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Does there exist an example of a commutative ring with unity $R$ and an $R$-module $M$ such that:

  • If $L \subseteq M$ is linearly independent, $|L| < \infty$.

  • For any $N \in \mathbb{N}$, there exists a linearly independent $L \subseteq M$ with $|L| \geq N$.

Equivalently, we can require:

  • There is no injection $R^{(I)} \hookrightarrow M$ for any infinite $I$

  • For any $N \in \mathbb{N}$, there exists $n \geq N$ and an injection $R^n \hookrightarrow M$

Stated more generally, considering the collection of cardinalities of the linearly independent subsets, does there exist an example where this collection does not attain its supremum?

Here are the cases which I've looked at which don't yield such an example:

If $R$ is a field, such an example cannot occur. If there are arbitrarily large linearly independent subsets, the dimension must be at least $|\mathbb{N}|$, meaning that there exists an infinite linearly independent subset, a contradiction.

If $M$ is finitely generated, such an example cannot occur, since there cannot be any infinite linearly independent set. See Infinite linear independent family in a finitely generated $A$-module .

Edit: In the following MathOverflow post, a large list of conditions for all maximal linearly independent subsets to have the same cardinality is exhibited. It seems like some counterexample from there could lead to a counterexample here: https://mathoverflow.net/q/30066/155881

Smiley1000
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    Which axiomatic theory of sets are you assuming? – David A. Craven Sep 26 '24 at 08:37
  • @DavidA.Craven ZFC – Smiley1000 Sep 26 '24 at 08:38
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    Is there a subtlety when doing the standard Zorn's lemma argument, like proving that every vector space has a basis? – David A. Craven Sep 26 '24 at 08:40
  • @DavidA.Craven Yes, the problem is that we don't assume that all finite subsets are linearly independent. We only assume that there are arbitrarily large finite linearly independent subsets. – Smiley1000 Sep 26 '24 at 08:41
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    Sure, not all finite subsets are linearly independent, that's also true for vector spaces. But you would let $X$ denote the set of all linearly independent subsets of $M$, not all finite subsets. $X$ is obviously a poset. Then the upper bound of a chain is the union of the chain, so ZL applies. – David A. Craven Sep 26 '24 at 08:43
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    @DavidA.Craven I don't see how to show that the maximal elements of $X$ are infinite. – Smiley1000 Sep 26 '24 at 08:45
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    So if $R$ is a domain you are now done, because all maximal linearly independent subsets have the same cardinality. This isn't true in general, and I'm wondering if some horrific $R$ like the direct sum of all $\mathbb Z/n\mathbb Z$ acting on an $\mathbb N$-generated module in some terrible way might do what you want, not have an infinite linearly independent set. – David A. Craven Sep 26 '24 at 09:02
  • @DavidA.Craven I just found https://mathoverflow.net/q/30066/155881 , where some large list of conditions for all maximal linearly independent subsets to have the same cardinality is exhibited. It seems like some counterexample from there could lead to a counterexample here. – Smiley1000 Sep 26 '24 at 09:04
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    Here's the kind of thing that will have weird properties. Think of infinitely many copies of $\mathbb Z$ labelled by $\mathbb N$ acting on itself by standard multiplication, plus the $i$th copy moves everything down $i$ places on the module, or something like this. Points in the first $n$ copies of $\mathbb Z$ look reasonable because most of the ring annihilates it, but as you try to increase the size of the linearly independent set you must leave the first $n$ copies, and then the rest of the module acts to push this new element into a linear combination. – David A. Craven Sep 26 '24 at 09:22
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    I'm not saying this works, I'm saying once you leave the realm of finitely generated you can do really ridiculous things. – David A. Craven Sep 26 '24 at 09:23
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    @David Unfortunately I don't understand your idea. Why ring do you take? Which module? – Martin Brandenburg Sep 26 '24 at 15:02
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    @Smiley1000 do I understand your question correctly, that you would like to have an example of a module $M$ over a commutative unital ring $R$ such that $R^n$ embedds into $M$ for each $n\geq 1$, but the free module $R^{(\mathbb{N})}$ of rank $\mathbb{N}$, does not? – Christian Lomp Oct 06 '24 at 11:45
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    This is indeed equivalent to the question. – Martin Brandenburg Oct 06 '24 at 11:48
  • @ChristianLomp Exactly – Smiley1000 Oct 06 '24 at 12:06
  • @ChristianLomp I've added your reformulation to the question. – Smiley1000 Oct 06 '24 at 14:16

1 Answers1

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Let $\mathbb{N}=\{1,2,3,\cdots\}$ denote the positive natural numbers, let $k$ be a field and let $R=k[x_i \vert i\in \mathbb{N}]\,\,/\,\langle x_ix_j \vert i,j\in \mathbb{N}\rangle$. So $R$ is the quotient of the polynomial ring in countably many variables, subject to the the relation that the product of any two variables is zero.

Let $M$ be a module generated over $R$ by $\{e_i\vert i\in \mathbb{N}\}$ subject to the relations $$e_ix_j=e_jx_i$$ whenever $j>2i$.

To build an intuitive picture of $M$, note that it has basis over $k$ given by the black terms:

$$\begin{array}{ccccccc}e_1&e_2&e_3&e_4&e_5&e_6&\cdots\\ e_1x_1&e_2x_1&e_3x_1&e_4x_1&e_5x_1&e_6x_1&\cdots\\ e_1x_2&e_2x_2&e_3x_2&e_4x_2&e_5x_2&e_6x_2&\cdots\\ \color{red}{e_1x_3}&e_2x_3&e_3x_3&e_4x_3&e_5x_3&e_6x_3&\cdots\\ \color{red}{e_1x_4}&e_2x_4&e_3x_4&e_4x_4&e_5x_4&e_6x_4&\cdots\\ \color{red}{e_1x_5}&\color{red}{e_2x_5}&e_3x_5&e_4x_5&e_5x_5&e_6x_5&\cdots\\ \color{red}{e_1x_6}&\color{red}{e_2x_6}&e_3x_6&e_4x_6&e_5x_6&e_6x_6&\cdots\\ \color{red}{\vdots} & \color{red}{\vdots} &\color{red}{\vdots}&\vdots&\vdots&\vdots&\ddots \end{array}$$

Here the red terms are equal to the black terms obtained by reflecting them in the main diagonal. In particular, any set of distinct terms of the form $e_ix_j$ are linearly independent over $k$, as long as it does not contain both $e_ix_j$ and $e_jx_i$ for some $j>2i$.

Thus for $i\in \mathbb{N}$ we have $e_i,e_{i+1},\cdots, e_{2i}$ linearly independent over $R$. It remains to show that $M$ contains no infinite linearly independent set.

Suppose we have a countable linearly independent set $u_1,u_2,\cdots$. We seek a contradiction. Let $m(u_i)$ denote the smallest integer $j$, such that the coefficient in $k$ on $e_j$ in $u_i$, with respect to the above (black) $k$-basis of $M$, is non-zero.

Note this integer exists as any element of an independent set must be $R$ torsion free, so cannot be just a linear combination of the $e_kx_l$.

By replacing the $u_i$ with finite $k$-linear combinations of the $u_i$, we obtain a new countable $R$-linearly independent set $v_1,v_2,\cdots$ with the added property that $m(v_i)$ is a strictly increasing sequence (with $m$ defined as before). This is essentially equivalent to row reducing a matrix to upper triangular form.

Now let $$v_1=\epsilon+\sum_{r=1}^n e_{i_r}\lambda_r,$$ with the $\lambda_r\in k$ and non-zero, the $i_r$ a strictly increasing sequence of integers and $\epsilon\in M\langle x_1,x_2,\cdots\rangle$.

By the increasing property of $m(v_i)$, we may pick an integer $j$, such that $m(v_j)>2i_n$. We have: $$v_j=\epsilon'+\sum_{s=1}^{n'} e_{j_s}\mu_s,$$ with the $\mu_s\in k$ and non-zero, the $j_s$ a strictly increasing sequence of integers and $\epsilon'\in M\langle x_1,x_2,\cdots\rangle$. Crucially we also have $j_1>2i_n$.

We will derive our contradiction by exhibiting a non-trivial linear relation between $v_1$ and $v_j$:

$$v_1\left(\sum_{s=1}^{n'}x_{j_s}\mu_s\right)\\=\sum_{r=1}^n\sum_{s=1}^{n'}e_{i_r}x_{j_s}\lambda_r\mu_s\\\qquad\qquad\qquad\qquad=\sum_{r=1}^n\sum_{s=1}^{n'}e_{j_s}x_{i_r}\lambda_r\mu_s \qquad {\rm as\,\, each\,\,} j_s>2i_r,\\=v_j\left(\sum_{r=1}^n x_{i_r}\lambda_r\right)$$

Thus the $v_i$ are not linearly independent over $R$, so the $u_i$ could not have been either, and we can conclude that $M$ does not contain any infinite linearly independent sets.

tkf
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