Maximum cardinal of a set of linearly independent vectors in a module 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.


*

*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?

*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?
 A: 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 $R$.


*

*Write $x_k = \Sigma r_{ij}^{(k)}m_{ij}$. Then $\{r_{ij}^{(k)}(0)\} \neq \{0\}$: otherwise $Tx_k=0$.

*$\{x_k\}$ 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 $k$.

*$\exists m$ 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 $x_{k_1}+x_{k_2}$.

*$\#(K) \leq m: \dim_{\mathbb{F}_2}\sum_nRm_{mn}$ mod $T = m$.

