(All my rings are unital.)
Suppose $R$ is a commutative ring and that $M$ is an $R$-module.
Definition. Call a subset $X \subseteq M$ pseudo-independent iff for all proper subsets $Y$ of $X,$ the $R$-module generated by $Y$ is a proper subset of the $R$-module generated by $X$.
Linear independence implies pseudo-independence. However, the converse often fails. For example, suppose that $R=\mathbb{Z}$ and $M = \mathbb{Z}/2\mathbb{Z}$. Then $\{1\}$ is a pseudo-independent subset of $M$, but not a linearly independent subset.
Now I suspect that the following are equivalent:
- $R$ is a field.
- $R$ is non-trivial, and for all $R$-modules $M$, every pseudo-independent subset of $M$ is linearly independent.
I'm working on the upward direction.
Question. Does the upward implication hold, and if so, how can we prove it?
Here's what I've got:
Suppose that $R$ is a non-trivial commutative ring, and that for all $R$-modules $M$, every pseudo-independent subset of $M$ is independent. Suppose for a contradiction that $R$ has a non-zero, non-maximum ideal $I$. Then $R/I$ is, of course, an $R$-module. Since $I$ is non-maximum, hence $R/I$ has an element $x$ distinct from $0$. It follows that $\{x\}$ is a pseudo-independent subset of $R/I$. Hence, it is independent.
Okay, now what? Ideas, anyone?
Since $I$ is nonzero, choose $0 \neq y \in I$. Then $y \cdot \bar x$ is a nontrivial linear combination of $\bar x$, so should be nonzero because $\{\bar x\} \subset R/I$ is independent. But this is absurd, because $yx \in I$ so $y \cdot \bar x = \overline{yx} = 0 \in R/I$.