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.

  • 2
    $\begingroup$ $\Bbb Z$ is a PID, so submodules of free modules are free. $\endgroup$ – Pedro Tamaroff Dec 8 '14 at 19:53

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, $x\otimes y-y\otimes x\mapsto x\otimes y-y\otimes x=1\otimes xy- 1\otimes yx=1\otimes xy-yx=0$.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.