Noetherian module and noetherian ring

Question

If $R$ is a noetherian ring then also $R[x]$ is a noetherian ring, i.e. $R[x]$ is noetherian as $R[x]$-module. Is $R[x]$ also noetherian as $R$-module?

Answer 1

$R[x]\cong \bigoplus_{i=0}^\infty R$ as $R$ modules, so obviously not.

Answer 2

Let $P_n$ be the $R$-sub-module of $R[x]$ whose elements are all polynomials of degree at most $n$.

Then you have the infinite chain $$P_0 \subset P_1 \subset P_2 \subset \cdots$$ with all inclusions proper.

Answer 3

A noetherian $R$-module has all its submodules finitely generated – in particular it has to be finitely generated as an $R$-module. If $R[x]$ were finitely generated, it would be generated by a finite number of polynomials as an $R$-module, so that polynomials would have a bounded degree.

Answer 4

Let $R$ be a commutative ring and $X$ an indeterminate. Then $R[X]$ is a noetherian $R$-module if and only if $R$ is the zero ring.

The other answers contain a proof (but not a statement) of this fact.

Edit. In the first version, I had written "$R[X]$ is a noetherian $R$-module if and only if $R$ is not the zero ring". Thanks to quid for having pointed out this mistake!