A student asked me this, and I can't believe I never knew the answer to this.


Let $R$ be a commutative ring, and $M$ be an $R$-module.


1. If $M$ has a set of $n$ linearly independent vector for each $n\in\mathbb{N}$, does that necessarily imply that $M$ has an infinite set of linearly independent vectors?
2. More generally, if $\kappa$ is the minimum cardinal such that there is no set of cardinality $\kappa$ of linearly independent vectors. Must $\kappa$ be always a successor cardinal?



Maximum cardinal of a set of linearly independent vectors in a module
Question 1. has a negative answer. 


Denote $A := \{(i,j) \in \mathbb{N}^2, j\leq i\}$ and $S := \{$finite subsets of A with at least 2 different first coordinates$\}$. Define the ring $$R := \mathbb{F}\_2[t\_s: s \in S]\,/\,\big(t\_{s\_1}t\_{s\_2}: s\_1,s\_2 \in S\big),$$ and the $R$-module
$$M := \bigoplus\_{(i,j) \in A}Rm\_{ij}\,/\,\left(t\_s\sum\_{(i,j)\in s}m\_{ij}: s \in S\right).$$


For example, $s\_0=\{(1,1),(2,1))\} \in S$ and $t\_{s\_0}(m\_{11}+m\_{21})=0.$


Then for each $m$, the subset $\{m\_{m1},..m\_{mm}\} \subset M$ is a linearly independent subset of size $m$.


Suppose that $(x\_k)\_{k\in K} \subset M$ is a linearly independent family and let us show that $K$ is finite.
Denote $T := (t\_s: s \in S)R$; this is a non-zero ideal of $R$.


1. Write $x\_k = \Sigma r\_{ij}^{(k)}m\_{ij}$. Then $\{r\_{ij}^{(k)}(0)\} \neq \{0\}$: otherwise $Tx\_k=0$.
2. $\{x\_k\}$ is linearly independent mod $T: \Sigma r\_kx\_k \in TM \Rightarrow \Sigma t\_{s\_0}r\_kx\_k =0 \Rightarrow t\_{s\_0}r\_k=0$ all $k \Rightarrow r\_k \in T$ all $k$.
3. $\exists m$ such that $\{x\_k\} \subset \sum\_nRm\_{mn}$ mod $T$: otherwise $\bigcup\_k\{(i,j):r\_{ij}(0) \neq 0\}$ has at least 2 different first coordinates and some $t\_s$ kills some $x\_k$ or some $x\_{k\_1}+x\_{k\_2}$.
4. $\#(K) \leq m: \dim\_{\mathbb{F}\_2}\sum\_nRm\_{mn}$ mod $T = m$.


