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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?

Thank you.

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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
  • $\begingroup$ @Gro-Tsen: there is an explicit example here which is also a free module; however we know our example cannot be free. $\endgroup$ – user105303 Feb 25 '17 at 17:25
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  1. No.

$A := \{(i,j) \in \mathbb{N}^2, j\leq i\}$

$S := \{$finite subsets of A with at least 2 different first coordinates$\}$

$R := \mathbb{F}_2[t_s: s \in S]/(t_{s_1}t_{s_2}: s_1,s_2 \in S)$

$M := \oplus_{(i,j) \in A}Rm_{ij}/(t_s\Sigma_{(i,j)\in s}m_{ij}: s \in S)$

For example, $s_0=\{(1,1),(2,1))\} \in S$ and $t_{s_0}(m_{11}+m_{21})=0.$

$\{m_{m1},..m_{mm}\} \subset M$ is a linearly independent set of size $m$.

Let $T := (t_s: s \in S)R$, non-zero ideal of $R$.

Suppose $\{x_k\} \subset M$ is a linearly independent set.

  1. Write $x_k = \Sigma r_{ij}m_{ij}$. Then $\{r_{ij}(0)\} \neq \{0\}$: 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 $\{x_k\} \subset \Sigma_nRm_{mn}$ mod $T$: otherwise $\cup_k\{(i,j):r_{ij}(0) \neq 0\}$ has at least 2 different first coordinates and some $t_s$ kills some $x_k$ or some $x_{k_1}+x_{k_2}$.
  4. $\#\{x_k\} \leq m: dim_{\mathbb{F}_2}\Sigma_nRm_{mn}$ mod $T = m$.
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