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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.

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    $\begingroup$ $\Bbb Z$ is a PID, so submodules of free modules are free. $\endgroup$ – Pedro Tamaroff Dec 8 '14 at 19:53
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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, $x\otimes y-y\otimes x\mapsto x\otimes y-y\otimes x=1\otimes xy- 1\otimes yx=1\otimes xy-yx=0$.

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    $\begingroup$ I like to think in finite dimensional algebras, and the example I would give is take $R$ the dual number $k[x]/(x^2)$. Obviously $R$ itself is free and projective, and the 1-dim $k$-space $k.x$ spanned by $x$ is a $R$-submodule of $R$, and $k.x\cong R/k.x$ as $R$-module. So we have non-split exact sequence $0\to k \to R\to k\to 0$ which says that $k$ is non-projective submodule of free module $R$. $\endgroup$ – Aaron Dec 9 '14 at 12:44
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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.

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  • $\begingroup$ This example doesn't work: $\Bbb Z/6 \cong \{0,2,4\} \oplus \{0,3\}$. (Actually $\Bbb Z/6$ is semisimple, so any module over $\Bbb Z/6$ is projective.) $\endgroup$ – MatheinBoulomenos Jun 10 '18 at 22:23

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