Problem: Let $R$ be a non-trivial ring and let $S = R \oplus R$. Show that $\mathfrak{a} = \left\{ (0, r) \mid r \in R \right\}$ is a projective $S$-module which is not free.
Attempt: I know that free $R$-modules are projective. Now $R$ as an $R$-module is free, so that $S = R \oplus R$ is free, and hence projective. But a submodule of a projective module need not necessarily be projective.
So I don't know how to show that $\mathfrak{a}$ is projective.
That it is not free as an $S$-module, is clear from this I believe: a basis would consist of $\left\{ (0, 1 ) \right\}$. But then $(1,0) \cdot (0, 1) = (0, 0)$ while the coefficient is non-zero.
Help is appreciated.
To show that $\mathfrak{a}$ is projective you need to find another module so that the direct sum of the two is free. Let me write $\mathfrak{b} = \{(r,0) \:|\: r \in R\}$, then we have $S = \mathfrak{a} \oplus \mathfrak{b}$ and so $\mathfrak{a}$ is a direct summand of $S$ which is free as $S$-module and hence $\mathfrak{a}$ is projective.
By the way, it is true that submodules of projective modules do not have to be projective, however, direct summands of projective modules are projective (this is an easy exercise).
To show that $\mathfrak{a}$ is not a free $S$-module, it suffices to show that $\mathfrak{a}$ is not a faithful $S$-module. In fact, it is easy to show that $Ann_S(\mathfrak{a}) = \mathfrak{b}$ whereas $Ann_S(M) = \{0\}$ for every free $S$-module $M$.
Note that a module being projective or free depends on the ring you consider. The way you phrased your attempt makes me believe that this is something you should think about a bit more: It is true that $R$ is $R$-free and thus also $R$-projective and thus $S$ is also $R$-free and $R$-projective as are $\mathfrak{a}$ and $\mathfrak{b}$ (as $R$-modules they are isomorphic to $R$). At the same time, $S$ is also $S$-free and $S$-projective but $\mathfrak{a}$ and $\mathfrak{b}$ are only $S$-projective and not $S$-free.