Is there a non-projective submodule of a free module? Is there an example of a commutative ring $R$ with identity such that there exists a free $R$ module $M$ that has a non-projective submodule?
I tried experimenting with modules over $\mathbb Z$ but it lead me nowhere.
 A: In fact all submodules of free $\mathbb{Z}$-modules are projective, even free, so the experimentation you're doing won't succeed. (This is because $\mathbb{Z}$ is a principal ideal domain.) Taking $(x,y)$ a submodule of $k[x,y]$ for $k$ some field will work better. $(x,y)$ is not projective, which we can show by showing it's not flat, since projective modules are flat.
There's an exact sequence $0\to (x,y)\to k[x,y]\to k\to 0$, where $k$ has the $k[x,y]$ module structure in which $x$ and $y$ act by $0$. Tensoring with $(x,y)$, we get the sequence $0 \to (x,y)\otimes_{k[x,y]}(x,y)\to k[x,y]\otimes_{k[x,y]}(x,y)\to (x,y)\otimes_{k[x,y]}k$, which is not exact. Indeed, $x\otimes y-y\otimes x\mapsto x\otimes y-y\otimes x=1\otimes xy- 1\otimes yx=1\otimes xy-yx=0$. 
A: Let $\mathbb Z/6$ be free $\mathbb Z/6$ module itself. Consider the set  $A$ = $\{0,2,4\}$ . We can easily make $A$ be the submodule of $\mathbb Z/6$ by conventional scalar multiplication. 
Consider this surjective homomorphism: $f:\mathbb Z/6 \to A$ such that $f(1):= 2$. 
Because $f$ is surjective, that means if $A$ is projective $\mathbb Z/6$ module we must have $A$ is direct summand of $\mathbb Z/6$. But it's not the case. 
