1
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ You should probably ask this on math.stackexchange.com... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 9, 2011 at 21:14
  • 3
    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Feb 9, 2011 at 21:33
  • 1
    $\begingroup$ I should add that Martin really means 'hard', as in gnarly problems out of Lang or similar. $\endgroup$
    – David Roberts ♦
    Commented Feb 9, 2011 at 23:00
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Feb 9, 2011 at 23:14
  • $\begingroup$ Hint: Möbius band. $\endgroup$
    – Pete L. Clark
    Commented Feb 10, 2011 at 5:07

0

Reset to default

Browse other questions tagged .