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?

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    $\begingroup$ Now I won't be able to sleep until I know the answer. :-( $\endgroup$ – Gro-Tsen Feb 24 '17 at 17:05
  • $\begingroup$ To start with, what are the standard examples/constructions of modules having maximal independent sets of different cardinalities? This question suggests that they are not easy to come by (I didn't look at the Lazarus paper, but "the examples (…) are highly nontrivial" doesn't seem too promising). $\endgroup$ – Gro-Tsen Feb 25 '17 at 0:03
  • $\begingroup$ The answer to 1 seems to be obviously yes. Let $\kappa$ be the cardinality of the set of all linearly independent vectors of $M$ and suppose $\kappa$ is finite. Then $\kappa\in\mathbb{N}$ contradicting that $M$ also has $\kappa+1$ linearly independent vectors. As for 2, I'm not a set theorist... $\endgroup$ – j0equ1nn Feb 25 '17 at 8:14
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    $\begingroup$ @j0equ1nn: "the set of all linearly independent vectors" doesn't make sense. $\endgroup$ – Laurent Moret-Bailly Feb 25 '17 at 10:38
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    $\begingroup$ (2) is answer negatively with $\kappa=\aleph_0$. Still, it would be interesting to ask (2) assuming that $\kappa>\aleph_0$. (And at the other extreme, ask whether every $\kappa$ can be achieved, or even whether there exists $R$ over which every $\kappa$ can be achieved.) $\endgroup$ – YCor Mar 13 '20 at 11:34

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 $T = m$.

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