# Projective but not free (exercise from Adkin - Weintraub) [closed]

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.

-

## closed as too localized by Martin Brandenburg, Andreas Thom, Yemon Choi, David Roberts, Chandan Singh DalawatFeb 10 '11 at 5:23

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

You should probably ask this on math.stackexchange.com... – Mariano Suárez-Alvarez Feb 9 '11 at 21:14
This site is not for asking the solutions of exercises (read the FAQ). There are some exceptions, for example when you give more background, show your results and in particular when the exercise is actually a hard problem. – Martin Brandenburg Feb 9 '11 at 21:33
I should add that Martin really means 'hard', as in gnarly problems out of Lang or similar. – David Roberts Feb 9 '11 at 23:00
Exercises in elementary graduate texts are usually (though not in absolutely all cases) just exercises. It's best to spend time thinking it through rather than asking other people. – Jim Humphreys Feb 9 '11 at 23:14
Hint: Möbius band. – Pete L. Clark Feb 10 '11 at 5:07