Let $R$ be a commutative ring with identity, and let $R^{N}$ be the free $R$-module of dimension $N$.

A set $\{e_{1},\ldots, e_{m}\}\subseteq R^{N}$ is linearly dependent if there exist $\alpha_{1},\ldots, a_{m}\in R$, not all zero, such that $\sum_{i=1}^{m}\alpha_{i}e_{i}=0$. If $\{e_{1},\ldots, e_{m}\}$ is not linearly dependent, it is linearly independent, and in this case $m\leq N$, by Section 1D in Lectures on Modules and Rings, by T.Y.~Lam (or by this answer).

What if we alter the definition of linear dependence? I will now declare $\{e_{1},\ldots, e_{m}\}$ to be pseudo linearly dependent if there exist $\alpha_{1},\ldots, \alpha_{m}\in R$, not all zero, such that each non-zero $\alpha_{i}$ is a unit, and such that $\sum_{i=1}^{m}\alpha_{i}e_{i}=0$. This is the same notion discussed here. If $\{e_{1},\ldots, e_{m}\}$ is pseudo linearly independent, is there an upper bound on $m$ in terms of $N$? Or are there examples showing that no interesting bound can exist?


1 Answer 1


You cannot bound $m$ in terms of $N$ in general. For instance, with $R=\mathbb{Z}$ and $N=1$, then the set of all powers of $2$ is an infinite pseudo-linearly independent set.

More generally, it is possible to have a pseudo-linearly independent set of arbitrarily large cardinality for $N=1$. For instance, for any set $S$, let $R$ be a polynomial ring over $\mathbb{Z}$ with the elements of $S$ as variables. Then $S$ will be pseudo-linearly independent as a subset of $R$ as an $R$-module.


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