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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 $F$?

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Is X finite? Then V generated by finite algebras works. –  Benjamin Steinberg Mar 24 '14 at 16:15
    
Yes $X$ is finite. –  M. Shahryari Mar 24 '14 at 16:26

1 Answer 1

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 2-sided ideal).

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