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Define $R:= \lbrace f \in C [0,1] \mid f(0)=f(1) \rbrace$, $V:= \lbrace f \in C [0,1] \mid f(0)=-f(1) \rbrace$.
$V$ is an $R$-module. I have no idea how to show it is not free.

I have tried to construct some function that hope can show $V$ has no independent set, but I find it hard to show it is continuous such that belongs to $R$. Is there any other way to show $V$ is not free? Thanks in advance!

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    $\begingroup$ This is a fun example... $\endgroup$
    – rschwieb
    Apr 11, 2022 at 17:01
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    $\begingroup$ Moreover, one can show that $V\oplus V\simeq R\oplus R$. $\endgroup$
    – user26857
    Apr 11, 2022 at 19:22

1 Answer 1

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If $f_1,f_2\in V$, then we have $(f_2^2)f_1-(f_1f_2)f_2=0$, so every two elements of $V$ are linearly dependent. Suppose that $V$ is free. Then $V$ is cyclic, that is, $V=Rf$. Since $f(0)f(1)\le0$ it follows that $f$ vanishes at a point $a\in[0,1]$. On the other side, $V$ contains functions that don't vanish at $a$.

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    $\begingroup$ Wow, so this example also involves some basic knowledge from analysis, that was fun! Thanks for your help $\endgroup$ Apr 12, 2022 at 4:35

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