The following is motivated by Silverman's exercise 6.7. Let $\Lambda = <1,\tau>\subset \mathbb{C}$ be a lattice and define the Weierstrass $\wp$-function to be \begin{align*} \wp(z) = \frac{1}{z^2} + \sum_{\omega\in \Lambda\setminus{(0,0)}} \bigg(\frac{1}{(z-\omega)^2} - \frac{1}{\omega^2}\bigg), \end{align*} so that $\wp$ is periodic with respect to $\Lambda$ and even, and $\wp'$ is periodic with respect to $\Lambda$ and odd. Then for $\Phi:\mathbb{C}/ \Lambda \rightarrow E$ where $E$ is the associated elliptic curve to $\Lambda$ and $\Phi(z)=(\wp(z),\wp'(z))$, $\Phi$ is a group isomorphism. If we assume $E$ is defined over $\mathbb{R}$ then we can show that $\Delta(E)\in \mathbb{R}$ and $\Lambda$ can be chosen so that $\Lambda=\overline{\Lambda}$. In this case, $\overline{\wp(z)}=\wp(\overline{z})$ for all $z\in \mathbb{C}/\Lambda$. Further, if $E[2]\subset E(\mathbb{R})$ then the cubic equation for $E$ has $3$ real roots and $\Delta(E)>0$. Now, I'm trying to prove:
If $E$ is defined over $\mathbb{R}$ and $E[2]\subset E(\mathbb{R})$ then there is some $z\not\in \mathbb{R}$ with $\text{Re}(z)=\frac{1}{2}$ and $\Phi(z)\in E[2]$.
Equivalently, I am trying to prove that under these same conditions, $\tau$ must be purely imaginary. I'm a bit stuck though -- any thoughts?
The one other thing I know already is that if $E$ is defined over $\mathbb{R}$, but $\tau$ isn't purely imaginary, then we can assume $\text{Re}(\tau)=\frac{1}{2}$. Somehow the condition $E[2]\subset\mathbb{R}$ is supposed to prevent this case from happening.