What Goes Wrong when $p=2$?

In the previous section, we set $p=3$ and constructed an abelian variety $A$ of dimension $p-1$ that (conjecturally) has nonsquare ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]$. We can construct an $A$ in an analogous way for any odd prime $p$, and the author expects that ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]$ is nonsquare in most cases. However, when $p=2$, the dimension of $A$ is $1$, so in that case $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})$ must be a perfect square.

What goes wrong? The problem lies in Theorem 3.1. The argument used to prove Theorem 3.1 at least provides a map

$$E(\mathbb{Q})/2 E(\mathbb{Q})\hookrightarrow\Vis_R(H^1(\mathbb{Q},A)).$$

When $p=2$, the condition $e<p-1$ is not satisfied, so the proof of Theorem 3.1 does not show that the image of $E(\mathbb{Q})/ 2 E(\mathbb{Q})$ is locally trivial at the prime $2$ (or at $\infty$). We thus only construct a subgroup of $H^1(\mathbb{Q},A)$ of nonsquare order, not of  ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})$. Thus even if two elliptic curves have the same $E[2]$, then can still possess very different Selmer groups.