Is $R[x]/(x)$ a free $R[x]$ module? Is $R[x]/(x)$ a free $R[x]$-module?
My thoughts are yes, it seems generated by {1+(x)}. Is this correct?
I am a bit confused since it seems to have torsion since $x\cdot f(x)=0$, and free modules can’t have torsion.
Secondly, what would be the rank of $R[x]\oplus R[x]/(x)$ as an $R[x]$-module. Is it 2?
Thanks.
 A: No, it is not possible for any ring $R$.
The right way to salvage the "torsion" idea is to point out that all nonzero free modules are faithful (have zero annihilator) but a quotient of a ring by a nonzero ideal $I$ is annihilated by $I$.
Actually you can go even further to show $(x)$ is not even projective. If it were, the following exact sequence splits:
$$
0\to (x)\to R[x]\to R[x]/(x)\to 0
$$
Then $R[x]=(x)\oplus N$ for some submodule $N$ of $R[x]$. Then there must be an idempotent $e\in R[x]$ such that $eR[x]=(x)$, so that $e$ acts as a left identity on $(x)$. But this is impossible since $ex\neq e$ no matter what $e$ is.
A: It is generated by $1+(x)$, but not freely: indeed
$$
x(1+(x))=0+(x)
$$
and $x\ne0$ in $R[x]$.
For an element $m$ of a basis in a free $A$-module $M$, the relation $am=0$ implies $a=0$.
By the same idea, no module of the form $A/I$, where $I$ is a nonzero ideal of $A$ is free, because every element of $A/I$ is annihilated by every nonzero element of $I$.
This assumes commutative rings, but switching to noncommutative ones is exactly the same.
A: It is not a free $R-$module because the $R-$module is not torsion free since $x$ is a element of $\operatorname{Tor}(M)$, but every free module only have a zero torsion element so that is a contradiction.
