I'm working on an exercise in Hungerford's Algebra:
A ring R with identity is a division ring iff every unitary R-module is free. (Hint: it suffices to show that R has no nonzero maximal left ideals. Note that every left ideal of R is a free R-module and hence a module direct summand of R.)
My attempt ($\Leftarrow$): Let I be a nonzero ideal in R.Since I is a free R-module, I is projective. And since every R-module is free, every R-module is projective and thus, every R-module is injective. So I is an injective R-module. Hence there's a R-module K such that $R$ $\cong$ $I$ $\oplus$ $K$. Then, there exists integers $m$,$n$ such that $I$ $\cong$ $R^m$, $K$ $\cong$ $R^n$, and $R$$\cong$ $R^{m+n}$. (If one of the $m$,$n$ is infinity, then R has the IBN property and the fact that R has a basis of cardinality one, the identity itself, will lead to a contradiction). On the other hand, since every R-module is projective and R has an idenity, R is (semisimple) Artinian. Thus, R has the IBN property (since every Noetherian ring does). Hence we have m+n=1 and since I is nonzero we must have $n=0$, i.e. K=0. So $I$=$R$.
Though there has been a question about this exercise on MSE: Every $R$-module is free $\implies$ $R$ is a division ring, it seems that no one has applied the method as Hungerfod suggests, so I'm not sure about my proof. Besides, this exercise appears in the first few sections of the module theory in the book and the Artin-Wedderburn theory has not been introduced, so I think there's a more direct solution to the problem with the author's hints. Thanks in advance for your help.
