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.