TITLE: Every $R$-module is free $\implies$ $R$ is a division ring
QUESTION [27 upvotes]: From Grillet's Abstract Algebra, section VIII.5.

Definitions. A division ring is a ring with identity in which every nonzero element is a unit. A vector space is a unital module over a division ring.
Theorem 5.2. Every vector space has a basis.
Exercises. 
(7.) Show that $R$ is a division ring if and only if it has no left ideal $L \neq 0, R$.
(*9.) Prove that $R$ is a division ring if and only if every left $R$-module is free. 

I'm trying to solve exercise 9. I suspect it should be formulated: 

Let  $R$ be a unital ring. Then $R$ is a division ring $\iff$ every unital left $R$-module is free.

Attempt of proof: $(\implies)$: Theorem 5.2.
$(\impliedby)$: By exercise 7, it suffices to show that every non-zero left ideal of $R$ is equal to $R$. Let $I\!\neq\!0$ be a left ideal. As a $R$-module, $I$ has a basis $B$. 
How can I show that $I=R$?

REPLY [5 votes]: Cite planetmath:

Assume now that every left $R$-module is free. In particular every
  left $R$-module is projective, thus $R$ is semisimple and therefore
  $R$ is Noetherian. This implies that $R$ has invariant basis number.
  Let $I\subseteq R$ be a nontrivial left ideal. Thus $I$ is a
  $R$-module, so it is free and since all modules are projective
  (because they are free), then $I$ is direct summand of $R$. If $I$ is
  proper, then we have a decomposition of a $R$-module $$R\simeq I\oplus
I',$$ but rank of $R$ is $1$ and rank of $I\oplus I'$ is at least $2$.
  Contradiction, because $R$ has invariant basis number. Thus the only
  left ideals in $R$ are $0$ and $R$. Now let $x\in R$. Then $Rx=R$, so
  there exists $\beta\in R$ such that $$\beta x=1.$$ Thus every element
  is left invertible. But then every element is invertible. Indeed, if
  $\beta x=1$ then there exist $\alpha\in R$ such that $\alpha\beta =1$
  and thus $$1=\alpha\beta=\alpha(\beta x)\beta=(\alpha\beta)x\beta=x\beta,$$ so $x$ is right invertible. Thus
  $R$ is a divison ring. $\square$