Let $R$ be a finite ring (with unit, possibly non-commutative), and $M$ a left module over $R$. Let $v,w\in M$. Then $$Rv = Rw \iff R^\times v = R^\times w.$$

This follows from Lemma 6.4 in Hyman Bass. K-theory and stable algebra. Publications Mathématiques de l'Institut des Hautes Études Scientifiques 22 (1964), 5–60.

The proof uses the Artin-Wedderburn classification theorem. My question is if there is a simpler proof. (The hard part is the direction $\Rightarrow$, of course.)

As a remark, the statement is not true any more for infinite rings. Counterexamples with $M={}_R R$ can be found on page 466 in Irving Kaplansky. Elementary divisors and modules. Transactions of the American Mathematical Society 66 (1949), 464–491. Counterexamle (b) is over a commutative Noetherian ring $R$.

semisimple. So I'd hope for a proof wich does not rely on the theory of semisimple rings/modules at all. $\endgroup$ – azimut Jul 16 '15 at 18:28