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yoyostein88
 Post subject: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 06:10:38 UTC 
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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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outermeasure
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 06:49:22 UTC 
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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.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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yoyostein88
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 13:43:41 UTC 
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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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outermeasure
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 14:22:34 UTC 
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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 --- (r_2)r_1+(-r_1)r_2=0 for any r_1,r_2\in R.

My other hint was for you to consider (\epsilon)\lhd\mathbb{F}_2[\epsilon] with \epsilon^2=0, 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.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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yoyostein88
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 14:52:34 UTC 
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Thanks a lot!
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yoyostein88
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 15:46:34 UTC 
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Hi,

Just one last question, what is \epsilon in (\epsilon)\lhd\mathbb{F}_2[\epsilon]?

Thank you so much.
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outermeasure
 Post subject: Re: Counterexamples to: Submodules of free modules are free
PostPosted: Fri, 30 Mar 2012 15:58:42 UTC 
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yoyostein88 wrote:
Hi,

Just one last question, what is \epsilon in (\epsilon)\lhd\mathbb{F}_2[\epsilon]?

Thank you so much.


\epsilon is just an infinitesimal that we adjoin to \mathbb{F}_2 to get the nonreduced ring \mathbb{F}_2[\epsilon], where \epsilon^2=0.

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

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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