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This question is a bit concerned with the Tate-Shaferevich group, lets start defining $C$ as $$C: X^2- \Delta Y^2=4$$

which are generally called as Pell-conics, so all in this question $K$ refers to a number field, so its well known that $Ш(C)\cong Cl(k)^2$ ( where $k$ is the quadratic number field ) . ( The above statement is referred from Prof.Franz Lemmermeyer and his paper on Pell-conics which is here ) . I understood that there is no similar interpretation of the cohomological definition of Shafarevich group on elliptic curves side (I am not sure, if there is one such definition quote a reference) .

So I started working out things on it and searching for the ingredients, but I was successful to the extent in stating that we can prove the following :

There is a map $$\alpha : Cl(K)\mapstoШ(K)$$ such that $\alpha $ is a homomorphism. And I think we can prove the following way :

Let $O_K$ be the ring of integers of $K$ then for each fractional ideal $I$ of $O_K$, the ideal $IO_H$ is principal, where $H$ is the the maximal abelian unramified extension of the field $K$ ( and we know that its degree is equal to class number of $K$ ) . Therefore, there exists $x \in H^*$ such that $IO_H = xO_H$.

For each $\sigma \in \rm{Gal(H/K)}$, we have $(IO_H)^\sigma = IO_H$ so $(x^\sigma) = (x)$ which means that $\sigma(x)/x \in O^{*}_{L}$ . Therefore, $x\in \bar K^*$ suchthat $\sigma(x)/x \in \bar O^*$ so the map $\sigma \mapsto \sigma(x)/x $ is a cocycle in $Z^1(K,\bar O^{*}_{K})$. If $y$ is another generator of $IO_H$, then $y/x \in O^{*}_{L}$ , so $\sigma(x/y)/(x/y)$ is a coboundary, so it is trivial in $H^1(K,\bar O^{*}_{K})$.

Clearly, the map $\alpha$ that takes $I$ to the cocycle $\alpha(I)(\sigma) = \sigma(x)/x$ is a homomorphism. To show that it induces the desired homomorphism, it is enough to check that if $I = (\beta)$ for $\beta \in K^*$, then $\alpha(I)$ is trivial. Then, we may choose $x =\alpha$ so $\alpha(I)(\sigma)= \sigma(\alpha)/\alpha = \alpha/\alpha= 1$, since $\alpha \in K$. Therefore, the cocycle is trivial.

So I am stuck in proving the isomorphism between the above groups which I will do it once I get the inverse of $\alpha$. But assume that isomorphism is proved and we can write $Cl(K)\cong Ш(K)$.

My question is that, if we have the above isomorphism with us then

" Why can't we write that $Cl(K)\cong Ш_E(K)$ for an elliptic curve $E$ defined over a number field $K$ ? "

So It satisfies the quest for finding a cohomological definition in elliptic curve sense .

P.S : I thank all members who suggested me to read basic things first, after reading basic things that the group suggested, I am sure that I improved a lot and I am ready to ask these types of questions and even eligible to hear answers.

Thank you all.

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What do you mean by ${\rm Sha}(K)$ (I don't want to make the cyrillic character)? When $K$ is a global field, I understand it to mean the kernel of the map $H^1(G_K,\overline K^\times)\rightarrow\prod_\nu H^1(G_{K_\nu},\overline K^\times_\nu)$. But $H^1(G_K,\overline K^\times)$ is trivial by Hilbert's Theorem 90, so ${\rm Sha}(K)$ is trivial. Also, why should ${\rm Cl}(K)\simeq {\rm Sha}(E/K)$? Do you think all elliptic curves should have the same Tate-Shaferevich group? – B R Dec 7 '11 at 22:03
Also, that quote that you attribute to Lemmermeyer doesn't appear anywhere in the paper you link to. – Dimitrije Kostic Dec 8 '11 at 1:49
@BR : Very nice point sir, I thought of giving the description but assumed that users can figure it out, I was talking about "Sha of number field $K$ which has the following definition $Ш(K)= \rm{Ker}(H^1(K,\bar O^*) \mapsto \prod _{v \nmid \infty} H^1(K_v,\bar O^{*}_{v}))$ . – Iyengar Dec 8 '11 at 4:53
@DimitrijeKostic : You are absolutely right sir, but it was an indirect implication, he talked that there is a good cohomological definition for pell-conics but didn't talk about the Sha of elliptic curves which means that there is no such definition on the other side, if there was a definition he would have surely presented it, as he gave an wonderful comparision between things – Iyengar Dec 8 '11 at 4:59
@iyengar: Please edit your question to remove that quotation, or otherwise rephrase the quotation as a question of your own. You cannot conclude that there is no cohomological definition of Sha for an elliptic curve simply because Lemmermeyer didn't say there was one in that paper. And you certainly cannot put a direct quotation into someone's mouth. – Dimitrije Kostic Dec 9 '11 at 1:58
up vote 4 down vote accepted

There is a relationship between Sha of an elliptic curve and class groups. To see this worked out explicitly in some examples, one can look at the series of papers of Cassels with the title "Arithmetic on curves of genus 1".

What is more accurate to say is that the $n$-Selmer group of an elliptic curve $E$ over a number field $K$ is a subgroup of $H^1(G_K, E[n])$ cut out by local conditions, and computing this cohomology group is essentially a problem of class field theory. E.g. if $E[n] = (\mathbb Z/n)\times (\mathbb Z/n)$ (i.e. all $n$-torsion points of $E$ are defined over $K$) then $H^1(G_K,E[n]) = Hom(G_K, \mathbb Z/n)^2$, and so desribing elements of the Selmer group is the same as describing certain abelian extensions of $K$ satisfying prescribed local conditions; this is then a question of class field theory.

The problem is that one doesn't know which part of the $n$-Selmer group comes from points, and which part comes from Sha.

In the case of the Pell question (or more generally, of units in a number field), one has Dirichlet's theorem which gives a precise description of the units. We are lacking a corresponding result describing the rational points of an elliptic curve.

In those situations where there is a description of points, e.g. by a Heegner point (as in the work of Gross--Zagier and Kolyvagin), then one can explicitly describe Sha, and the situation is closely analogous to the situation of units and class groups of a number field.

Another way to think about the difference between number fields and elliptic curves is that for number fields, the order of vanishing of the $\zeta$-function at every point is known, and just depends on the number of real and complex embeddings of the field.

On the other hand, for an elliptic curve, the order of vanishing of $L(E,s)$ at $1$ is (according to BSD) the rank of the group of rational points, which can (presumably) be anything. It is not controlled by any of the other, more basic, invariants of the curve. (If the rank were always only zero or one, so that it was simply dictated by the sign of the functional equation, then we could always apply the results of Gross--Zagier and Kolyvagin, and BSD and the problem of computing the group of rational points, and Sha, would essentially be solved.)

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Infinitely many thanks to you sir – Iyengar Dec 21 '11 at 1:00

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