June 03, 2026
Let \(\mathbb{F}_q\) be a finite field of characteristic \(p\) and \(\pi\colon Y\to X\) be a finite \(\mathbb{F}_q\)-morphism of separated \(\mathbb{F}_q\)-schemes of finite type. Suppose \(\pi\) is generically Galois with group \(G\) of prime order \(r\neq p\). We determine the mod-\(r\) reduction of the zeta function of \(Y\) in terms of the zeta function of \(X\) and the branch locus \(Z\subset X\) of \(\pi\). We give applications to curves and to numerators of hyperelliptic/superelliptic curves.
2026-06-12
Let \(p\in\mathbb{N}\) be a prime and \(\mathbb{F}_q\) be a finite extension of \(\mathbb{F}_p\). For each separated \(\mathbb{F}_q\)-scheme \(X\) of finite type, let \(|X|\) be the set of closed points, that is, the \(\mathop{\mathrm{Gal}}(\bar\mathbb{F}_q/\mathbb{F}_q)\)-orbits of \(X(\bar\mathbb{F}_q)\), and \[\zeta(X,T) := \exp\left(\sum_{n=1}^\infty|X(\mathbb{F}_{q^n})|\frac{T^n}{n}\right) = \prod_{x\in|X|}\det(1-T^{\deg(x)})^{-1} \in 1+T\cdot\mathbb{Z}[[T]]\] be the (Hasse-Weil) zeta function of \(X\). Recall \(\zeta(X,T)\in\mathbb{Q}(T)\) (see [1] and [2]).
Theorem 1. Let \(\mathbb{F}_q\) be a finite field of characteristic \(p\) and \(\pi\colon Y\to X\) be a finite \(\mathbb{F}_q\)-morphism of separated \(\mathbb{F}_q\)-schemes of finite type. Let \(Z\subset X\) be the branch locus of \(\pi\). Suppose \(\pi\) is generically Galois with group \(G\) of prime order \(r\neq p\). Then \[\label{eqn:main-identity} \zeta(Y,T) \equiv \zeta(X,T)^r \zeta(Z,T)^{1-r} \bmod r.\qquad{(1)}\]
A striking aspect of ?? is that the analogous identity over \(\mathbb{Z}\) is not necessarily true (because of weight mismatches).
Theorem ?? is inspired by [3]. We give a naive proof in Section 2 and a cohomological proof in Section 4. The latter uses a cohomological formula given in Section 3 for the reduction of \(\zeta(X,T)\) modulo a prime \(\ell\neq p\). We also give two corollaries (for curves) in Section 5.
An impetus for this paper is question of Richard Griffon: when is the numerator of the zeta function of a curve over a finite field a trinomial? A necessary condition is that the numerator is congruent to a trinomial modulo an integer \(N>1\). In Section 6, we apply the results of Section 5 (where \(N=r\)) to numerators of hyperelliptic (\(r=2\)) and superelliptic (\(r>2\)) curves.
Consider the Euler-product expansion \[\label{eqn:euler-product-expansion} \zeta(Y,T) = \prod_{y\in|Y|}(1-T^{\deg(y)})^{-1} = \prod_{x\in|X|}\prod_{y\in|Y|:\pi(y)=x}(1-T^{\deg(y)})^{-1}.\tag{1}\] Suppose \(x\in|X|\), and let \[\label{eqn:defn-of-Y95x-and-L40Y95x44T41} |Y_x|:=\{y\in|Y|:\pi(y)=x\}\text{ and } L(Y_x,T):=\prod_{y\in |Y_x|}(1-T^{\deg(y)}).\tag{2}\] Suppose \(y\in|Y_x|\), and let \(I_y\subseteq D_y\subseteq G\) be the inertia and decomposition groups of \(y\).
Recall that \(\pi\) is étale over \(X\smallsetminus Z\) and totally ramified of prime degree \(r\) over \(Z\) and that \(G\cong\mathbb{Z}/r\) acts transitively on \(|Y_x|\). Deduce \[\label{eqn:triple-classification} (|I_y|,[D_y:I_y],|Y_{x}|) = \begin{cases} (1,1,r)\text{ or }(1,r,1) & x\in |X|\smallsetminus|Z| \\ (r,1,1) & x\in |Z| \end{cases}.\tag{3}\] Observe that \[(1-T^{\deg(x)})^r\equiv (1-T^{r\deg(x)})\bmod r,\] and 3 imply \[\label{eqn:euler-congruence} L(Y_x,T) \equiv \begin{cases} (1-T^{\deg(x)})^r & x\in |X|\smallsetminus|Z| \\ 1-T^{\deg(x)} & x\in |Z| \end{cases}.\tag{4}\] Combining 1 , 2 , and 4 gives \[\zeta(Y,T) \equiv \prod_{x\in|X|}(1-T^{\deg(x)})^{-r}\prod_{z\in|Z|}(1-T^{\deg(z)})^{r-1} \equiv \zeta(X,T)^r \zeta(Z,T)^{1-r} \bmod r\] as desired.
Theorem 2. Let \(X\) be a separated \(\mathbb{F}_q\)-scheme of finite type and \(\bar{X}\) be its base change to \(\bar\mathbb{F}_q\). Let \(\phi\in\mathop{\mathrm{Gal}}(\bar\mathbb{F}_q/\mathbb{F}_q)\) be the geometric Frobenius and \(\ell\neq p\) be a rational prime. Then \[\label{eqn:zeta-mod-ell-as-alternating-product} \zeta(X,T) \equiv \prod_i\det(1-\phi\,T \mid H^i_c(\bar{X},\mathbb{F}_\ell))^{(-1)^{i+1}} \bmod\ell.\qquad{(2)}\]
Proof. A theorem of Grothendieck (see [2] or [4]) implies that \[\zeta(X,T)=\prod_i\det(1-\phi\,T \mid H^i_c(\bar{X},\mathbb{Q}_\ell))^{(-1)^{i+1}}\] (since \(X\) is \(\mathbb{F}_q\)-compactifiable). We must analyze a product with \(\mathbb{F}_\ell\) in lieu of \(\mathbb{Q}_\ell\). This requires taking into account that \(M_i:=H^i_c(\bar{X},\mathbb{Z}_\ell)\) is a finitely generated (see [5]) and not necessarily free (compare [6]).
Let \(T_i\subseteq M_i\) be the torsion submodule and \(F_i:=M_i/T_i\) be the maximal free quotient so that \[\det(1 - \phi\,T \mid H^i_c(\bar{X},\mathbb{Q}_\ell)) = \det(1 - \phi\,T \mid F_i) \in 1+T\cdot\mathbb{Z}_\ell[T]\] and thus \[\label{eqn:zeta-over-Z95ell-as-alternating-product} \zeta(X,T)=\prod_i\det(1-\phi\,T \mid F_i)^{(-1)^{i+1}}.\tag{5}\]
Consider the exact sequence of étale \(\mathbb{Z}_\ell\)-sheaves on \(X\) given by \[0\longrightarrow\mathbb{Z}_\ell\overset{\times\ell}\longrightarrow\mathbb{Z}_\ell\longrightarrow\mathbb{F}_\ell\longrightarrow 0.\] Its long exact cohomology sequence breaks into short exact sequences of \(\mathbb{F}_\ell[\phi]\)-modules \[0 \longrightarrow H^i_c(\bar{X},\mathbb{Z}_\ell)\otimes\mathbb{F}_\ell \longrightarrow H^i_c(\bar{X},\mathbb{F}_\ell) \longrightarrow H^{i+1}_c(\bar{X},\mathbb{Z}_\ell)[\ell] \longrightarrow 0\] or equivalently \[\label{lem:short-exact-with-F95ell-cohomology} 0 \longrightarrow(T_i\oplus F_i)\otimes\mathbb{F}_\ell \longrightarrow H^i_c(\bar{X},\mathbb{F}_\ell) \longrightarrow T_{i+1}[\ell] \longrightarrow 0.\tag{6}\]
For each finite \(\mathbb{F}_\ell[\phi]\)-module \(M\), let \[\Lambda(M):=\det(1-\phi\,T\mid M)\in 1+T\cdot \mathbb{F}_\ell[T].\] Observe that the exactness of 6 implies that \[\Lambda(H^i_c(\bar{X},\mathbb{F}_\ell)) = \Lambda((T_i\oplus F_i)\otimes\mathbb{F}_\ell)\cdot \Lambda(T_{i+1}[\ell]) = \Lambda(F_i\otimes\mathbb{F}_\ell)\cdot \Lambda(T_i\otimes\mathbb{F}_\ell)\cdot \Lambda(T_{i+1}[\ell]).\] Moreover, Proposition 3 (in Appendix 3.1) implies that \[\Lambda(T_{i}\otimes\mathbb{F}_\ell) = \Lambda(T_{i}[\ell]).\]
Deduce that \[\prod_i(\Lambda(T_i\otimes\mathbb{F}_\ell)\cdot\Lambda(T_{i+1}[\ell]))^{(-1)^{i+1}} = \prod_i\Lambda(T_i[\ell])^{(-1)^{i+1}}\cdot\prod_i\Lambda(T_{i+1}[\ell])^{(-1)^{i+1}} = 1\] (since the product telescopes) and \[\label{eqn:comparing-F95ell-and-Z95ell-parts} \prod_i\Lambda(H^i_c(\bar{U},\mathbb{F}_\ell))^{(-1)^{i+1}} = \prod_i\Lambda(F_i\otimes\mathbb{F}_\ell)^{(-1)^{i+1}} \equiv \prod_i\det(1-\phi\,T\mid F_i)^{(-1)^{i+1}} \bmod\ell.\tag{7}\] The desired identity ?? now follows from 5 and 7 . ◻
Proposition 3. Suppose \(\mathbb{Z}_\lambda\) is a finite extension of \(\mathbb{Z}_\ell\) and \(A\) is a finite \(\mathbb{Z}_\lambda[\phi]\)-module. Let \(\lambda\in\mathbb{Z}_\lambda\) be a uniformizer and \(\mathbb{F}_\lambda:=\mathbb{Z}_\lambda/\lambda\mathbb{Z}_\lambda\) be the residue field. Let \(A[\lambda]\subseteq A\) be the \(\mathbb{Z}_\lambda[\phi]\)-submodule annihilated by \(\lambda\). Then \(A[\lambda]\) and \(A\otimes\mathbb{F}_\lambda:=A/\lambda A\) are finite \(\mathbb{F}_\lambda[\phi]\)-modules, and \[\label{eqn:A[lambda]-vs-A/lambdaA} \det(1-\phi\,T\mid A[\lambda]) = \det(1-\phi\,T\mid A\otimes\mathbb{F}_\lambda).\qquad{(3)}\]
Proof. Suppose \(n\geq 0\) and \(\lambda^n\mathbb{Z}_\lambda\subseteq\mathbb{Z}_\lambda\) is the annihilator of \(A\).
Observe that the proposition is true for \(n=0\) since \(A[\lambda]=A\otimes\mathbb{F}_\lambda=0\).
Suppose that \(n>0\) and that \[\label{eqn:inductive-identity} \det(1 - \phi\,T\mid B[\lambda])=\det(1 - \phi\,T\mid B\otimes\mathbb{F}_\lambda)\tag{8}\] for every finite \(\mathbb{Z}_\lambda[\phi]\)-module \(B\) annihilated by \(\lambda^{n-1}\mathbb{Z}_\lambda\).
Observe that \(B:=\lambda A\subseteq A\) and \(\lambda B\subseteq B\) are \(\mathbb{Z}_\lambda[\phi]\)-invariant. Moreover, 8 holds since \(\lambda^{n-1}\mathbb{Z}_\lambda\) annihilates \(B\).
Consider the commutative diagram of \(\mathbb{Z}_\lambda[\phi]\)-modules with injective/surjective morphisms and exact rows/columns \[\xymatrix{ B[\lambda]\ar@{^{(}->}[r]\ar@{^{(}->}[d] & A[\lambda]\ar@{->>}[r]\ar@{^{(}->}[d]\ar@{-->}[dr] & A[\lambda]/B[\lambda]\ar@{^{(}->}[d] \\ B\ar@{^{(}->}[r]\ar^{\times\lambda}@{->>}[d] & A\ar@{->>}[r]\ar^{\times\lambda}@{->>}[d] & A\otimes\mathbb{F}_\lambda\ar^{\times\lambda}@{->>}[d] \\ \lambda B\ar@{^{(}->}[r] & B\ar@{->>}[r] & B\otimes\mathbb{F}_\lambda }.\] Applying the snake lemma to the bottom two rows yields \[\label{eqn:result-of-snake-lemma} 0\longrightarrow B[\lambda]\longrightarrow A[\lambda]\longrightarrow A\otimes\mathbb{F}_\lambda\longrightarrow B\otimes\mathbb{F}_\lambda\longrightarrow 0\tag{9}\] is an exact sequence of \(\mathbb{F}_\lambda[\phi]\)-modules. In particular, 8 and 9 imply that \[\label{eqn:herbrand-quotients-wrt-A-and-B} \frac{\det(1 - \phi\,T\mid A\otimes\mathbb{F}_\lambda)}{\det(1 - \phi\,T\mid A[\lambda])} = \frac{\det(1 - \phi\,T\mid B[\lambda])\cdot\det(1 - \phi\,T\mid A\otimes\mathbb{F}_\lambda)}{\det(1 - \phi\,T\mid A[\lambda])\cdot \det(1 - \phi\,T\mid B\otimes\mathbb{F}_\lambda)} = 1\tag{10}\] as desired. ◻
Let \(j\colon U\to X\) be the inclusion of \(U:=X\smallsetminus Z\) and \(V:=\pi^{-1}(U)\). Observe that \[\label{eqn:Z40X44T41-as-product} \zeta(X,T)=\zeta(U,T)\zeta(Z,T)\tag{11}\] since \(X=U\sqcup Z\) and that \[\label{eqn:Z40Y44T41-as-product} \zeta(Y,T)=\zeta(V,T)\zeta(Z,T)\tag{12}\] since \(\pi\) induces an isomorphism \(\pi^{-1}(Z)\to Z\).
Let \(\phi\in\mathop{\mathrm{Gal}}(\bar\mathbb{F}_q/\mathbb{F}_q)\) be the geometric Frobenius. Recall that \(r:=|G|\) is prime and \(r\neq p\). Observe that \(\mathcal{P}_r:=\pi_*\mathbb{F}_r\) is a constructible sheaf of \(\mathbb{F}_r[G]\)-modules and that \[H^i_c(\bar{V},\mathbb{F}_r)\cong H^i_c(\bar{U},\mathcal{P}_r)\] as \(\mathbb{F}_r[\phi]\)-modules since \(\pi\) is finite (see [4]), hence \[\label{eqn:zeta40V44T41-mod-r} \zeta(V,T) \equiv \prod_i\det(1-\phi\,T\mid H^i_c(\bar{U},\mathcal{P}_r))\bmod r\tag{13}\] by Theorem 2.
Observe that \(j^*\mathcal{P}_r\) is a lisse \(\mathbb{F}_r[G]\)-sheaf on \(U\) of rank \(r\) and that its \(G\)-semisimplification \(\mathcal{S}_r\) satisfies \[\label{eqn:passage-to-semisimplification} \det(1-\phi\,T\mid H^i_c(\bar{U},\mathcal{P}_r)) = \det(1-\phi\,T\mid H^i_c(\bar{U},\mathcal{S}_r))\tag{14}\] for every \(i\). Also, \(\mathbb{F}_r\) is the only simple \(\mathbb{F}_r[G]\)-module, hence \(\mathcal{S}_r\cong\mathbb{F}_r^r\) as \(\mathbb{F}_r[G]\)-sheaves and \[\label{eqn:passage-to-power-of-trivial} \det(1-\phi\,T\mid H^i_c(\bar{U},\mathcal{S}_r)) = \det(1-\phi\,T\mid H^i_c(\bar{U},\mathbb{F}_r))^r\tag{15}\] for every \(i\). In particular, combining 13 , 14 , and 15 yields \[\zeta(V,T) \equiv \prod_i\det(1-\phi\,T\mid H^i_c(\bar{U},\mathbb{F}_r))^{r(-1)^{i+1}} \equiv \zeta(U,T)^r\bmod r\] which combines with 11 and 12 to yield \[\label{eqn:main-identity-restated} \zeta(Y,T) \equiv \zeta(U,T)^r\zeta(Z,T) \equiv \zeta(X,T)^r\zeta(Z,T)^{1-r} \bmod r\tag{16}\] as desired.
Suppose \(X,Y\) are proper, smooth, and geometrically connected curves over \(\mathbb{F}_q\). Let \[L(C,T):=\zeta(X,T)(1-T)(1-qT)\in 1+T\mathbb{Z}[T]\] be the numerator of \(\zeta(C,T)\) for each \(C\in\{X,Y\}\), and recall \[\deg(L(C,T))=2\cdot\mathop{\mathrm{genus}}(C).\] Observe that ?? (restated in 16 ) is equivalent to \[\label{eqn:L40Y44T41-modulo-r} L(Y,T)\equiv L(X,T)^r(1-T)^{1-r}(1-qT)^{1-r}\prod_{z\in|Z|}(1-T^{\deg(z)})^{r-1}\bmod r.\tag{17}\] Moreover, taking degrees of both sides of ?? yields the Riemann-Hurwitz formula: \[\label{eqn:riemann-hurwitz} 2\cdot\mathop{\mathrm{genus}}(Y) - 2 = r\cdot(2\cdot\mathop{\mathrm{genus}}(X)-2)+(r-1)\cdot\deg(Z).\tag{18}\]
Let \(\mathbb{F}_q\) be a finite field, \(\pi_1,\ldots,\pi_d\in\mathbb{F}_q[x]\) be distinct monic irreducibles, \(e_1,\ldots,e_d\in\mathbb{N}\) be positive, and \[f:=\prod_{i=1}^m\pi_i^{e_i},\;\mathop{\mathrm{rad}}(f):=\prod_{i=1}^m\pi_i\in\mathbb{F}_q[x].\] Let \(Z\subset\mathbb{A}^1\) be the zero locus of \(f\), and observe \(\deg(Z)=\deg(\mathop{\mathrm{rad}}(f))\).
Let \(r\) be a prime divisor of \(q-1\) and \(A/\mathbb{F}_q\) be the affine curve \(y^r=f(x)\). The group \(G:={\boldsymbol{\mu}}_r\subseteq\mathbb{F}_q^\times\) acts faithfully on \(A\) via \(\zeta(x,y)=(x,\zeta y)\) for \(\zeta\in G\), and the morphism \(A\to\mathbb{A}^1\) given by \((x,y)\mapsto x\) is the quotient map \(A\to A/G\).
Let \(A^\nu\to A\) be the normalization of \(A\) and \(A^\nu\to Y\) be the smooth completion of \(A^\nu\). The action \(G\curvearrowright A\) extends uniquely to (faithful) actions \(G\curvearrowright A^\nu\) and \(G\curvearrowright Y\) such that the morphisms \(A^\nu\to A\) and \(A^\nu\to Y\) are \(G\)-equivariant (see Propositions 5 and 6 in Appendix 6.2). The composed morphism \(A^\nu\to A\to\mathbb{A}^1\) is the quotient \(A^\nu\to A^\nu/G\). It extends uniquely to a morphism \(\pi\colon Y\to\P^1\), the quotient \(Y\to Y/G\).
Suppose \(f\) that is \(r\)th power free, that is, \(0<e_i<r\) for \(1\leq i\leq m\) so that \(N\to A\) is bijective (on points). Suppose that \(r\mid\deg(f)\) so that \(Z\subset\P^1\) is the ramification locus of \(\pi\colon Y\to\P^1\). Then \[\label{eqn:superelliptic-L-polynomial-modulo-r} L(Y,T)\equiv (1-T)^{2-2r}\prod_{i=1}^m(1-T^{\deg(\pi_i)})^{r-1}\bmod r\tag{19}\] by 17 since \(L(\P^1,T)=1\) and \(q\equiv 1\bmod r\). Moreover, \[\label{eqn:superelliptic-genus-formula} 2\cdot\mathop{\mathrm{genus}}(Y) = 2 + r(-2)+(r-1)\deg(Z) = (r-1)(\deg(\mathop{\mathrm{rad}}(f))-2)\tag{20}\] by 18 .
Suppose \(r=2\) (hence \(q\) is odd), and let \(g:=\mathop{\mathrm{genus}}(Y)\). Observe that \[\label{eqn:product-of-cyclotomics} (1-T)^2L(Y,T) \equiv \prod_{i=1}^m(1-T^{\deg(\pi_i)}) \equiv \zeta(Z,T)^{-1} \bmod 2\tag{21}\] by 19 (compare [7]) and \[\deg(f) = \deg(\mathop{\mathrm{rad}}(f)) = 2g + 2\] by 20 .
Proposition 4. Let \(\ell\in\mathbb{N}\) be a prime and \(\mathcal{D}\) be a finite submultiset of \(\mathbb{N}_{\ell'}:=\mathbb{N}\smallsetminus\ell\mathbb{N}\). Let \(Z\) be a finite \(\mathbb{F}_q\)-scheme and \(Z_n\) be its base change to \(\mathbb{F}_{q^n}\). Let \[e:=\mathop{\mathrm{lcm}}(\mathop{\mathrm{ord}}_\ell(\deg(z):z\in|Z|))\] and \(d:=\ell^e\). Then
\(\deg(z_d)\in\mathbb{N}_{\ell'}\) for every \(z_d\in|Z_d|\);
\(\zeta(Z,T)^{-1}\equiv\prod_{d\in\mathcal{D}}(1-T^d)\bmod \ell\) if and only if \(\mathcal{D}=\{\deg(z_d):z_d\in|Z_d|\}\) as multisets;
\(\mathop{\mathrm{ord}}_{T=1}(\zeta(Z,T)^{-1}\in\mathbb{F}_\ell[T])=||Z_d||\).
Proof. If \(z\in|Z|\) and \(e':=\mathop{\mathrm{ord}}_\ell(\deg(z))\), then \(z\) splits into \(\ell^{e'}\) points in \(Z_{\ell^e}\), all of degree \(\frac{1}{\ell^{e'}}\deg(z)\in\mathbb{N}_{\ell'}\), so [item:deg40z95d41-is-coprime-to-ell] holds. Proposition 7 (in Appendix 6.3) and Proposition 8 (in Appendix 6.4) then imply that part [item:detecting-Z40Z44T41] holds. Finally, \(\mathop{\mathrm{ord}}_{T=1}(1-T^m)=1\) for \(m\in\mathbb{N}_{\ell'}\), hence part [item:ord-T611-of-Z40Z44T41] holds. ◻
We give necessary and sufficient conditions for \[\label{eqn:trinomial-condition} L(Y,T) \equiv 1 + aT^g + q^gT^{2g}\in\mathbb{F}_2[T].\tag{22}\]
\(a=0\): Let \(2g=o2^e\) be the unique factorization in \(\mathbb{Z}\) with \(o\) odd. Observe \(e\geq 1\) and 22 is equivalent to \[(1+T)^2L(Y,T) \equiv (1+T)^2(1+T^{2g}) \equiv (1+T)^2(1+T^o)^{2^e}\bmod 2.\] This holds (by Proposition 4) if and only if \(f\) splits over \(\mathbb{F}_{q^{2^e}}\) into a product of two linears and \(2^e\) irreducibles of (odd) degree \(o\).
\(a=1\): Let \(g=o2^e\) be the unique factorization in \(\mathbb{Z}\) with \(o\) odd. Observe \(e\geq 0\) and 22 is equivalent to \[\label{eqn:L40Y44T41-mod-2-for-a611} (1+T)^2L(Y,T) \equiv (1+T)^2(1+T^g+T^{2g}) \equiv (1+T)^2(1+T^o+T^{2o})^{2^e}\bmod 2.\tag{23}\] Moreover, \(\mathop{\mathrm{ord}}_{T=1}(1+T^o+T^{2o})=0\) since \(o\) is odd. Therefore 21 and Proposition 4 imply that 23 holds if and only if \(f\) factors over \(\mathbb{F}_{q^{2^e}}\) as a product of two irreducibles of odd degrees \(d_1\leq d_2\) and \[\begin{align} (1+T^2)(1+T^g+T^{2g}) & \equiv & 1+T^2+T^g+T^{g+2}+T^{2g-2}+T^{2g} \\ &\equiv & 1+T^{d_1}+T^{d_2}+T^{d_1+d_2}\bmod 2. \end{align}\] The last equivalence holds if and only if \((g,e,d_1,d_2)\) equals \((1,0,1,3)\) or \((2,1,3,3)\).
Proposition 5. Let \(S\) be an integral scheme and \(\nu\colon S^\nu\to S\) be a normalization morphism. There is a unique homomorphism \(\nu^*\colon\mathop{\mathrm{Aut}}(S)\to\mathop{\mathrm{Aut}}(S^\nu)\) given by \(\alpha\mapsto\alpha^\nu\) and making \[\label{eqn:normalization-square} \xymatrix{ S^\nu\ar[d]_\nu & S^\nu\ar[l]_{\alpha^\nu}\ar[d]^\nu \\ S & S\ar[l]_{\alpha} }\qquad{(4)}\] commute for every \(\alpha\in\mathop{\mathrm{Aut}}(S)\).
Proof. By definition of a normalization morphism (see [8]), \(S^\nu\) is normal and every dominant morphism \(t\colon T\to S\) with \(T\) normal factors uniquely through \(\nu\): \[\xymatrix{ S^\nu\ar[d]_\nu & T\ar[l]_{t^\nu}\ar[dl]^t \\ S }\] In particular, if \(\alpha\in\mathop{\mathrm{Aut}}_k(S)\) and \(t:=\alpha\circ\nu\colon S^\nu\to S\), then \(\alpha^\nu:=t^\nu\colon S^\nu\to S^\nu\) is the unique morphism making ?? commute. Uniquenesses forces \(\mathop{\mathrm{od}}_S^\nu=\mathop{\mathrm{od}}_{S^\nu}\) and \((\beta^{-1})^\nu\circ\alpha^\nu=(\beta^{-1}\circ\alpha)^\nu\) for each \(\alpha,\beta\in\mathop{\mathrm{Aut}}(S)\). Also, if \(\alpha\in\mathop{\mathrm{Aut}}(S)\), then \(\alpha^\nu\in\mathop{\mathrm{Aut}}(S^\nu)\) since \((\alpha^{-1})^\nu\circ\alpha^\nu=\mathop{\mathrm{od}}_{S^\nu}=\alpha^\nu\circ(\alpha^{-1})^\nu\). Therefore \(\alpha\mapsto\alpha^\nu\) gives a group homomorphism \(\mathop{\mathrm{Aut}}(S)\to\mathop{\mathrm{Aut}}(S^\nu)\). ◻
Proof. ◻
Proposition 6. Let \(k\) be a field and \(S\) be a smooth connected \(k\)-curve. Let \(S\to S^\kappa\) be the smooth completion of \(S^\nu\). There is a unique homomorphism \(\kappa^*\colon\mathop{\mathrm{Aut}}_k(S)\to\mathop{\mathrm{Aut}}_k(S^\kappa)\) given by \(\alpha\mapsto\alpha^\kappa\) and making \[\xymatrix{ S^\kappa\ar[d]_\kappa & S^\kappa\ar[l]_{\alpha^\kappa}\ar[d]^\kappa \\ S & S\ar[l]_{\alpha} }\] commute for every \(\alpha\in\mathop{\mathrm{Aut}}_k(S)\).
Proof. Suppose \(\alpha,\beta\in\mathop{\mathrm{Aut}}_k(S)\). Observe that \(\alpha\) represents a unique birational \(k\)-map \([\alpha]\colon S^\kappa\dashrightarrow S^\kappa\) since \(S,S^\kappa\) are birational. Moreover, \([\alpha]\) extends uniquely to a \(k\)-morphism \(\alpha^\kappa\in S^\kappa\to S^\kappa\) since \(S^\kappa\) is smooth and complete (see [9] or [10]). Uniquenesses forces \(\mathop{\mathrm{od}}_S^\kappa=\mathop{\mathrm{od}}_{S^\kappa}\) as well as \((\beta^{-1})^\kappa\circ\alpha^\kappa=(\beta^{-1}\circ\alpha)^\kappa\) and \(\alpha^\kappa\in\mathop{\mathrm{Aut}}_k(S^\kappa)\). Therefore \(\alpha\mapsto\alpha^\kappa\) gives a group homomorphism \(\mathop{\mathrm{Aut}}_k(S)\to\mathop{\mathrm{Aut}}_k(S^\kappa)\). ◻
Proposition 7. Let \(\mathbb{F}_q\) be a finite field characteristic \(p\) and \(\ell\neq p\) be a rational prime. Let \(S\) be a separated \(\mathbb{F}_q\)-scheme of finite type and \(S_n\) be its base change to \(\mathbb{F}_{q^n}\). Then \(S=S_1\) and \[\zeta(S_\ell,T)\equiv\zeta(S_1,T)\bmod\ell.\]
Proof. Let \(\phi\in\mathop{\mathrm{Gal}}(\bar\mathbb{F}_q/\mathbb{F}_q)\) be the geometric Frobenius and \[\zeta_i(S_n,T) := \det(1 - \phi^n\,T\mid H^i_c(\bar{S},\mathbb{F}_\ell))\in 1+T\cdot\mathbb{F}_\ell[T]\] so that Theorem 2 implies \[\label{eqn:base-change-zeta-as-product} \zeta(S_n,T) \equiv \prod_i\zeta_i(S_n,T)^{(-1)^{i+1}}\bmod\ell.\tag{24}\] Let \(d_i:=\deg(\zeta_i(S_1,T))\) and \[\zeta_i(S_1,T)=\prod_{j=1}^{d_i}(1-\alpha_{i,j}T)\in\bar\mathbb{F}_\ell[T]\] be a factorization. Observe that \[\label{eqn:congruence-of-factors} \zeta_i(S_\ell,T) \equiv \prod_{j=1}^{d_i}(1-\alpha_{i,j}^\ell T) \equiv \prod_{j=1}^{d_i}(1-\alpha_{i,j} T) \equiv \zeta_i(S_1,T) \bmod\ell\tag{25}\] since \(\psi_i(S_\ell,T)\in\mathbb{F}_\ell[T]\). Then 24 and 25 imply \[\zeta(S_\ell,T) \equiv \prod_i\zeta_i(S_\ell,T)^{(-1)^{i+1}} \equiv \prod_i\zeta_i(S_1,T)^{(-1)^{i+1}} \equiv \zeta(S_1,T) \bmod\ell\] as desired. ◻
Proposition 8. Let \(\ell\in\mathbb{N}\) be a prime, \(\mathcal{D},\mathcal{E}\) be finite submultisets of \(\mathbb{N}_{\ell'}:=\mathbb{N}\smallsetminus\ell\mathbb{N}\), and \[\psi_\mathcal{D}:=\prod_{d\in\mathcal{D}}(1-T^{d}) \text{ and } \psi_{\mathcal{E}}:=\prod_{e\in\mathcal{E}}(1-T^{e})\] If \(\psi_\mathcal{D}=\psi_E\), then \(\mathcal{D}=\mathcal{E}\) as multisets.
Proof. We induct on \(|\mathcal{D}|,|\mathcal{E}|\). For the base case, observe \(\mathcal{D}=\emptyset\) if and only if \(\mathcal{E}=\emptyset\). Suppose that \(\mathcal{D},\mathcal{E}\) are both nonvoid and that \(\psi_{\mathcal{D}}=\psi_{\mathcal{E}}\). We show that \(d:=\max(\mathcal{D})\) equals \(e:=\max(\mathcal{E})\) and induct.
Let \(\psi_n\in\mathbb{F}_r[T]\) be the cyclotomic polynomial \[\psi_n:=\prod_{m\mid n}(1-T^m)^{\mu(n/m)}.\] Observe that \(\psi_m,\psi_n\) are coprime when \(m,n\in\mathbb{N}_{\ell'}\) are distinct, hence \[d=\max\{n\in\mathbb{N}_{r'}:\gcd(\psi_n,\psi_{\mathcal{D}})\neq 1\}.\] Deduce \(d\geq\max(\mathcal{E})\) since \(\psi_d\mid\psi_{\mathcal{E}}=\psi_{\mathcal{D}}\) for \(e\in\mathcal{E}\). A symmetric argument implies that \[e=\max(\mathcal{E})\geq\max(\mathcal{D})=d,\] hence \(d=e\).
Let \(\mathcal{D}':=\mathcal{D}\smallsetminus\{d\}\) and \(\mathcal{E}':=\mathcal{E}\smallsetminus\{e\}\). Observe that the identities \[\psi_{\mathcal{D}'}\cdot(1-T^d) = \psi_{\mathcal{D}} = \psi_{\mathcal{E}} = \psi_{\mathcal{E}'}\cdot(1-T^e)\] imply \(\psi_{\mathcal{D}'}=\psi_{\mathcal{E}'}\). Deduce \(\mathcal{D}'=\mathcal{E}'\) as multisets (by induction) and \(\mathcal{D}=\mathcal{E}\) as multisets. ◻