Selmer group

Given an elliptic curve $E$ we can define two very interesting andimportant groups, the Selmer group and theTate-Shafarevich group, which together provide a measure ofthe failure of the Hasse principle for elliptic curves, bymeasuring whether the curve is everywhere locally soluble. Here wepresent the construction of these groups.

Let $E,E^{\prime }$ be elliptic curves defined over $\mathbb{Q}$ and let$\overline{\mathbb{Q}}$ be an algebraic closure of $\mathbb{Q}$. Let$\varphi :E\to E^{\prime }$ be an non-constant isogeny (for example, wecan let $E=E^{\prime }$ and think of $\varphi$ as being the “multiplicationby $n$” map, $[n]:E\to E$). The following standard resultasserts that $\varphi$ is surjective over $\overline{\mathbb{Q}}$:

Theorem 1.

Let $C_{\mathrm{1}}\mathrm{,}{C}_{\mathrm{2}}$ be curves defined over an algebraically closed field$K$ and let

$$\psi :C_1\to C_2$$

be a morphism (oralgebraic map) of curves. Then $\psi$ is either constant orsurjective.

Proof.

See [4], Chapter II.6.8.∎

Since $\varphi :E(\overline{\mathbb{Q}})\to E^{\prime }(\overline{\mathbb{Q}})$ isnon-constant, it must be surjective and we obtain the followingexact sequence:

$$0\to E(\overline{\mathbb{Q}})[\varphi ]\to E(\overline{\mathbb{Q}})\to E^{\prime }(\overline{\mathbb{Q}})\to 0\mathit{\hspace{1em}\hspace{1em}}(1)$$

where $E(\overline{\mathbb{Q}})[\varphi ]=\mathrm{Ker}\varphi$. Let$G=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, theabsolute Galois group of $\mathbb{Q}$, and consider the$i^{th}$-cohomology group $H^i(G,E(\overline{\mathbb{Q}}))$ (weabbreviate by $H^i(G,E)$). Using equation $(1)$ we obtain thefollowing long exact sequence (see Proposition 1 in groupcohomology):

$$0\to H^0(G,E(\overline{\mathbb{Q}})[\varphi ])\to H^0(G,E)\to H^0(G,E^{\prime })\to H^1(G,E(\overline{\mathbb{Q}})[\varphi ])\to H^1(G,E)\to H^1(G,E^{\prime })\mathit{\hspace{1em}\hspace{1em}}(2)$$

Note that

$$H^0(G,E(\overline{\mathbb{Q}})[\varphi ])={(E(\overline{\mathbb{Q}})[\varphi ])}^G=E(\mathbb{Q})[\varphi ]$$

and similarly

$$H^0(G,E)=E(\mathbb{Q}),H^0(G,E^{\prime })=E^{\prime }(\mathbb{Q})$$

From $(2)$ we can obtain an exact sequence:

$$0\to E^{\prime }(\mathbb{Q})/\varphi (E(\mathbb{Q}))\to H^1(G,E(\overline{\mathbb{Q}})[\varphi ])\to H^1(G,E)[\varphi ]\to 0$$

We could repeat the same procedure but this time for $E,E^{\prime }$defined over ${\mathbb{Q}}_p$,for some prime number $p$, and obtaina similar exact sequence but with coefficients in ${\mathbb{Q}}_p$which relates to the original in the following commutative diagram(here $G_p=\mathrm{Gal}(\overline{{\mathbb{Q}}_p}/{\mathbb{Q}}_p)$):

$0\to E^{\prime }(\mathbb{Q})/\varphi (E(\mathbb{Q}))\to$$H^1(G,E(\overline{\mathbb{Q}})[\varphi ])$$\to H^1(G,E)[\varphi ]\to 0$

$\downarrow$$\downarrow$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\downarrow$

$0\to E^{\prime }({\mathbb{Q}}_p)/\varphi (E({\mathbb{Q}}_p))\to$$H^1(G_p,E(\overline{{\mathbb{Q}}_p})[\varphi ])$$\to H^1(G_p,E)[\varphi ]\to 0$

The goal here is to find a finite group containing$E^{\prime }(\mathbb{Q})/\varphi (E(\mathbb{Q}))$. Unfortunately$H^1(G,E(\overline{\mathbb{Q}})[\varphi ])$ is not necessarily finite. Withthis purpose in mind, we define the $\varphi$-Selmer group:

$$S^{\varphi }(E/\mathbb{Q})=\mathrm{Ker}\left(H^1(G,E(\overline{\mathbb{Q}})[\varphi ])\to \prod _p{H}^1(G_p,E)\right)$$

Equivalently, the $\varphi$-Selmer group is the set ofelements $\gamma$ of $H^1(G,E(\overline{\mathbb{Q}})[\varphi ])$ whoseimage ${\gamma }_p$ in $H^1(G_p,E(\overline{{\mathbb{Q}}_{𝕡}})[\varphi ])$ comesfrom some element in $E({\mathbb{Q}}_p)$.

Finally, by imitation of the definition of the Selmer group, wedefine the Tate-Shafarevich group:

$$TS(E/\mathbb{Q})=\mathrm{Ker}\left(H^1(G,E)\to \prod _p{H}^1(G_p,E)\right)$$

The Tate-Shafarevich group is precisely the group that measuresthe Hasse principle in the elliptic curve $E$. It is unknown ifthis group is finite.