# 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!

## 1 Answer

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, $Va\tilde{b}=\left\{0_V\right\}$.

• Thanks for the answer, but where did you use the "infinite dimension-ness" of $V$? – karparvar Aug 10 '15 at 5:12
• @karparvar Why do we need to use every given information to solve a problem? – Batominovski Aug 10 '15 at 5:20