Selmer group associated to the Chow group of certain codimension two cycles

TL;DR

This paper investigates the Galois-invariant algebraically trivial cycles on certain surfaces over number fields, demonstrating that a specific quotient related to these cycles is a torsion group, thus contributing to the understanding of their arithmetic properties.

Contribution

It proves that the Galois-invariant cycles' quotient group is torsion for surfaces with geometric genus and irregularity zero over number fields.

Findings

The quotient of Galois-invariant cycles is torsion.

The result applies to surfaces with geometric genus and irregularity zero.

Provides insight into the arithmetic of algebraically trivial cycles.

Abstract

Let $X$ be a surface with geometric genus and irregularity zero which is defined over a number field $K$. Let $\mathscr{X}$ denote a smooth spread of $X$ over the spectrum of a Zariski open subset in the spectrum of the ring of integers and $A^2$ stands for the group of algebraically trivial cycles on schemes modulo rational equivalence. If $j^*: A^2(\mathscr{X})\to A^2(X)$ be the flat pull-back corresponding to the embedding $j:X\hookrightarrow \mathscr{X}$ then we prove that $\im(j^*)(K)/A^2(\mathscr{X})(K)$ is a torsion group. Here $\im(j^*)(K)$, $A^2(\mathscr{X})(K)$ stand for the cycles fixed under the action of the absolute Galois group.

Equations

$$\Gamma .(x_0{\times }_{{\mathcal{O}}_K}\mathcal{X})-\Gamma .(x_1{\times }_{{\mathcal{O}}_K}\mathcal{X})=z_1-z_2.$$

$$X{\times }_K\bar{K}$$

$${\mathcal{X}}_{\bar{K}}:=\mathcal{X}{\times }_{{\mathcal{O}}_K}\times \overline{{\mathcal{O}}_K}.$$

$$j^{\ast }:A^2({\mathcal{X}}_{\bar{K}})\to A^2(X_{\bar{K}})$$

$$j^{\ast }:A^2({\mathcal{X}}_{\bar{K}})(K)\to A^2(X_{\bar{K}})(K).$$

$$j:X_{\bar{K}}\to {\mathcal{X}}_{\bar{K}}.$$

$$H^1(G,{\Sigma }_{\bar{K}}[6N])\to H^1(I_v,{\Sigma }_v[6N])$$

$$\mathrm{im}(j^{\ast }[6N])(K)/A^2({\mathcal{X}}_{\bar{K}})[6N](K)$$

$$\mathrm{Pic}(X)\to \mathrm{Pic}(X_{\bar{K}})\to \mathrm{NS}(X_{\bar{K}})$$

$$H^2({\mathcal{X}}_{{\mathcal{O}}_K[1/f],\mathfrak{p}},{\mathcal{O}}_{{\mathcal{X}}_{{\mathcal{O}}_K[1/f]},\mathfrak{p}})=0$$

$$X(K)\ne \varnothing .$$

$$\mathrm{im}(j^{\ast })(K)/A^2({\mathcal{X}}_{\overline{{\mathcal{O}}_K[1/f]}})(K)$$

$$A^2({\mathcal{X}}_{\overline{{\mathcal{O}}_K[1/f]}})(K){\otimes }_{\mathbb{Z}}\mathbb{Q}=\{0\}.$$

$${\mathcal{E}}_{{\mathcal{O}}_K[1/6N]}\times {\mathcal{E}}_{{\mathcal{O}}_K[1/6N]}$$

$$0\to {\Sigma }_K\to A^2(\mathcal{X})\to A^2(X).$$

$$0\to {\Sigma }_{\bar{K}}\to A^2({\mathcal{X}}_{\bar{K}})\to A^2(X_{\bar{K}}).$$

$$0\to {\Sigma }_{\bar{K}}^G\to A^2({\mathcal{X}}_{\bar{K}}{)}^G\to \mathrm{im}(j^{\ast }{)}^G\to H^1(G,{\Sigma }_{\bar{K}})\to H^1(G,A^2({\mathcal{X}}_{\bar{K}}))\to H^1(G,\mathrm{im}(j^{\ast })).$$

$$\{m\in M|g.m=m,\forall g\in G\}.$$

$$A^2(X_{\bar{K}}{)}^G,A^2({\mathcal{X}}_{\bar{K}}{)}^G,\mathrm{im}(j^{\ast }{)}^G$$

$$A^2(X_{\bar{K}})(K),A^2({\mathcal{X}}_{\bar{K}})(K),\mathrm{im}(j^{\ast })(K)$$

$$0\to \mathrm{im}(j^{\ast })(K)/A^2({\mathcal{X}}_{\bar{K}})(K)\to H^1(G,{\Sigma }_{\bar{K}})\to H^1(G,A^2({\mathcal{X}}_{\bar{K}}))\to H^1(G,\mathrm{im}(j^{\ast })).$$

$$H^1(G,{\Sigma }_{\bar{K}})\to H^1(G_v,A^2({\mathcal{X}}_{\overline{{\mathcal{O}}_{K_v}[1/6N]}})).$$

$$\mathrm{im}(j^{\ast })(K)/A^2({\mathcal{X}}_{\bar{K}})(K)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H^1(G,{\Sigma }_{\bar{K}})\to \prod _v{H}^1(G_v,A^2({\mathcal{X}}_{{\bar{K}}_v}))$$

$$0\to {\Sigma }_{\bar{K}}[n]\to A^2({\mathcal{X}}_{\bar{K}})[n]\to A^2(X_{\bar{K}})[n].$$

$$H^1(G,{\Sigma }_{\bar{K}}[n])\to \prod _v{H}^1(G_v,A^2({\mathcal{X}}_{{\bar{K}}_v}))[n].$$

$$j^{\ast }:A^2({\mathcal{X}}_{\bar{K}})[n](K)\to A^2(X_{\bar{K}})[n](K)$$

$$H^1(G,Alb(X_{\bar{K}})[n])\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$S^n(Alb(X_{\bar{K}})/K):=\ker (H^1(G_K,Alb(X_{\bar{K}})[n])\to \prod _v{H}^1(G_{K_v},Alb(X_{{\bar{K}}_v})[n]))$$

$$S^n(A^2(X_{\bar{K}})/K):=\ker (H^1(G_K,A^2(X_{\bar{K}})[n])\to \prod _v{H}^1(G_{K_v},A^2(X_{{\bar{K}}_v})[n])).$$

Full text

1. Introduction

Suppose $X$ be a smooth projective surface defined over a number field $K$ and assume that it can be spread out to a smooth projective scheme ${\mathscr{X}}$ over an affine open subset of the spectrum of the number ring ${\mathscr{O}}_{K}$. Let $A^{2}(X)$ denote the group of algebraically trivial cycles of codimension $2$ modulo rational equivalence on $X$.

We recall the definition of algebraic equivalence over ${\mathscr{O}}_{K}$, which will be used in the sequel. Let us consider the free abelian group of codimension two cycles on ${\mathscr{X}}$. Two cycles $z_{1},z_{2}$ are said to be algebraically equivalent if there exists a smooth projective curve $C$ defined over ${\mathscr{O}}_{K}$, two scheme theoretic points $x_{0},x_{1}$ on $C$ and a relative correspondence $\Gamma$ on $C\times_{{\mathscr{O}}_{K}}{\mathscr{X}}$, such that the intersection

[TABLE]

Here . denotes the relative intersection product in the sense of [Fu].

Let $\bar{K}$ denote the algebraic closure of $K$ and $G=$ Gal $({\bar{K}}/K)$ be the absolute Galois group. Also, $X_{\bar{K}}$ denotes the surface

[TABLE]

and

[TABLE]

Here $\overline{{\mathscr{O}}_{K}}$ denotes the integral closure of ${\mathscr{O}}_{K}$ in $\bar{K}$.

Let $A^{2}({\mathscr{X}}_{\bar{K}})(K),A^{2}(X_{\bar{K}})(K)$ be the $G$-fixed part of the action of $G$ on $A^{2}({\mathscr{X}}_{\bar{K}}),A^{2}(X_{\bar{K}})$ respectively, and $j$ be the embedding of $X_{\bar{K}}$ into ${\mathscr{X}}_{\bar{K}}$. If one considers the flat pullback

[TABLE]

of codimension $2$-cycles, it gives a map

[TABLE]

Then a general question is: what is the cokernel of this homomorphism?

Mildenhall [Mil] studied this flat pull-back for $X$ to be the self-product of an elliptic curve admitting a complex multiplication. He has shown that the kernel of this flat-pullback over any number field $K$ is finite. We use Mildenhall’s result to derive the information about the quotient ${\rm im}(j^{*})(K)[n]/A^{2}({\mathscr{X}})(K)[n]$ at the level of $n$-torsions of this homomorphism for the case where $X$ is the self-product of a CM elliptic curve.

Let $E$ be an elliptic curve with complex multiplication by the ring of integers of a number field $K$ and $N$ be it’s discriminant (for this we fix an Weirstrass equation for $E$ once and for all). We denote $E\times E$ by $X$ and suppose that there exists a smooth spread of $X$, say ${\mathscr{X}}$ defined over ${\mathscr{O}}_{K}[1/6N]$. Let $j^{*}$ be the flat pull-back at the level of $A^{2}$ induced by the embedding

[TABLE]

Then the main result is:

Theorem 2.2: The group $S^{6N}(\Sigma_{\bar{K}})$ is a colimit of the unramified cohomology groups

[TABLE]

here $G$ is the Galois group $Gal(\bar{K}/L)$ for a finite extension $L$ of $K$ and $I_{v}$ is the inertia subgroup of the Galois group $G$ corresponding to a finite place $v$ and hence

[TABLE]

is a colimit of unramified cohomology groups.

The main tools used here are the Galois module structure of the Chow group of $X_{\bar{K}}$ and that of ${\mathscr{X}}_{\bar{K}}$ and the Galois cohomology of the groups $A^{2}({\mathscr{X}}_{\bar{K}})$ and that of $A^{2}(X_{\bar{K}})$. The proof involves similar techniques as to show that the Selmer group of an abelian variety defined over a number field is finite.

Coombes [Co] proved that for a surface $X$ with geometric genus and irregularity zero, $A^{2}(X)$ is finite under the assumption that $A^{2}(X_{\bar{K}})=0$.

Towards proving our main result Theorem 2.2, we start with a surface $X$ of geometric genus and irregularity zero defined over $K$ which satisfies the condition that the map:

[TABLE]

is surjective,

[TABLE]

for all $\mathfrak{p}$ in ${\rm{Spec}}({\mathscr{O}}_{K}[1/f])$, here $f$ is some element in ${\mathscr{O}}_{K}$ and

[TABLE]

Then we prove by using the result of [CTR][lemma 3.3], that:

Theorem 2.6: The group

[TABLE]

is a torsion group.

Here we do not assume the vanishing of $A^{2}(X_{\bar{K}})$ as in [Co]. The above theorem is important to prove the finiteness or triviality of $A^{2}(X_{\bar{K}})(K)$. It says that, atleast to prove that ${\rm im}(j^{*})(K)\otimes_{{\mathbb{Z}}}{\mathbb{Q}}=\{0\}$ it is enough to prove that

[TABLE]

This gives some information on how to prove $A^{2}(X)$ is finite.

Acknowledgements: The authors thank department of atomic energy (DAE) for funding this project and for the hospitality of Harish-Chandra Research Institute, India, where the work has been done. The first author also thanks VIT University Chennai for hosting this project.

2. Proof of the theorems

Let $E$ be as before having complex multiplication by ${\mathscr{O}}_{K}$ and $X=E\times E$. Let us fix a Weirstrass equation for the elliptic curve once and for all. Let us consider a spread ${\mathscr{E}}_{{\mathscr{O}}_{K}[1/6N]}$ of $E$ over ${\mathscr{O}}_{K}[1/6N]$ which is smooth and denote

[TABLE]

by ${\mathscr{X}}$. Then we consider the restriction homomorphism from $A^{2}({\mathscr{X}})\to A^{2}(X)$. It is known due to Mildenhall’s result that the kernel of this restriction map is finite and we name it $\Sigma_{K}$. Then we have the exact sequence

[TABLE]

Now consider the sequence at the level of $\bar{K}$, namely

[TABLE]

Note that $G$ acts naturally on each member of the above short exact sequence. Also if the inclusion of $X_{\bar{K}}\hookrightarrow{\mathscr{X}}_{\bar{K}}$ be denoted by $j$, then the pullback map $j^{*}$ is from $A^{2}({\mathscr{X}}_{\bar{K}})$ to $A^{2}(X_{\bar{K}})$.

Therefore one has the natural long exact sequence on the group cohomology level of $G$ for these Galois modules,

[TABLE]

Here $M^{G}$, for a $G$-module $M$, denotes the group of $G$-invariants in $M$, i.e.

[TABLE]

Let us denote the groups

[TABLE]

as

[TABLE]

respectively. Also for notational convenience we continue to denote $\Sigma_{\bar{K}}^{G}$ as $\Sigma_{K}$. Then we have the following exact sequence

[TABLE]

Let $v$ be a place of $K$ and $K_{v}$ be the completion of $K$ at $v$. Let $\bar{K}_{v}$ be the algebraic closure of $K_{v}$ and we embed $\bar{K}$ into $\bar{K}_{v}$. This embedding gives an injection of ${\rm{Gal}}(\bar{K}_{v}/K_{v})=G_{v}$ into ${\rm{Gal}}(\bar{K}/K)=G$ and consequently a homomorphism (considering the Galois cohomology)

[TABLE]

Again for notational convenience we write $A^{2}({\mathscr{X}}_{\overline{{\mathscr{O}}_{K_{v}}[1/6N]}})$ in the above as $A^{2}({\mathscr{X}}_{\bar{K}_{v}})$. Then we have the following commutative diagrams:

[TABLE]

Let us now focus on

[TABLE]

and consider the sequence of $n$-torsion subgroups of $\Sigma_{\bar{K}},A^{2}({\mathscr{X}}_{\bar{K}}),A^{2}(X_{\bar{K}})$ given by:

[TABLE]

Here $A[n]$ for an abelian group $A$, denotes the group of $n$-torsions of $A$. Then considering the above groups as $G$-modules we have a homomorphism at the level of Galois cohomology given by:

[TABLE]

Definition 2.1**.**

The kernel of this map is defined to be the $n$-Selmer group associated to the restriction homomorphism $A^{2}({\mathscr{X}}_{\bar{K}})\to A^{2}(X_{\bar{K}})$, at the level of $n$-torsions in the group of algebraically trivial codimension $2$-cycles and it is denoted by $S^{n}(\Sigma_{\bar{K}})$.

Let $Alb(X_{\bar{K}})$ be the Albanese variety such that there exists a natural (universal) homomorphism of abelian groups from $A^{2}(X_{\bar{K}})$ to $Alb(X_{\bar{K}})$. Now $X_{\bar{K}}$ is an abelian variety $Alb(X_{\bar{K}})\cong X_{\bar{K}}$. Since the following argument is more general in nature, that is, it works for any smooth projective $X_{\bar{K}}$ and for its albanese variety $Alb(X_{\bar{K}})$, provided the kernel of

[TABLE]

is finite, we do not use the isomorphism $Alb(X_{\bar{K}})\cong X_{\bar{K}}$. Specifically this is required to prove the analogous result as stated in Remark 2.5.

Let’s consider the commutative diagram:

[TABLE]

Now by Roitman’s theorem [R2], the groups $Alb(X_{\bar{K}})[n]$ and $A^{2}(X_{\bar{K}})[n]$ are isomorphic as Galois modules and therefore the group cohomologies are isomorphic. Thus the left vertical arrow in the above diagram is an isomorphism. Let

[TABLE]

here $v$ varies over all finite places of $K$. Similarly

[TABLE]

If we take an element in $S^{n}(Alb(X_{\bar{K}})/K)$, then by the commutativity of the above diagram, the image of the element under the left vertical homomorphism is in $S^{n}(A^{2}(X_{\bar{K}})/K)$. Now we prove our main result which has already been stated in the introduction.

Theorem 2.2**.**

The group $S^{6N}(\Sigma_{\bar{K}})$ is a colimit of the unramified cohomology groups

[TABLE]

here $G$ is the Galois group $Gal(\bar{K}/L)$ for a finite extension $L$ of $K$ and $I_{v}$ is the inertia subgroup of the Galois group $G$ corresponding to a finite place $v$ and hence

[TABLE]

is a colimit of unramified cohomology groups.

Proof.

Let $n$ be a positive integer. Let us consider the diagram

[TABLE]

Suppose that some element is there in $S^{n}(\Sigma_{\bar{K}}/K)$.

Consider the following commutative squares:

[TABLE]

Let $\eta$ be in the kernel of

[TABLE]

Then it follows by the exactness of the sequence, induced by the long exact sequence corresponding to the short exact sequence of Galois modules given as :

[TABLE]

[TABLE]

that there exists an element $z$ in $A^{2}({\mathscr{X}}_{K_{v}})(K_{v})$ such that

[TABLE]

for all $\sigma$ in $G_{v}$. Here $\phi$ is from $Z^{1}(G,\Sigma_{\bar{K}}[n])$ to $Z^{1}(G_{v},A^{2}({\mathscr{X}}_{\bar{K}_{v}})[n])$ and $\eta$ is a co-cycle such that it’s cohomology class is in the kernel the homomorphism induced by $\phi$ at the level of cohomology. For simplicity as before we denote ${\mathscr{X}}_{\bar{K}_{v}},\Sigma_{\bar{K}_{v}}$ by ${\mathscr{X}}_{v},\Sigma_{v}$ respectively. In particular for all $\sigma$ in the inertia group $I_{v}$, we have

[TABLE]

Let $v$ be a finite place such that $v$ does not divide $n$ and $Alb(X_{\bar{K}}),X_{\bar{K}}$ have good reduction at $v$. We consider the specialization homomorphism from $A^{2}(X_{v})$ to $A^{2}(X_{v}^{\prime})$, where $X_{v}^{\prime}$ is the reduction of ${\mathscr{X}}_{v}$ at $v$. Then it follows that the image of

[TABLE]

in $A^{2}(X_{v})$ goes to zero under the specialization homomorphism for all $\sigma$ in $I_{v}$. But on the other hand

[TABLE]

is an $n$-torsion for each $\sigma$ in $G_{v}$ (as $\eta$ is an $n$-torsion), so by Roitman’s theorem on torsion, the image of the element $\sigma.z-z$ in $A^{2}(X_{v})$ corresponds to an $n$-torsion on $Alb(X_{v})$. By the previous argument this $n$-torsion on $Alb(X_{v})$ is mapped to zero under $Alb(X_{v})\to Alb(X_{v}^{\prime})$. But we know that the $n$-torsions of $Alb(X_{v})$ are embedded in $Alb(X_{v}^{\prime})$ (for $v$ which does not divide $n$, this follows from the theory of formal groups over $v$-adic numbers). Therefore this $n$-torsion on $A^{2}(X_{v})$ is zero (this is because of the injectivity of the albanese map on $n$-torsions) and consequently

[TABLE]

for all $\sigma\in I_{v}$ and

[TABLE]

in $\Sigma_{v}[n]$, for all $\sigma\in I_{v}$ (where $v$ does not divide $n$).

This implies $S^{n}(\Sigma_{\bar{K}})$ consists of elements which are unramified for all but finitely many places $v$, i.e., the image of the elements in $S^{n}(\Sigma_{\bar{K}})$ under the map

[TABLE]

is zero for all but finitely many places $v$.

Hence the following variance of lemma [Sil][lemma 4.3, chapter X] tells us that $S^{6N}(\Sigma_{\bar{K}})$ is a colimit of this unramified cohomology groups..

Lemma 2.3**.**

[Sil]**[Lemma 4.3, Chapter 10] Let $L$ be a finite extension of the number field $K$. Let $M$ be the finite $G={\rm{Gal}}(\bar{K}/L)$ module $\Sigma_{L}$ and $S$ be a set of finitely many places in $L$. Consider

[TABLE]

consisting of all elements $\eta$ in $H^{1}(G,M)$, which are unramified outside $S$, that is in the kernel

[TABLE]

here $M_{v}=\Sigma_{v}$, for a place in $L$. Then $H^{1}(G,M;S)$ is finite.

Thus this result, applied to $M=\Sigma_{L}[6N]$ for any finite extension $L$ over $K$ (is finite by Theorem 1.1 in [CTR]), we have that $S^{6N}(\Sigma_{L})$ is finite. Also observe that

[TABLE]

[TABLE]

Therefore $H^{1}(G,\Sigma_{\bar{K}}[6N])$ is isomorphic to the colimit of

[TABLE]

Hence $S^{6N}(\Sigma_{\bar{K}})$ is isomorphic to the colimit of $S^{6N}(\Sigma_{L})$, all of which are finite unramified cohomology groups. Consequently

[TABLE]

is a colimit of unramified cohomology groups. The proof actually follows from the finiteness of $S^{6N}(\Sigma_{L})$. ∎

Remark 2.4**.**

In the previous theorem 2.2, it is interesting to see whether the groups $S^{6N}(\Sigma_{L})$ are subgroups of the group $S^{6N}(\Sigma_{\bar{K}})$.

Remark 2.5**.**

The analogue of Mildenhall’s result was proved for a Fermat quartic surface in [Ot]. Thus for the Fermat quartic surface too, Theorem 2.2 is true.

Now by lemma 3.3 in [CTR], if ${\mathscr{O}}_{L}[1/f]$ is such that ${\mathscr{X}}_{L}:={\mathscr{X}}_{{\mathscr{O}}_{L}[1/f]}$ is smooth and the following conditions are true:

[TABLE]

then for a finite extension $L$ of $K$, $A^{2}({\mathscr{X}}_{L})\to A^{2}(X_{L})$ has finite kernel. This leads us to prove:

Theorem 2.6**.**

Under the above conditions

[TABLE]

is torsion and is described by a colimit of unramified cohomology groups

[TABLE]

Proof.

The proof goes verbatim as Theorem 2.2. ∎

Remark 2.7**.**

For a surface with geometric genus and irregularity zero, such that the above conditions as in Theorem 2.2 are satisfied, one has that

[TABLE]

is torsion. Therefore tensoring with ${\mathbb{Q}}$ gives

[TABLE]

is surjective. Therefore to prove the triviality for ${\rm im}(j^{*})(K)\otimes{\mathbb{Q}}$, it is enough to prove that

[TABLE]

is trivial. This gives some information about the structure of of $A^{2}(X_{\bar{K}})$.