TITLE: Rings over which every module is free
QUESTION [19 upvotes]: We know that modules over skewfields are free. Is the converse true? In other words, is it true that a nontrivial ring over which every module is free is a skewfield?
If the ring A is commutative, then writing that for any proper ideal I of A, the A-module A/I is free yields the result. What about the general case?

REPLY [14 votes]: Caution: the zero ring is not a division ring, but all of its modules are free!  But other arguments already given here show that any nonzero ring over which every right module is free is a division ring.
I'll pipe in with one more approach, a personal favorite.  Assume that every right R-module is free and that R ≠ 0.  Let S be a simple right R-module (at least one exists because R is nonzero: choose any maximal right ideal M of R and let S = R/M).  Then S is free, so choose a basis for S.  Pick any element x of the basis.  Because S is simple and x ∈ S∖{0} we have S = xR, and because x was a basis element we have xR ≅ R as right R-modules.  In particular, RR ≅ SR is simple, from which it follows that R is a division ring.