Let $V$ be a variety and $F$ be a relatively free algebra in $V$. Suppose $X$ is a minimal generating set for $F$. Under what conditions we can deduce that $X$ is a free basis of $F$?

$\begingroup$ Is X finite? Then V generated by finite algebras works. $\endgroup$ – Benjamin Steinberg Mar 24 '14 at 16:15

$\begingroup$ Yes $X$ is finite. $\endgroup$ – M. Shahryari Mar 24 '14 at 16:26
This is a partial answer.
The question has at least two possible meanings. Q1 = the question above with "minimal generating set" interpreted to mean "generating set of least cardinality". Q2 = the question above with "minimal generating set" interpreted to mean "generating set that is minimal under inclusion".
The answer to both questions for varieties of $R$modules is easy to work out.
A1: The variety of all $R$modules has the property that any least cardinality generating set $X'$ in a finitely generated free module $F_R(X)$ is a free generating set iff $M_n(R)$ is Dedekind finite for each $n$. Such rings are called "stably finite".
A2: The variety of all $R$modules has the property that any $\subseteq$minimal generating set $X'$ in a finitely generated free module $F_R(X)$ is a free generating set iff $R$ is a local ring (the nonunits form a 2sided ideal).