# Constructing a homomorphism from class group to Sha.

In response to my previous question I got a wonderful answer from Prof.Emerton explaining about the similarities between $Ш$ and class group. In order to add something the comments I got from Mr. B R were not sufficient in figuring out the complete idea of the map I was expecting, I got a half way solution to prove that there exists an homomorphism from class group to Ш which can be as follows

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 can anyone offer still more easier proof or any alternate proof. ( as I think that there is some problem with the proof )

Are there any known proofs that shows the isomorphism between Sha and Class group ?

P.S : Mr.B R remarked that there can be a nicer way to prove the homomorphism, but he didn't give complete details ( he indeed gave some details but those are not enough to study completely ) , nor any references. So I thought I can ask a separate question, so that the same user or the other users can help.

Thanks a lot.

Yours truly.

Iyengar.

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Merry Christmas to all – Iyengar Dec 24 '11 at 16:05