Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 1 / a

The assertion is false. The integers $\mathbb{Z}$ form a Noetherian ring, since every ideal is principal, but this ring is not an Artinian ring: the strictly descending chainnever stabilizes.

The assertion is true. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

The assertion is false. The module over a ring $\mathbb{Z}$ over itself is Noetherian, because its submodules are the principal ideals $n\mathbb{Z}$, but the descending chainshows that it is not an Artinian module.

The assertion is false. For a prime number $p$, the Prüfer p-group $C_{p^{\infty }}$ is an Artinian $\mathbb{Z}$-module: every proper subgroup is a finite cyclic group, so no infinite strictly descending chain of subgroups exists. It is not Noetherian because its cyclic subgroups form the strict ascending chain