A simple divisible module Let $k$ be division ring and $V$ be a left $k$-vector space of infinite dimension. Let $E=\text{End}_k(V)$ be the ring of right linear transformations on $V$. My questions are:
(1) Is $V$ a simple right $E$-module?
(2) If $u\in V$ and $a\in E$ with $u\notin Va$, how could we define an endomorphism $x\in E$ that vanishes on $Va$ but not on $u$? (Therefore, $(Va)x=0 $ gives $ax=0$, but since $ux$ is not zero, we conclude that $V$ is a divisible $E$-module.)
For (2), I think we must choose a suitable basis for $V$. 
Thanks for any help!

For (1), I think it is easy to show that, if $v\in V$ is nonzero and $w\in V$,  there exists $a\in E$ such that $va=w$.  If $w\in \text{span}_k(v)$, then $a$ is just a scalar multiplication.  If $w\notin\text{span}_k(v)$, then you can extend the $k$-linearly independent set $\{v,w\}$ to a basis of $V$, and then the rest should be trivial.  Hence, $V$ is a simple right $E$-module.
For (2), let $W$ be a $k$-vector subspace of $V$ such that $V=Va\oplus W$ (since vector spaces over $k$ are projective unitary $k$-modules, $W$ exists).  Now, any endomorphism $b$ of $W$ can be uniquely extended to $\tilde{b}\in \text{End}_k(V)$ such that $Va\subseteq \ker\left(\tilde{b}\right)$ via the definition $(x+y)\tilde{b}:=yb$ for all $x\in Va$ and $y\in W$.  Since $u \notin Va$, $u=x+y$ for some $x\in Va$ and $y \in W$ with $y\neq 0_V$.  Choose $b\in\text{End}_k(W)$ to be the identity map, then $u\tilde{b}=y \neq 0_V$.  However, $Va\tilde{b}=\left\{0_V\right\}$.
