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Let's define associated elements of a module over a commutative ring as elements that are scales of each other (Associated elements in groups and modules).

Let's say two bases of a free module are associated if each element of one basis is associated with an element from another one.

If a module has a basis, all associated bases are given "for free": there is no need to search them.

How do we know if a free module has a basis that is not associated with the current one?

Does it make sense to group bases of a free module into classes of associates?

Example of associated bases in $\mathbb Z \times \mathbb Z$:

  • $B_1 = \{[(1,0)], [(0,1)]\}$, where $[(1,0)] = \{(1,0), (-1,0)\}$ and $[(0,1)] = \{(0,1), (0,-1)\}$ is one set of associated bases;
  • $B_2 = \{[(1,0)], [(1,1)]\}$, where $[(1,1)] = \{(1,1), (-1,-1)\}$ is another set of associated bases that is not associated with $B_1$.
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  • $\begingroup$ If it has rank $\ge2$? $\endgroup$ Sep 29, 2019 at 17:19
  • $\begingroup$ @LordSharktheUnknown: I meant an arbitrary rank (including infinite - not sure if this is a correct usage of "rank"). But, of course, ranks $0$ and $1$ are not that interesting. $\endgroup$
    – Alex C
    Sep 29, 2019 at 17:23

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Let $R$ be a commutative ring. Let $M$ be a free module (so it has a basis). The rank of $M$ is defined to be the cardinality of any basis. (This is well-defined since all bases have the same cardinality: see here for an argument that reduces this statement to the analogous statement for vector spaces over a field, and here for the proof for vector spaces.)

Let $I$ be a set whose only purpose is to index the elements of a basis of $M$. (You said in comments you want $M$ to be able to have arbitrary rank, so $I$ is arbitrary.) Let $(m_i)_{i \in I}$ and $(m_i')_{i\in I}$ be two bases of $M$ over $R$. (Since all bases have the same cardinality, per above, it was reasonable for me to use $I$ to index both bases.)

If I understand your definition right, you say that $(m_i)_{i \in I}$ is associated to $(m_i')_{i\in I}$ if there exists a permutation $\sigma:I\rightarrow I$ and two sets of elements of $R$, $(r_i)_{i\in I}$ and $(r_i')_{i\in I}$, such that

$$ m_{\sigma(i)}' = r_im_i \text{ and } m_i = r_i'm_{\sigma(i)}' $$

for all $i\in I$. (Do you agree this is your definition? I really just wrote it down to check that I understand your definition right. It's saying: the two bases are "associated" if they can be reindexed so that corresponding elements are scalar multiples of each other. The permutation $\sigma$ denotes the reindexing that might be required, and the $r_i$ and $r_i'$ are the needed scalars.)

I will take your second question first because the answer allows me to reword your first question in a way that makes it slightly clearer to me.

Second question: Does it make sense to group bases of a free module into classes of associates?

If you mean "does it make sense to...?" in a strictly mathematical way, I think the answer is yes. The relation "associated" that you've defined is an equivalence relation, and it therefore partitions the collection of all bases into equivalence classes of associated bases, by the fundamental theorem of equivalence relations.

To make sure your "associated" is an equivalence relation, we need to check it is reflexive, symmetric, and transitive. I think it is clear it is reflexive and symmetric from your definition, and nearly clear that it is transitive. I spent some time just now writing out a proof using my formal rephrasing of your definition, given above, but in the end I think it doesn't really add any clarity to do that. My version of the definition, in order to be able to write symbols, canonized one of the bases as "first" and the other as "second," and this introduced an asymmetry that's not really there. Proving the relation is symmetric meant fighting against this chimeric asymmetry and this felt like tilting at windmills :)

That said, your definition (if I understand it right) says two bases are associated if they can be reindexed so that corresponding basis elements are scalar multiples of each other. Once this reindexing is done, we have to check that "scalar multiples of each other" is an equivalence relation separately on each index. But this just comes down to these three simple things:

  • Reflexivity - a basis element is a scalar multiple of itself. (Need to assume $R$ is a ring and not a rng for this!)
  • Symmetry - $m,m'\in M$ being scalar multiples of each other is, on its face, a symmetric relation between $m$ and $m'$, i.e. it doesn't change meaning if you switch $m$ and $m'$.
  • Transitivity - if $m,m'$ are scalar multiples of each other, and $m'$ and $m''$ are also scalar multiples of each other, then $m$ and $m''$ are also scalar multiples of each other, as you can check by writing it out.

Since this shows your "associated" relation is an equivalence relation, it makes sense at least from a purely mathematical standpoint to speak of classes of associated bases.

First question: How do we know if a free module has a basis that is not associated with the current one?

Now that we know associated bases come in classes, I can interpret this question as: "How do we know if there are at least two classes?" (Let me know if this is not what you meant.)

The answer is that if the rank of $M$ is at least two, there are always at least 2 classes!

Indeed, let $(m_i)_{i\in I}$ be any basis.

If the cardinality of $I$ is at least 2, then we can take two distinct elements $\alpha,\beta\in I$. Then I claim that replacing $m_\alpha$ with $m_\alpha+m_\beta$ creates a new basis that is not associated to the old one. (More formally, the new basis is

$$(n_i)_{i\in I},$$

where $n_\alpha = m_\alpha+m_\beta$, and $n_i=m_i$ for all $i\neq \alpha$.) I leave the proof to you! The fact that the new basis is not associated to the old follows from the linear independence of $m_\alpha$ and $m_\beta$ over $R$.

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  • $\begingroup$ By associated bases I meant bases on the same set of classes of associated elements. The definition is different from the one in the answer. My question is similar to this one: "We know all classes of associates for the current basis; do we know if an arbitrary class of associates is also in some bases?". My guess, the fact that a certain class is a "basis class" or not is a fundamental property that defines the structure itself. It cannot follow from other properties. $\endgroup$
    – Alex C
    Feb 11, 2020 at 13:31
  • $\begingroup$ I do not understand your definition. What do you mean "bases on the same set of classes of associated elements"? (I am not sure what could be meant by "a basis on a class.") Maybe can you illustrate with an example? $\endgroup$ Feb 11, 2020 at 19:54
  • $\begingroup$ In $\mathbb Z \times \mathbb Z$: $B_1 = \{[(1,0)], [(0,1)]\}$, where $[(1,0)] = \{(1,0), (-1,0)\}$ and $[(0,1)] = \{(0,1), (0,-1)\}$ is one set of associated bases. $B_2 = \{[(1,0)], [(1,1)]\}$, where $[(1,1)] = \{(1,1), (-1,-1)\}$ is another set of associated bases that is not associated with $B_1$ since it has different classes of associates. Is it more clear? $\endgroup$
    – Alex C
    Feb 11, 2020 at 20:16
  • $\begingroup$ I added the example into the question. $\endgroup$
    – Alex C
    Feb 11, 2020 at 20:56
  • $\begingroup$ As far as I can tell, your notion of associated bases does coincide with the interpretation of it in my answer. $\endgroup$ Feb 12, 2020 at 3:04

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