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Let $R$ be an integral domain and $M$ be a free module over $R$.

My question is this:

Is the intersection of two free submodules of $M$ a free submodule?

P.S: We need the intersection of two free submodules is not $\{0\}$.

rschwieb
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Mod.esty
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  • Take a ring $R$ and two principal ideals $(a), (b) \subseteq R$. Is their intersection always principal? –  Apr 30 '20 at 17:21
  • @PaulK Yes, this is what I have thought and I know it is not true. But I don't construct a specific example. – Mod.esty Apr 30 '20 at 17:29
  • @PaulK Especially $R$ is an integral domain. – Mod.esty Apr 30 '20 at 17:34
  • P.P.S. The zero module is a free module on zero generators.... – rschwieb Apr 30 '20 at 17:58
  • @rschwieb Yes, but I think zero module don't have basis because 0 is linear dependent. I think this definition is a little insufficient. – Mod.esty Apr 30 '20 at 18:05
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    @algebra.And.analysis Why would $0$ be in the basis? I said it has zero generators. $0\notin \emptyset$ – rschwieb Apr 30 '20 at 18:17
  • @rschwieb Sorry...in the book I saw, the definition of free module is that it has a basis...I really don't know what is zero generator......I think every free submodule and its basis relate to $R$. – Mod.esty Apr 30 '20 at 18:52
  • @algebra.And.analysis I mean that the generating set for the $R$ module ${0}$ is $\emptyset$, which has zero elements. It is vacuously linearly independent. – rschwieb Apr 30 '20 at 18:56
  • @rschwieb I understand what you said, thank you. But I still feel that an $\emptyset$ can generate an element zero is a little strange. – Mod.esty Apr 30 '20 at 19:31
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    @algebra.And.analysis The relevant convention is that the sum of nothing is $0$, so the empty set is a basis for the trivial module because it spans the module (the sum of zero elements from the set is $0$) and is linearly independent (since there's no dependence relation since there are no elements). – Matt Samuel Apr 30 '20 at 19:53
  • @PaulK There has since been a conversation about the intersection of principal ideals question at this post – rschwieb May 01 '20 at 14:24

1 Answers1

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Let $P$ be a projective module which is not free over a domain $R$. (There are such domains.) We know that we may complement $P$ with another module so that $P\oplus N$ is a free module.

Consider the module $M=N\oplus P\oplus N\oplus P$ which is an (external) direct sum of two free $R$ modules, so it is free. Clearly $N\oplus P\oplus\{0\}\oplus\{0\}\cong \{0\}\oplus P\oplus N\oplus\{0\}$ are free modules. But their intersection is, you guessed it, $\{0\}\oplus P\oplus\{0\}\oplus\{0\}\cong P$, not free.

rschwieb
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  • I think we can consider $M = N \bigoplus P \bigoplus N \bigoplus P$. I don't know whether $M = N \bigoplus P \bigoplus N $ is a free module... – Mod.esty May 01 '20 at 01:44
  • Because $P \bigoplus N$ is a free module, there is a basis $S_1$. $N \bigoplus N$ is also a free module which has a basis $S_2$. $(0, 1,0)$ can be linear expressed by $S_1$ and $(0, -1,0)$ can be linear expressed by $S_2$. Thus $S_1 \bigcup S_2$ is linearly dependent. – Mod.esty May 01 '20 at 01:52
  • @algebra.And.analysis Yes, sorry, I was just reading your title and forgot that you later also asked for $M$ to be free. It is easily adjusted as you noticed. – rschwieb May 01 '20 at 01:56
  • @rschwieb Sorry, $P \bigoplus N$ is also a free module. – Mod.esty May 01 '20 at 01:56
  • @algebra.And.analysis As for what you wrote , $S_1\cap S_2$ doesn’t even have to be a subset of $P$, and moreover it may even be empty. From what you’re saying you sound like you expect it to be a basis of $P$ but you are nowhere near to proving that. – rschwieb May 01 '20 at 02:06
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    @algebra.And.analysis By the way, I would encourage you to append your thoughts to your original post. As it is, the post has been closed for looking like a problem statement, and may eventually be deleted. From what you've said in comments I believe you've put some work into it, so it's in your best interest to share ( in your post, not comments.) – rschwieb May 01 '20 at 12:45
  • @algebra.And.analysis I don't think it's "too simple." Coming up with counterexamples is an art form that can be very easy or incredibly hard, depending on the question and the person asking. Please do consider transferring some of the thoughts you processed in the thread here into the original post. – rschwieb May 01 '20 at 13:44
  • @algebra.And.analysis The reason this question is not too easy is that it goes beyond principal ideal domains. For principal ideal domains (like $\mathbb Z$ or $F[x]$ for a field $F$) submodules of free modules are free modules, so you could not find a counterexample there. So you do need to know more advanced things (in this case projective modules) to come up with an example. Many people with less experience would have a hard time going beyond PIDs. – rschwieb May 01 '20 at 14:18
  • @algebra.And.analysis I like the question but I understand why people have voted to close it. It would be a shame to lose it. Fortunately it is not too late to improve it, and I hope you are able to. – rschwieb May 01 '20 at 14:30