Dimension of a module

Question

Let $M$ be a $R$-module. If $a_1, \dots, a_m\in M$ are linearly independent and $b_1, \dots, b_n \in M$ spans $M$, is it true that $m\leq n$?

If $R$ is a field, I can prove it.

Answer 1

Note that if a ring $R$ fails to have the invariant basis number property, then we can easily generate an example of a module over $R$ such that both $\{a_i\}$ and $\{b_i\}$ are bases. One example of such a ring and associated module is given in exercise VI.1.12 in Aluffi's chapter 0, as is discussed in detail in this MSE post.

If $R$ does have the IBN property (for instance, if $R$ is a commutative ring), then I suspect your statement will hold. Here is an attempted proof:

Suppose that $b_1,\dots,b_n$ span the module $M$. Then $M$ is isomorphic to the free module on $b_1,\dots,b_n$ modulo some relations. Now, $a_1,\dots,a_m$ are linearly independent. Thus, there exists $\tilde a_1,\dots \tilde a_m$ within the free module on $b_1,\dots,b_n$ such that $\pi(\tilde a_i) = a_i$ (where $\pi$ denotes the quotient map from the free module on $b_1,\dots,b_n$ to $M$).

Because $a_i$ are linearly independent, $\tilde a_i$ must be linearly independent. Because $\tilde a_1,\dots,\tilde a_m$ form the basis of a submodule of the free module generated by $b_1,\dots,b_n$, the IBN property implies that $m \leq n$, as desired.