(u)=\frac{1}{2}u^r+O(u)\), so the ramification index at infinity is again \(r\). The remaining points over \(0\) and \(\infty\) are the simple roots of \(u^{r-1}+2\) and \(2u^{r-1}+1\), respectively, and are unramified.

In the second case, write \[f(u)=u^{-r}\frac{1+2u^{r+1}}{2+u^{r+1}}.\] The derivative of the factor \(u^{-r}\) vanishes in characteristic \(p\), while \((u^{r+1})'=u^r\). Hence \[f'(u)=\frac{3}{(2+u^{r+1})^2}\] away from poles. Thus there are no affine critical points away from the poles. At \(u=0\) the function has a pole of order \(r\), so the local ramification index is \(r\). At infinity, with \(v=1/u\), one has \(f(u)=2u^{-r}+O(u^{-2r-1})\), equivalently \(f=v^r(2+O(v^{r+1}))\), so the local ramification index at infinity is also \(r\). The other zeros and poles are simple. Hence the only branch values are \(0\) and \(\infty\). ◻

Proof. Suppose, to the contrary, that \(f=g\circ h\) with \[\deg h=m>1, \qquad \deg g=\ell>1.\] Let \(R\) be a pole of \(g\) of multiplicity \(A\). If \(S\in h^{-1}(R)\) has ramification index \(B=e_h(S)\), then \(S\) is a pole of \(f\) of multiplicity \(AB\).

All pole multiplicities of \(f\) are equal to \(1\), except for the unique pole \(P\) of multiplicity \(r\). Let \(R_0=h(P)\) and let \(A\) be the pole multiplicity of \(g\) at \(R_0\); write \(B=e_h(P)\). Then \[AB=r.\] If \(A>1\), then every point in \(h^{-1}(R_0)\) would give a pole of \(f\) of multiplicity at least \(A\). Since \(P\) is the only nonsimple pole, \(P\) is the only point over \(R_0\). Hence \(m=B=r/A\) divides \(r\). But \(m\) also divides \(n=\deg f\), and \(\gcd(r,n)=1\), so \(m=1\), contradiction.

Thus \(A=1\) and \(B=r\). Therefore \(m\ge r\). Since \(\ell\ge2\), one has \[m\le \frac{n}{2}.\] If \(n=2r-1\), then \(n/2<r\), contradiction. If \(n=2r+1\), then \(m\le r\); hence \(m=r\), but \(r\nmid n\) by hypothesis. This is again impossible. Therefore no nontrivial decomposition exists.

Finally assume \(f\) is separable. For a separable cover of curves, decomposability of the rational function is equivalent to imprimitivity of the geometric monodromy action: an intermediate rational function is the same thing as an intermediate function field between \(k(f)\) and \(k(u)\), hence a nontrivial block system. Thus the monodromy action is primitive. ◻

Proof. Use the normal forms of 25. If \(r\equiv1\pmod3\), then, up to a nonzero scalar, \[\Phi_{2r+1}(u)=u^r\frac{u^{r-1}+2}{2u^{r-1}+1}\] has degree \(n=2r-1\). Its pole divisor is \[r[\infty]+\sum_{2\alpha^{r-1}+1=0} [\alpha].\] The roots of \(2u^{r-1}+1\) are distinct because \(p\nmid r-1\). Hence the pole divisor has one pole of multiplicity \(r\) and \(r-1=n-r\) simple poles.

If \(r\equiv2\pmod3\), then, again up to a nonzero scalar, \[\Phi_{2r+1}(u)=u^{-r}\frac{1+2u^{r+1}}{2+u^{r+1}}\] has degree \(n=2r+1\). Its pole divisor is \[r[0]+\sum_{2+\alpha^{r+1}=0} [\alpha],\] and the roots of \(2+u^{r+1}\) are distinct because \(p\nmid r+1\). Thus the same pole-divisor pattern holds, with \(n-r=r+1\) simple poles.

In both cases \(\gcd(r,n)=1\). The derivative computations in 25 show that the normal forms are separable, hence so is \(h_d\). Applying 10 proves indecomposability, and its separable monodromy clause gives primitivity. ◻

Proof. By 26, \(G_{h_d}\) is primitive. By 25, over one of the two branch values there is exactly one ramified point of ramification index \(p\), while all other points over that branch value are unramified. Since \(p\) is prime, the final clause of 8 gives an actual \(p\)-cycle in the inertia image on the sheets specializing to the ramified point, and this element fixes the remaining \(n-p\) sheets. Here \(n-p\ge3\) for \(p\ge5\).

Now apply 24 with \(m=p\). This gives \(A_n\le G_{h_d}\le S_n\), and the final assertion follows from the two-transitive collision obstruction, 11. ◻

Proof. Consider first the case \(r\equiv1\pmod3\). Up to a nonzero scalar, the local expression at \(u=0\) is \[t=f(u)=u^r\frac{u^{r-1}+2}{2u^{r-1}+1}.\] The expansion begins with \(2u^r\), so the ramification index is \(r\). The derivative computed in 25 is \[f'(u)=\frac{3u^{2r-2}}{(2u^{r-1}+1)^2},\] so \(v_u(f')=2r-2\). At \(u=\infty\), set \(v=1/u\) and use the target parameter \(t=1/f\). Then \[t=v^r\frac{2+v^{r-1}}{1+2v^{r-1}},\] and differentiation gives valuation \(2r-2\) again.

If \(r\equiv2\pmod3\), the normal form is \[f(u)=u^{-r}\frac{1+2u^{r+1}}{2+u^{r+1}}\] up to a nonzero scalar. At \(u=0\) the branch value is \(\infty\), so use \(t=1/f\): \[t=u^r\frac{2+u^{r+1}}{1+2u^{r+1}}.\] Since \(r=0\) and \(r+1=1\) in the ground field, differentiating gives \[\frac{dt}{du}=\frac{-3u^{2r}}{(1+2u^{r+1})^2},\] so the different exponent is \(2r\). At \(u=\infty\), with \(v=1/u\) and target parameter \(t=f\), one obtains \[t=v^r\frac{2+v^{r+1}}{1+2v^{r+1}},\] and the same derivative valuation \(2r\). In each of the four local forms just displayed, \(t\) has the form \(u^r\) times a unit, or the same expression in the parameter \(v=1/u\). Hence \(k[[u]]\) is integral and finite over \(k[[t]]\): equivalently, by Weierstrass preparation, the equation \(f(U)-t=0\) gives a distinguished monic polynomial for the chosen uniformizer over \(k[[t]]\) after multiplying by a unit. The extension is separable because the displayed derivative is not identically zero. Thus the completed source ring is monogenic over the completed base, and \[\Omega_{k[[u]]/k[[t]]} \cong k[[u]]/(dt/du)\,du.\] The standard monogenic different formula for complete discrete valuation rings, applied to this finite separable extension, therefore gives the different exponent as \(v_u(dt/du)\); see, for example, [15]. This proves the claim. ◻

Proof. The reduced quotient has degree at most \(m\) precisely when it can be represented as \(P/Q\) with \(\deg P,\deg Q\le m\) and \(Q\not=0\). Since \(H_d^{\rm raw}=N_d/D_d\) as a rational function, this is equivalent to \[Q(z)N_d(z)=P(z)D_d(z)=P(z)(z^d-1).\] Comparing coefficients of \(z^j\) gives ?? . Conversely, any solution with \(Q\ne0\) gives the displayed polynomial identity and hence a representative of degree at most \(m\) after removing any common factor of \(P\) and \(Q\). A solution with \(Q=0\) would force \(P(z)(z^d-1)=0\), hence \(P=0\) in the polynomial ring; thus no nonzero kernel vector is lost by requiring \(Q\ne0\). The last assertion is immediate. ◻

Proof. We use least-significant-digit first base-\(p\) words, read in parallel with zero padding. Integers are represented canonically by finite words with no trailing zero in this convention, except that \(0\) is represented by the one-letter word \(0\); allowing additional padded zeros does not change any recognized relation and is used for parallel reading of tuples such as \((d,j,t)\).

By 15, the condition \(\deg h_d\le m\) over \(\bar{\mathbb{F}}_p\) is equivalent to the existence, after scalar extension to \(\bar{\mathbb{F}}_p\), of a kernel vector whose \(Q\)-part is nonzero. Since the matrix entries lie in \(\mathbb{F}_p\), both the full matrix rank and the rank after imposing \(q_0=\cdots=q_m=0\) are unchanged by field extension; therefore such a vector exists over \(\bar{\mathbb{F}}_p\) if and only if one exists over \(\mathbb{F}_p\). Thus it is enough to test the finitely many coefficient vectors \[(p_0,\ldots,p_m,q_0,\ldots,q_m)\in\mathbb{F}_p^{2m+2}\] with nonzero \(Q\)-part satisfying the coefficient equations \[\sum_{i=0}^m q_i n_{j-i}=p_{j-d}-p_j \qquad(0\le j\le d+m),\] where \[n_t=(-1)^d\binom dt-\mathbf{1}_{t=d}\] with the convention \(n_t=0\) for \(t\notin[0,d]\). Since the vector space \(\mathbb{F}_p^{2m+2}\) is finite, we may test the finitely many coefficient vectors with nonzero \(Q\)-part one at a time and take their union.

For a fixed such coefficient vector, each displayed equation is a first-order condition in the variables \(d\) and \(j\) over the natural numbers with addition, order, base-\(p\) digit predicates, and the relation \[\binom dt\equiv c\pmod p \qquad(c\in\mathbb{F}_p).\] Lucas’ theorem expresses this relation as \[\binom dt\equiv \prod_{\nu\ge0}\binom{d_\nu}{t_\nu}\pmod p,\] where \(d=\sum d_\nu p^\nu\) and \(t=\sum t_\nu p^\nu\). Thus the relation is recognized by a finite automaton reading the base-\(p\) expansions of \(d\) and \(t\) in parallel: the automaton keeps only the running product in \(\mathbb{F}_p\) and moves to a zero state if some \(t_\nu>d_\nu\). The sign \((-1)^d\), the boundary tests \(t=d\), \(t\in[0,d]\), and the shifted symbols \(p_{j-d}\) and \(p_j\) are also finite-state conditions. More explicitly, because \(m\) is fixed, the alternatives \(j-d=t\) with \(0\le t\le m\), \(j-i=t\) with \(0\le i\le m\), and \(0\le j\le d+m\) are recognized by the standard addition automata for base-\(p\) words with constants and order. Thus the variable upper bound and all shifted indices in ?? are encoded inside the same automatic structure; no unbounded memory is required beyond the usual carry state.

Operationally, for a fixed such coefficient vector the finite state records only: the vector itself, the carries for the additions and subtractions defining \(j-d\), \(j-i\), and \(d+m\), the running Lucas product in \(\mathbb{F}_p\) for each binomial coefficient queried by the equation, the comparison states for endpoint tests such as \(0\le j\le d+m\), and the single-field-element accumulator for the coefficient equation. This makes the asserted effectivity a finite construction rather than an abstract existence statement.

For a fixed such coefficient vector, build the automatic relation \(E(d,j)\) asserting that the coefficient equation at index \(j\) holds. The bounded universal quantifier over all \(j\) with \(0\le j\le d+m\) is implemented by first recognizing the existential counterexample relation \(\exists j\,(0\le j\le d+m\;\wedge\;\neg E(d,j))\), then projecting away \(j\), and finally complementing the resulting language of bad \(d\). These closure properties are standard in the first-order theory of automatic structures over \((\mathbb{N},+,<,V_p)\) and finite automata over numeration systems; see Büchi [22] and Allouche–Shallit [23]. Therefore, for each fixed such coefficient vector, the set of \(d\) satisfying all equations is regular. A finite union over the finitely many coefficient vectors with nonzero \(Q\)-part gives a regular language for the exponents with \(\deg h_d\le m\) before imposing the tame condition. Intersect this language with the regular language of canonical least-significant-digit-first base-\(p\) words for positive integers whose first digit is nonzero. Because the words are read least-significant digit first, this first digit is the residue of \(d\) modulo \(p\); requiring it to be nonzero imposes exactly \(p\nmid d\) and gives the regular language for \(\mathcal{D}_{p,\le m}\). Finally, \[\mathcal{D}_{p,m}=\mathcal{D}_{p,\le m}\setminus \mathcal{D}_{p,\le m-1},\] with the evident convention for \(m=0\), so the exact-degree stratum is regular as well. ◻

Proof. If \(x\in\Gamma_{n,q}^*\), then all its conjugates are nonzero and their sum is zero, so the displayed point lies in \(U_n\). Frobenius sends it to its cyclic shift, hence it is fixed by \(\Theta_{n,q}\). Multiplication of \(x\) by an element of \(K^*\) does not change the projective point.

The map is injective on \(K^*\)-orbits. Indeed, if \[[x:\sigma x:\cdots:\sigma^{n-1}x] =[y:\sigma y:\cdots:\sigma^{n-1}y],\] then \(y=cx\) for some \(c\in L^*\). Comparing the second coordinates gives \(\sigma(c)=c\), hence \(c\in K^*\).

For surjectivity, let \(P=[X_0:\cdots:X_{n-1}]\in U_n(\overline{\mathbb{F}}_q)^{\Theta_{n,q}}\). Since \(\Theta_{n,q}^n=\mathop{\mathrm{Frob}}_{q^n}\), the point \(P\) is defined over \(\mathbb{F}_{q^n}\) after a projective scaling. Choose coordinates \(X_i\in\mathbb{F}_{q^n}^*\) for \(P\). The relation \(\mathop{\mathrm{Frob}}_q(P)=\rho(P)\) gives a scalar \(c\in\mathbb{F}_{q^n}^*\) with \[X_i^q=cX_{i+1} \qquad(i\bmod n).\] Iterating around the cycle gives \(\mathop{\mathrm{Nm}}_{\mathbb{F}_{q^n}/\mathbb{F}_q}(c)=1\). By multiplicative Hilbert 90 there is \(b\in\mathbb{F}_{q^n}^*\) with \(b^{q-1}=c^{-1}\). Replacing \(X_i\) by \(bX_i\) makes the scalar equal to \(1\), so the new coordinates satisfy \(X_i^q=X_{i+1}\). Thus they are \[[x:\sigma x:\cdots:\sigma^{n-1}x]\] with \(x=X_0\). Since the coordinates sum to zero, \(x\in\Gamma_{n,q}^*\). ◻

Proof. Let \[S=k[X_0,\ldots,X_{n-1}]/(X_0+\cdots+X_{n-1})\] and let \(I\subset S\) be the homogeneous ideal generated by the coordinates of 14 , equivalently by \[X_1^d-X_0^d,\;X_2^d-X_1^d,\ldots,\;X_0^d-X_{n-1}^d.\] The base locus is \(\operatorname{Proj}(S/I)\), with the usual saturation by the irrelevant ideal. If a homogeneous prime \(\mathfrak p\) in this Proj contains one coordinate \(X_i\), then the equations in \(I\) force \(X_{i+1},X_{i+2},\ldots\) cyclically into \(\mathfrak p\); hence all coordinates lie in \(\mathfrak p\), contradicting the condition that \(\mathfrak p\) is a point of Proj. Thus the support of the saturated base locus is contained in the torus open \(X_0\cdots X_{n-1}\ne0\). Let \(J\) be the saturation of \(I\) by the irrelevant ideal. Because no associated geometric support remains on the complement of this torus open, the saturated closed subscheme is obtained scheme-theoretically by contraction from the torus localization: \[J=S\cap I\,S[(X_0\cdots X_{n-1})^{-1}].\] This saturation-contraction step preserves possible nilpotent structure; in particular, no reduction is being taken when \(p\mid d\).

On the affine chart \(X_0=1\) inside the torus, put \(u_i=X_i/X_0\). The localized equations are exactly \[u_i^d-1=0\qquad(1\le i\le n-1),\] and the trace-zero hyperplane equation becomes \[1+u_1+\cdots+u_{n-1}=0.\] Therefore, as a closed subscheme after saturation by the irrelevant ideal, the base locus is \[\operatorname{Spec} k[u_1^{\pm1},\ldots,u_{n-1}^{\pm1}]/ (u_1^d-1,\ldots,u_{n-1}^d-1,1+u_1+\cdots+u_{n-1}),\] which is precisely the stated torsion scheme. This argument also covers the nonreduced case when \(p\mid d\). ◻

Proof. The zero element of \(\Gamma_{n,q}\) is fixed. On \(\Gamma_{n,q}^*\), scalar multiplication by \(c\in K^*\) changes \(P_d(x)\) by the factor \(c^d\). Hence the map on each \(K^*\)-fiber is bijective precisely when \(\gcd(d,q-1)=1\).

By 16, the quotient of \(\Gamma_{n,q}^*\) by \(K^*\) is the twisted fixed set \(U_n^{\Theta_{n,q}}\). Formula 14 is obtained by applying \(\sigma^i\) to \(P_d(x)\): \[\sigma^i(P_d(x))=\sigma^{i+1}(x)^d-\sigma^i(x)^d.\] Thus \(\Psi_{n,d}\) is the quotient map wherever the image is nonzero. The image is zero exactly when all displayed differences vanish, i.e. exactly at the underlying geometric point set of the base locus of 17. Moreover, for a twisted fixed point coming from \(x\in\Gamma_{n,q}^*\), the target coordinates are the conjugates \(\sigma^i(P_d(x))\). If any one of these coordinates is zero, then \(P_d(x)=0\), so all of them are zero and the point is in the base locus. Hence every non-base image of a twisted fixed point again lies in \(U_n\). Therefore condition (ii) is the higher-dimensional denominator condition, and condition (iii) is the quotient bijectivity condition. Only the underlying geometric points of \(B_{n,d}\) matter here: the criterion evaluates the rational map on actual twisted fixed points, so nilpotent structure in a nonreduced base locus cannot create additional inputs. Combining the fiber and quotient conditions proves the criterion.

When \(n=3\), the hyperplane \(X_0+X_1+X_2=0\) with \(z=X_1/X_0\) gives \(X_2/X_0=-1-z\). The quotient map 14 becomes the same rational map as 2 , and the base-locus condition becomes \(\mu_d\cap\Lambda_q=\varnothing\). Thus the result recovers precisely the fixed-exponent specialization \(r=d\), \(B(z)=z^d-1\) of 2, not the full generality of that proposition for arbitrary \(r\) and \(B\). ◻

Proof. The fixed points of \(\Theta_{n,q}\) on \(C_f\) are exactly ordered pairs of distinct twisted fixed points with the same image. The first formula is Grothendieck–Lefschetz. If \(r=0\), then \(C_f\) is finite etale over the chosen locus and the trace formula is simply the exact count of fixed geometric points, which is the stated definition of \(a_f(q)\). If \(r\ge1\), then 6 makes \(C_f\) smooth and equidimensional of dimension \(r\), and the top compactly supported cohomology contributes one copy of \(\mathbb{Q}_\ell(-r)\) for each \(r\)-dimensional irreducible component fixed by \(\Theta_{n,q}\), giving the main term \(a_f(q)q^r\).

It remains to justify the error term for the twisted operator. The cyclic shift \(\rho\) has order \(n\) and commutes with Frobenius, so \(\Theta_{n,q}^n=\mathop{\mathrm{Frob}}_{q^n}\) on \(C_f\) and on compactly supported cohomology. If \(\alpha\) is an eigenvalue of \(\Theta_{n,q}\) on \(H_c^i(C_{f,\overline{\mathbb{F}}_p},\mathbb{Q}_\ell)\), then \(\alpha^n\) is an eigenvalue of \(\mathop{\mathrm{Frob}}_{q^n}\) on the same space. Deligne’s weight bounds [12] give \(|\alpha^n|\le q^{ni/2}\), hence \(|\alpha|\le q^{i/2}\). All terms with \(i<2r\) therefore contribute \(O_f(q^{r-1/2})\). The final assertion for \(r\ge1\) follows exactly as in 21. ◻

Proof. First note the additive Hilbert–90 exactness used in the statement. For every \(x\in L\), \[\mathop{\mathrm{Tr}}_{L/K}(x^q-x)=0,\] so \(\delta(L)\subseteq\Gamma_q\). Also \(\ker(\delta)=K\), hence \[|\delta(L)|=|L|/|K|=q^2.\] Since \(L/K\) is finite and separable, the trace map \(\mathop{\mathrm{Tr}}_{L/K}:L\to K\) is a nonzero \(K\)-linear functional, hence is surjective; therefore \(|\Gamma_q|=q^2\). It follows that \(\delta(L)=\Gamma_q\), and the fibers are exactly the cosets \(u+K\).

Now apply the usual fiber criterion. If \(F\) is bijective, then it is injective on fibers and the commutative diagram forces \(G\) to be bijective on the quotient. Conversely, if \(G\) is bijective and \(F\) is injective on every fiber, then two points with the same image under \(F\) must have the same image under \(G\circ\delta\), hence lie in the same fiber of \(\delta\), where injectivity applies. Since \(L\) is finite, injectivity implies bijectivity. ◻

Proof. Let \(\delta=X^q-X\). The first term factors through \(\delta\), and \[\delta((\delta X)^d)=(\delta X)^{dq}-(\delta X)^d=P_d(\delta X).\] Thus it induces the trace-zero map of 6. The remaining trace-linear term lies in \(K\), so it is killed by \(\delta\). On a fiber \(u+t\) with \(t\in K\), the trace-linear term changes by \[\mathop{\mathrm{Tr}}_{L/K}(c(u+t)^{2^m})-\mathop{\mathrm{Tr}}_{L/K}(cu^{2^m}) =\mathop{\mathrm{Tr}}_{L/K}(c)t^{2^m}.\] Since \(t\mapsto t^{2^m}\) is a bijection of \(K\), the fiber map is injective if and only if \(\mathop{\mathrm{Tr}}_{L/K}(c)\ne0\). Apply 11. ◻

Proof. Write \[(X^q+X)^d=\sum_{j=0}^d X^{e_j}, \qquad e_j=q(d-j)+j.\] All binomial coefficients are \(1\) in characteristic two because \(d=2^a-1\). Under the stable hypothesis all exponents appearing below are positive and not congruent to \(0\) modulo \(M\), so the canonical exponent convention agrees with ordinary residue computations in \(\mathbb{Z}/M\mathbb{Z}\). For a coefficient \(\alpha\in\mathbb{F}_{q^3}\), the operator \(\delta\) sends a monomial \(\alpha X^e\) to \(\alpha^qX^{qe}+\alpha X^e\). Thus coefficients may change by Frobenius, but exponents remain inside the orbit of \(e\) under multiplication by \(q\) on \(\mathbb{Z}/M\mathbb{Z}\).

Because \(d<q-1\), the exponents \(e_j\) are distinct, nonzero, and less than \(q^2+q+1\); hence none is fixed by multiplication by \(q\) modulo \(M=q^3-1\). We next check the orbit collisions. If \(e_i\equiv qe_j\pmod M\), then in fact \(qe_j<M\), since \(e_j\le qd<q(q-1)\); hence equality holds in the integers. The equation \(e_i=qe_j\) is \[qd-(q-1)i=q(qd-(q-1)j),\] which reduces to \(i=q(j-d)\). Thus the only solution with \(0\le i,j\le d\) is \((i,j)=(0,d)\), giving \[qe_d=e_0=qd.\] The case \(e_i\equiv q^2e_j\pmod M\) is equivalent, after multiplying by \(q\), to \(qe_i\equiv e_j\pmod M\), and the same argument gives the same collision. Therefore the \(d+1\) exponents \(e_0,\ldots,e_d\) occupy exactly \(d\) distinct \(q\)-orbits.

Thus the support of \[P_d(\delta X)=(X^q+X)^{dq}+(X^q+X)^d\] meets exactly \(d\) distinct \(q\)-orbits: one orbit contributes the two exponents \(d\) and \(q^2d\), and for each \(1\le j\le d-1\) one orbit contributes the two exponents \(e_j\) and \(qe_j\).

The target has no contribution on a \(q\)-fixed exponent class, and the support analysis above placed every nonzero target contribution on a length-three orbit. A monomial \(aX^e\) contributes only to the \(q\)-orbit of the exponent class of \(e\), namely as \(a^qX^{qe}+aX^e\) in characteristic two, with the same canonical reduction of positive exponents. Thus no monomial from one orbit can help produce a nonzero target contribution on another orbit. Consequently every preimage of \(P_d(\delta X)\) must contain at least one monomial from each of these \(d\) nonzero orbits, independently of the coefficient field. This proves the lower bound.

Finally, \[\delta\bigl(X^{q^2d}\bigr)=X^d+X^{q^2d}\] as functions on \(\mathbb{F}_{q^3}\), and for \(1\le j\le d-1\), \[\delta\bigl(X^{e_j}\bigr)=X^{e_j}+X^{qe_j}.\] Adding these identities gives \(\delta(F_0)=P_d(\delta X)\), so the bound is sharp. ◻

Proof. The functional reduction is immediate from \(M=q^3-1\) together with the canonical positive-exponent convention. If \(b>0\), then positive exponents congruent modulo \(M\) define the same monomial function on \(\mathbb{F}_{q^3}\), so \(X^{d_a}=X^{2^b-1}\) and likewise for the \(q\)-multiple exponent. If \(b=0\), then \(d_a\) is a positive multiple of \(M\), so both \(X^{d_aq}\) and \(X^{d_a}\) are represented by the same nonconstant monomial function \(X^M\); they also agree at \(0\). Hence \(P_{d_a}=0\).

It remains to count the minimum number of terms in a primitive for the additive Hilbert–90 operator \(\delta(F)=F^q-F=F^q+F\). Write \[(X^q+X)^{2^b-1}=\sum_{j=0}^{2^b-1}X^{e_j}, \qquad e_j=q(2^b-1)-(q-1)j.\] The nonzero exponent class of \(e_j\) modulo \(M=(q-1)S\) is determined by the residue of \(j\) modulo \(S\). Multiplication of exponents by \(q\) sends the residue \(j\) to \[T_{b,k}(j)=qj-q(2^b-1)\pmod S,\] and \(T_{b,k}^3=1\) because \(q^3\equiv1\pmod S\) and \(1+q+q^2=S\).

Only the parity of the number of occurrences of each residue matters, since the characteristic is two. The interval \(0\le j\le 2^b-1\) has length \(2^b\). Modulo \(S\), its parity support differs from \(I_R\) at most by the full set \(\mathbb{Z}/S\mathbb{Z}\); the full set is fixed by \(T_{b,k}\) and therefore cancels in the symmetric difference. Hence the support of \[P_{2^b-1}(X^q-X)=\delta\bigl((X^q+X)^{2^b-1}\bigr)\] on the set of \(q\)-orbits is represented exactly by \[I_R\triangle T_{b,k}(I_R).\] On a \(q\)-orbit of exponent classes of length three, each monomial term contributes to two adjacent classes in that orbit, with Frobenius-conjugate coefficients. In the present target all coefficients are in \(\mathbb{F}_2\). For an orbit written as \(\{e,qe,q^2e\}\), the three possible two-position supports are produced by \[\delta(X^e)=X^e+X^{qe},\qquad \delta(X^{qe})=X^{qe}+X^{q^2e},\qquad \delta(X^{q^2e})=X^{q^2e}+X^e.\] Hence every nonzero two-term target vector on a length-three orbit is produced by one monomial term, and no monomial from another orbit can contribute to it. A length-three orbit on which the target vector is zero requires no term in a term-minimal primitive: after combining like canonical monomials, any terms supported there contribute an element of \(\ker\delta\) on that orbit and may be deleted without changing the target.

The same omission argument is needed on \(q\)-fixed exponent classes, including the class \(0\bmod M\) represented by \(X^M\). If such a class has combined coefficient \(\alpha\) in a primitive, then its contribution is \((\alpha^q+\alpha)X^e\). The target coefficient on every \(q\)-fixed class is zero, so necessarily \(\alpha^q+\alpha=0\); the term is therefore in \(\ker\delta\) and can be omitted from a term-minimal primitive. Therefore the minimum number of monomial terms in a primitive is one for each nonzero length-three orbit, namely \[\frac{1}{2}|I_R\triangle T_{b,k}(I_R)|.\] If \(0<b\le k\), then \(R=2^b\le q\). The two intervals \(I_R\) and \(T_{b,k}(I_R)\) meet in the single residue \(0\), so the displayed formula gives \(\nu_{b,k}=R-1=2^b-1\). ◻

Proof. This is a direct restatement of the exact orbit count. The function \(P_{2^a-1}\) depends only on the residue of \(a\) modulo \(3k\), and the least number of quotient monomials in an additive primitive is exactly \(\nu_{b,k}\). Thus any sparse unstable representative is sparse because its exponent has collapsed to a sparse residue class in the function algebra of \(\mathbb{F}_{q^3}\). ◻