Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 (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
By the way...if no hints are given, this seems like a pretty tough exercise! – Pete L. Clark Feb 16 '11 at 4:13
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 $h$.

share|cite|improve this answer
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.

share|cite|improve this answer
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
(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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.