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 Post subject: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 06:10:38 UTC
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Joined: Fri, 9 Mar 2012 04:50:42 UTC
Posts: 15
Hi,

Does anyone have a (simplest possible) counterexample to the false statement that:
"Submodules of free modules are free"?

Thanks a lot!

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 06:49:22 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6009
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
yoyostein88 wrote:
Hi,

Does anyone have a (simplest possible) counterexample to the false statement that:
"Submodules of free modules are free"?

Thanks a lot!

The easiest is probably to take ideals of a domain. Obviously you want a non-PID, and what is the simplest example you can think of?

Alternatively, play with non-domains and think about the annihilator of some zerodivisor.

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 13:43:41 UTC
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Joined: Fri, 9 Mar 2012 04:50:42 UTC
Posts: 15
Consider Z[X]

I suspect the ideal (2,X) is not a free submodule?

What would be a quick way to see that it is not free?

Thanks.

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 14:22:34 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
yoyostein88 wrote:
Consider Z[X]

I suspect the ideal (2,X) is not a free submodule?

What would be a quick way to see that it is not free?

Thanks.

It is not projective.

If you don't know that, then note that any generating set must have at least 2 elements (the ideal is not principal --- I trust you know how to prove this?), but now the ring structure tells you such a set cannot be linearly independent --- for any .

My other hint was for you to consider with , which would count as an example of being (one of) the simplest when defined as the example with the smallest number of elements in your ring.

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 14:52:34 UTC
 Member

Joined: Fri, 9 Mar 2012 04:50:42 UTC
Posts: 15
Thanks a lot!

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 15:46:34 UTC
 Member

Joined: Fri, 9 Mar 2012 04:50:42 UTC
Posts: 15
Hi,

Just one last question, what is in ?

Thank you so much.

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 Post subject: Re: Counterexamples to: Submodules of free modules are freePosted: Fri, 30 Mar 2012 15:58:42 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6009
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
yoyostein88 wrote:
Hi,

Just one last question, what is in ?

Thank you so much.

is just an infinitesimal that we adjoin to to get the nonreduced ring , where .

By the way, the other ring of 4 elements also works, and that too I'll leave as an exercise for you.

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