Is $(x)$ as a $k[x]$ module free?

Question

Is $(x)$ as a $k[x]$ module free?

I think it is free because it seems the basis element is $x$ and it is not annihilated by any element of $k[x]$.

Thanks!

Answer 1

Let $k$ be a field and let $k[x]$ denote the polynomial ring. Let $\varphi:k[x]\to k[x]$ be the multiplication by $x$. Then $\varphi$ is an injective $k[x]$-module homomorphism whose image is $\langle x\rangle=xk[x]$ that's the ideal generated by $x$. Consequently, $\langle x\rangle\cong k[x]$ as $k[x]$-module, hence $\langle x\rangle$ is a free $k[x]$-module of rank $1$.

The same arguments holds over any commutative ring (with identity).

Answer 2

Yes. More generally, if $A$ is a commutative ring with $1$ and $a \in A$ not a zero divisor, then $(a)$ is free of rank $1$.