Let $A$ be a commutative ring, $(I, \leq)$ some partially ordered set, and $V$ a free $A$-module with basis $\{e_i\}_{i \in I}$. Suppose that there is a set of elements $\{x_i\}_{i \in I} \subseteq V$ satisfying $$ x_i = e_i + \sum_{j > i} a_{ij} e_j$$ where finitely many of the $a_{ij}$ are zero for each fixed $i$. I am happy to assume that the set $I$ is countable, if this matters.
Is the map $X: V \to V$, $Xe_i = x_i$ invertible? Equivalently, is the set $\{x_i\}_{i \in I}$ a basis of $V$?
If $I$ is finite, then the above conditions mean that the matrix of $X$ in the $e_i$ basis is triangular with $1$'s along the diagonal, and so is invertible by virtue of having determinant $1$.
If $I$ is a bounded-above poset, in the sense that for any $J \subseteq I$, $J$ has a maximal element, then $\{x_i\}_{i \in I}$ is a basis. The proof is by an application of Zorn's lemma:
- Define $\mathscr{X}$ to be the set of subsets $J \subseteq I$ satisfying the two conditions
- If $a \in J$, $b \in I$, and $a \leq b$, then $b \in J$. ($J$ is "upwardly closed" in the partial order).
- $\{x_j\}_{j \in J}$ is a basis of $\mathrm{span} \{e_j\}_{j \in J}$.
- $\mathscr{X}$ is nonempty, since taking $m \in I$ to be a maximal element, $\{m\} \in \mathscr{X}$ (we must have $x_m = e_m$ in this case). By Zorn's lemma, $\mathscr{X}$ (with the inclusion ordering) has a maximal element $M$.
- If $M \neq I$, then let $i_0$ be maximal in $I \setminus M$. Then $J' = J \cup \{i_0\}$ also satisfies the requirements to be in $\mathscr{X}$, and hence we have a contradiction. So $J = I$.
The key part of this last step is the equality $$ e_{i_0} = x_{i_0} - \underbrace{\sum_{j > i_0} a_{i_0 j} e_j}_{\in \mathrm{span}\{e_j \mid j \in J\}}$$ from which we can see that $\{x_j\}_{j \in J} \cup \{x_{i_0}\}$ is linearly independent and spanning.
So, does this hold more generally?
This is not true in general. For instance, let $I=\mathbb{N}$ and let $x_i=e_i-e_{i+1}$ for each $i$. Note then that each $x_i$ has the property that the sum of its coefficients is $0$, and so this property also holds for every element of the submodule generated by the $x_i$. So, the $x_i$ do not generate $V$ (unless $A=0$).
More generally, if $J\subseteq I$ is any nonempty subset which does not have a greatest element, for each $i\in J$ let $x_i=e_i-e_j$ for some $j>i$ in $J$ and let $x_i=e_i$ if $i\not\in J$. Then every element of the submodule generated by the $x_i$ has the sum of its coefficients in $J$ equal to $0$. So, the condition that $I$ is co-well-founded is both necessary and sufficient for this result to hold.