# Showing that $\mathfrak{a}$ is a projective $S$-module which is not free

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.

• Do you know the characterization of projective modules as those which are direct summands of free modules? That should help you show that $\mathfrak{a}$ is projective. As for showing that $\mathfrak{a}$ is not free, your proof is incomplete but has the right idea. You can't just show that a particular spanning set is not a basis. On the other hand, your proof shows that every element $x \in \mathfrak{a}$ admits a nonzero $s \in S$ such that $sx = 0$, and so every nonempty subset $X \subset \mathfrak{a}$ is $S$-linearly dependent. Commented Jul 26, 2019 at 8:17
• You mean this: An $R$-module $M$ is projective if there exists another $R$-module $N$ such that $M \otimes N$ is a free $R$-module? Yes I know this, but not sure how it can help me. I would say $\mathfrak{a} \oplus R \cong R \oplus R$, and since $S$ is projective, so is $\mathfrak{a}$. Commented Jul 26, 2019 at 8:37
• You have not proved it is not free; you have just proved that $\{(0,1)\}$ is not a basis. There might be a different basis. Commented Jul 26, 2019 at 15:55

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.
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.