Is Hom scheme between projective curves of large genus finite etale?












2














Let $K$ be a number field, $T$ be a finite set containing some finite places of $K$, and $S=operatorname{Spec} O_{K,T}$. If $X,Y$ are two projective smooth curves over $S$ with genus large than $1$ and $0$ respectively. Then is the Hom scheme $Hom(X,Y)$ finite etale over $S$?



Motivation: Let $E$ be an elliptic curve over $mathbb Q$, then $E$ can be defined over $mathbb Z[1/N]$ where $N$ is the conductor of $E$. We know there is a non-constant map $phi:X_0(N) rightarrow E$ over $mathbb Q$ where the modular curve $X_0(N)$ is also defined over $mathbb Z[1/N]$, then can $phi$ also be defined over $mathbb Z[1/N]$?






number-theory algebraic-geometry arithmetic-geometry







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  • If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
    – Ariyan Javanpeykar
    Dec 12 at 19:58










  • @AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
    – zzy
    Dec 12 at 20:08












  • You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
    – Ariyan Javanpeykar
    Dec 12 at 20:15










  • What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
    – Ariyan Javanpeykar
    Dec 12 at 20:16










  • @AriyanJavanpeykar Thank you！The reference is useful.
    – zzy
    Dec 12 at 22:33
















2














Let $K$ be a number field, $T$ be a finite set containing some finite places of $K$, and $S=operatorname{Spec} O_{K,T}$. If $X,Y$ are two projective smooth curves over $S$ with genus large than $1$ and $0$ respectively. Then is the Hom scheme $Hom(X,Y)$ finite etale over $S$?



Motivation: Let $E$ be an elliptic curve over $mathbb Q$, then $E$ can be defined over $mathbb Z[1/N]$ where $N$ is the conductor of $E$. We know there is a non-constant map $phi:X_0(N) rightarrow E$ over $mathbb Q$ where the modular curve $X_0(N)$ is also defined over $mathbb Z[1/N]$, then can $phi$ also be defined over $mathbb Z[1/N]$?






number-theory algebraic-geometry arithmetic-geometry







share|cite|improve this question
























  • If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
    – Ariyan Javanpeykar
    Dec 12 at 19:58










  • @AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
    – zzy
    Dec 12 at 20:08












  • You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
    – Ariyan Javanpeykar
    Dec 12 at 20:15










  • What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
    – Ariyan Javanpeykar
    Dec 12 at 20:16










  • @AriyanJavanpeykar Thank you！The reference is useful.
    – zzy
    Dec 12 at 22:33














2












2








2


1





Let $K$ be a number field, $T$ be a finite set containing some finite places of $K$, and $S=operatorname{Spec} O_{K,T}$. If $X,Y$ are two projective smooth curves over $S$ with genus large than $1$ and $0$ respectively. Then is the Hom scheme $Hom(X,Y)$ finite etale over $S$?



Motivation: Let $E$ be an elliptic curve over $mathbb Q$, then $E$ can be defined over $mathbb Z[1/N]$ where $N$ is the conductor of $E$. We know there is a non-constant map $phi:X_0(N) rightarrow E$ over $mathbb Q$ where the modular curve $X_0(N)$ is also defined over $mathbb Z[1/N]$, then can $phi$ also be defined over $mathbb Z[1/N]$?






number-theory algebraic-geometry arithmetic-geometry







share|cite|improve this question















Let $K$ be a number field, $T$ be a finite set containing some finite places of $K$, and $S=operatorname{Spec} O_{K,T}$. If $X,Y$ are two projective smooth curves over $S$ with genus large than $1$ and $0$ respectively. Then is the Hom scheme $Hom(X,Y)$ finite etale over $S$?



Motivation: Let $E$ be an elliptic curve over $mathbb Q$, then $E$ can be defined over $mathbb Z[1/N]$ where $N$ is the conductor of $E$. We know there is a non-constant map $phi:X_0(N) rightarrow E$ over $mathbb Q$ where the modular curve $X_0(N)$ is also defined over $mathbb Z[1/N]$, then can $phi$ also be defined over $mathbb Z[1/N]$?






number-theory algebraic-geometry arithmetic-geometry




number-theory algebraic-geometry arithmetic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 at 20:52

























asked Dec 9 at 19:08









zzy

2,3571419




2,3571419












  • If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
    – Ariyan Javanpeykar
    Dec 12 at 19:58










  • @AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
    – zzy
    Dec 12 at 20:08












  • You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
    – Ariyan Javanpeykar
    Dec 12 at 20:15










  • What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
    – Ariyan Javanpeykar
    Dec 12 at 20:16










  • @AriyanJavanpeykar Thank you！The reference is useful.
    – zzy
    Dec 12 at 22:33


















  • If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
    – Ariyan Javanpeykar
    Dec 12 at 19:58










  • @AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
    – zzy
    Dec 12 at 20:08












  • You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
    – Ariyan Javanpeykar
    Dec 12 at 20:15










  • What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
    – Ariyan Javanpeykar
    Dec 12 at 20:16










  • @AriyanJavanpeykar Thank you！The reference is useful.
    – zzy
    Dec 12 at 22:33
















If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
– Ariyan Javanpeykar
Dec 12 at 19:58




If $Y$ is an elliptic curve, then this is false. Let $f:Xto Y$ be a morphism. Now, any isogeny $phi:Yto Y$ defines a new morphism $phi circ f:Xto Y$ and the degree of this map increases. Thus, $Hom(X,Y)$ is not even finite etale over $mathbb{C}$. Anyway, if genus(Y) > 1, then $Hom(X,Y)$ is finite (assuming $X$ and $Y$ are projective smooth curves of genus >1). It is not etale in general, as there are maps $Xto Y$ over $mathbb{F}_p$ which can't be lifted in general. (Hint: find a curve $X$ and an automorphism over $mathbb{F}_p$ of order $> 84(g-1)$.)
– Ariyan Javanpeykar
Dec 12 at 19:58












@AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
– zzy
Dec 12 at 20:08






@AriyanJavanpeykar Oh, thank you! What I thought before is if $Hom$ is finite etale over $O_{K,T}$, the existence of a $K$ point will imply $Hom$ has an irreducible component isomorphic to $O_{K,T}$ thus $Hom$ has a $O_{K,T}$ point.
– zzy
Dec 12 at 20:08














You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
– Ariyan Javanpeykar
Dec 12 at 20:15




You're welcome. I think that the hom-scheme will complicate things here a bit. It is not of finite type over $O_{K,T}$. However, the hom-scheme satisfies the valuative criterion for properness. Indeed, any morphism $X_Kto Y_K$ with $X$ and $Y$ smooth proper curves of genus $>0$ over $O_{K,T}$ extends (uniquely) to a morphism $Xto Y$. You can read Liu-Tong's paper on Neron models for a proof.
– Ariyan Javanpeykar
Dec 12 at 20:15












What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
– Ariyan Javanpeykar
Dec 12 at 20:16




What you ask in your motivation has nothing to do with the "finite etaleness" of the Hom-scheme. The extension you desire exists by the Neron mapping property of a smooth proper model of E over ℤ[1/N] (and the fact that X0(N) is smooth over ℤ[1/N]). (Thus: the hom-scheme satisfies the valuative criterion of properness. It is not necessarily proper though, because it is not of finite type.)
– Ariyan Javanpeykar
Dec 12 at 20:16












@AriyanJavanpeykar Thank you！The reference is useful.
– zzy
Dec 12 at 22:33




@AriyanJavanpeykar Thank you！The reference is useful.
– zzy
Dec 12 at 22:33















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