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?



Rings over which every module is free
The non-commutative case isn't very different. Assume for reasonableness that your ring is Noetherian (I refuse to think about non-Noetherian rings on principle). Your argument gives that all two-sided ideals just be the whole algebra or trivial. So, your algebra is simple. 


Another consequence is that every exact sequence splits. Thus, if you have a pair of left ideals $I \subset J$, you can choose left ideals $K\_1,K\_2$ such that $J=I \oplus K\_1$ and $R=J \oplus K\_2$. Thus, if you had a infinite descending chain of ideals, you would have a complementary ascending chain of ideals, which is impossible, since your ring is Noetherian. Thus, your ring is Artinian.


By Artin-Wedderburn, your algebra is a matrix algebra over a skew field. But every matrix ring except 1 x 1 has a non-free projective module. So A is a skew-field.


