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This is exercise 38 from Chapter 3 (Modules and Vector Spaces) in Algebra by Adkins and Weintraub (GTM). How do you solve this problem?

Let \begin{equation*} R = \lbrace f : [0, 1] \to \Re : f \;\text{ is continuous and} \; f (0) = f (1) \rbrace \end{equation*} and let \begin{equation*} M = \lbrace f : [0, 1]\to \Re : f \;\text{is continuous and} \; f (0) = - f (1) \rbrace. \end{equation*} Then $R$ is a ring under addition and multiplication of functions, and $M$ is an $R$-module. Show that $M$ is a projective $R$-module that is not free.

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While browsing T.Y. Lam's excellent text Lectures on Modules and Rings, I found that this problem is treated as an example on p. 28. (His solution is very similar to Mariano's, and Swan's Theorem is not mentioned or used.) –  Pete L. Clark Feb 13 '11 at 7:38
    
@Pete: nice catch. He adds a pretty proof that $P^2=R$. –  Mariano Suárez-Alvarez Feb 16 '11 at 0:21
    
@Mariano: this is such a nice example that probably a lot of papers and texts have their own distinctive take on it. Including a paper of mine: see the discussion following Theorem 12 in math.uga.edu/~pete/ellipticded.pdf. (And note that this is all but completely irrelevant to the main result of the paper, but I couldn't resist!) –  Pete L. Clark Feb 16 '11 at 4:12
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By the way...if no hints are given, this seems like a pretty tough exercise! –  Pete L. Clark Feb 16 '11 at 4:13

2 Answers 2

up vote 6 down vote accepted

I'll consider the interval $[0,2\pi]$ for notational simplicity. Consider the matrix $$ A = \left( \begin{array}{cc} \sin ^2\tfrac{\theta }{2} & - \sin \tfrac{\theta }{2} \cos \tfrac{\theta }{2} \\ -\sin \tfrac{\theta }{2}\cos \tfrac{\theta }{2} & \cos ^2\tfrac{\theta }{2} \end{array} \right), $$ which defines an $R$-linear map $p:R^2\to R^2$. Computing $A^2$ we see that $p^2=p$, so $p$ is idempotent, and its kernel is a projective $R$-module $P$.

Now consider the map $$ \phi : f\in M \mapsto(f(\theta)\cos \tfrac{\theta }{2},f(\theta)\sin \tfrac{\theta }{2}) \in R^2. $$ It is clearly an $R$-linear injective map, whose image is precisely the kernel $P$ of $p$. It follows that $M\cong P$, and this shows projectivity.

Non-freeness is more subtle...

There is a morphism of rings $\varepsilon:R\to\mathbb R$ given by evaluation at $0$. One can see that $P\otimes_R\mathbb R$ is of dimension $1$ over $\mathbb R$, so that if $P$ is free, then it is free of rank $1$. In that case, $M$ would be free of rank $1$: suppose so, and let $h\in M$ be a generator. It is immediate then that every element of $M$ has to vanish where $h$ vanishes. But one can easily find an element of $M$ whose only zero is not a zero of $h$.

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One might mention that the key point in the last paragraph is that $h$ must have at least one zero (because of the condition that $h(0) = - h(1)$), and that this is the difference between $R$ and $M$. From the point of view of Pete Clark's answer, there is no non-vanishing section of the Mobius band over the circle, and so it defines a line bundle that can't be trivialized. –  Matt E Feb 12 '11 at 3:56
    
It seems this answer is geared towards the interval $[0,2\pi]$ and the correct argument for the interval $[0,1]$ would be $\pi\theta$ instead of $\theta/2$? –  joriki Feb 16 '11 at 0:09
    
@joriki: indeed. –  Mariano Suárez-Alvarez Feb 16 '11 at 0:20
    
I'm not sure about the notational simplicity :-). For notational simplicity $[0,\pi]$ would have been best. –  joriki Feb 16 '11 at 5:33

I made a similar comment on MO where this question was first posted. Here is an elaboration:

Since the circle $S^1$ can be thought of as the unit interval $[0,1]$ with the two endpoints identified, $R$ may be viewed as the ring of all real-valued continuous functions on $S^1$.

My hint is to view $M$ as the module of global sections of the Möbius band. For this, think about building the Möbius band as an identification space of $[0,1] \times \mathbb{R}$: you glue the two ends together with a half-twist.

It is only fair to mention that I have implicitly in mind the celebrated theorem of Richard Swan which gives an equivalence between vector bundles over a compact base and modules over the ring of continuous functions on the base: see e.g. Chapter 6 of these notes. Perhaps it is possible to give a more elementary solution of this problem: I would be happy to see one myself.

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Considering the case of polynomials instead of arbitrary functions might help, too. –  Mariano Suárez-Alvarez Feb 11 '11 at 19:31
    
The Swan theorem became much more intuitive to me when I found out that it was a topological analog of the algebraic result for varieties: projective modules over a noetherian ring $R$ correspond to vector bundles (a.k.a. locally free sheaves) over $\mathrm{Spec}(R)$. –  Akhil Mathew Feb 16 '11 at 0:45
    
@Akhil: sure; many people lump these results together as the "Serre-Swan Theorem". I actually find that a little strange, and moreover think that -- your own experience notwithstanding -- students of algebraic geometry would benefit from knowing about topological vector bundles before they start learning about algebraic vector bundles. (Thus in my commutative algebra notes I treat Swan's theorem but not Serre's: the latter is a super-important result, but it will get its place in algebraic geometry courses.) –  Pete L. Clark Feb 16 '11 at 4:08
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(I also think that the name "Serre-Swan Theorem" is meant to suggest that Serre -- who after all began his career as a topologist -- also had the topological case in mind but chose not to enunciate it. This may be unfair to Richard Swan...but I wouldn't feel too bad for him: there are plenty more "Swan's Theorem"s!) –  Pete L. Clark Feb 16 '11 at 4:12

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