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Free modules over a commutative ring $R$ with $1$ have well-defined rank.

I have been wondering if there is a ring $R$ such that there are free modules $M'\subset M$ with $\operatorname{rank}(M')>\operatorname{rank}(M)$ for a while, but can't come up with examples or a proof of the contrary in general.

All I know is that this never happens if $R$ is a domain, since you can consider $M'$ and $M$ as vector spaces over the fraction field of $R$.

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marked as duplicate by YACP, Amzoti, Daniel Rust, Davide Giraudo, Branimir Ćaćić Aug 16 '13 at 20:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This has been asked at least 5 times on math.SE. –  Martin Brandenburg Aug 16 '13 at 13:09
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@YACP: Must everyone be familiar with all 163,765 (as of writing) questions posted on this site lest they incur your downvoting wrath? FYI, rschweib has started a meta thread about this. –  Rahul Aug 16 '13 at 19:50
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@YACP: I might see your point if any of the people who answered this question also answered, or even commented on, the other, but as far as I can see, none did. Questions slip by without people seeing them, so I see no reason to penalize the people who answered here. –  robjohn Aug 16 '13 at 19:50
    
@YACP Among commutative algebraists, maybe, and those who can recall it from the thousands of other questions they've read. The question can still be novel for those of us who have never cracked open Atiyah-Macdonald and/or didn't absorb the question at the time. –  rschwieb Aug 21 '13 at 12:45

2 Answers 2

up vote 7 down vote accepted

This can't happen. See Corollary 5.11 of Keith Conrad's notes on exterior powers:

http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf

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Please try to describe as much here as possible in order to make the answer self-contained. Links are fine as support, but they can go stale and then an answer which is nothing more than a link loses its value. –  robjohn Aug 16 '13 at 20:01

Lemma

One way to see why free submodules of free modules over a commutative ring have to have lesser or equal rank uses this lemma. The lemma is shown here, that surjections of finitely generated modules over commutative rings are necessarily isomorphisms.

If $R^m$ were isomorphic to a submodule of $R^n$ and $m>n$, you would easily be able to construct a projection of that submodule onto $R^n$ which would then necesarily be an isomorphism, but that is impossible since we have decided ranks of f.g. free modules are well-defined for commutative rings.


Bonus (=ignorable)

Rings which have well defined ranks for their finitely generated free modules are said to have the IBN property. This property is: "If $R^n$ and $R^m$ are isomorphic as right $R$ modules, then $m=n$."

The proof that a commutative ring $R$ has the IBN is pretty easy if you can make a few jumps. First of all, after picking a basis for $R^n$ and $R^m$, one can see this amounts to finding an $n\times m$ matrix $A$ and an $m\times n$ matrix $B$, both over $R$, such that $AB=I_n$ and $BA=I_m$. The second observation is that we can pick a maximal ideal $M$ of $R$ and project from $R$ onto $R/M$, a field. But if you apply this projection to the entries of $A$ and $B$, you wind up with two matrices which are mutually inverse over a field: but that implies $m=n$ since we know dimension is well-defined for vector spaces.

There are examples for rings in general, but not for commutative rings. The familiar example for rings with ranks which aren't well defined are that of full linear rings.

Take $R$ to be the ring of transformations of a countable infinite dimensional vector space. It's not hard to show that $R\cong R^2$ as $R$ modules, and hence by induction $R\cong R^n$ for any $n\in \Bbb N$, and then $R^n\cong R^m$ for any positive integers $n,m$.

If by "rank" you are thinking of the $n$ and $m$ above, then this example shows that "rank" of a free module isn't always well defined.

Moreover, if this didn't convince you already, take $R^2$ and look at the submodule $R\times \{0\}$. The submodule is still isomorphic to $R^m$ for whatever positive integer $m$ you like.

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Hi, thanks for all this but the first sentence in my post says that all commutative rings have IBN; your example in the non-commutative case is the usual one that motivates proving this fact. –  JessicaB Aug 16 '13 at 12:34
    
Dear @JessicaB : I was not done editing. In the interval I had added the proof for commutative rings. Hope you find the outline of the proof helpful. I think I will reorganize it to prioritize the proof. –  rschwieb Aug 16 '13 at 12:42
    
Dear @rschwieb, I don't think the OP was asking for a proof that rank is well-defined for free modules over commutative rings with identity. She was asking if it's possible to have a free submodule with larger rank than the ambient free module. –  Keenan Kidwell Aug 16 '13 at 12:43
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Dear @KeenanKidwell : The problem is that nobody is willing to wait more than 15 minutes for an anwer to be complete. The solution I just finished writing does show that. Of course, I'm never going to win a speed race against answers that are just links to complete solutions :) –  rschwieb Aug 16 '13 at 12:50
    
@KeenanKidwell Your eyes aren't fooling you though, I did initially key off the first sentence as a question rather than a statement, which led me down this path. In light of that, I should probably reduce that cruft. –  rschwieb Aug 16 '13 at 12:54

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