Extension of isomorphism of submodules If $V$ is an $n$-dimensional vector space over $\mathbb{F}_q$, and $V_1$ and $V_2$ are subspaces of $V$ (both k-dimensional), there exists an isomorphism $L: V \to V$, where if $u \in V_1$, then $L(u) \in V_2$.
So, if now, instead of $\mathbb{F}_q$ we have a finite ring $R$, there is some condition on $R$ so the analogous for $R$-submodules of $R^{n}$ is still valid? i.e. every isomorphism of submodules of $R^{n}$ extend to an isomorphism of $R^{n}$. For instance if every submodule has a basis and every basis of a submodule can be extended to a basis of $R^{n}$ we can use the same proof in the case of the finite field, but in other case i dont know. Anyone can help me at least pointing me on the right direction about some literature? thanks

This property (isomorphisms between submodules of a module extend to an automorphism of the module) is possessed by any finitely generated projective module over a quasi-Frobenius ring. (See Lectures on modules and rings by T. Y. Lam pg 415).
$R^n$ is finitely generated and projective, of course, and quasi-Frobenius rings are always Artinian, which relaxes the finiteness condition a lot.
I find myself wondering now if the condition characterizes QF rings or not... I don't know offhand.
