Ok!

correct.is a Euclidean domain, hence it is principal ideal domain and therefore noetherian, because every principal ideal domain is a noetherian ring.

you've only proved that your ring is Noetherian -module. you haven't proved that it's a Noetherian "ring" yet.Or another idea: is isomorphic to as -module. The -module is noetherian (e.g. is noetherian ring) and therefore is also a noetherian -module.

there are at least three other ways to prove that is Noetherian:

1) is a Noetherian subring of and is finitely generated over thus is a Noetherian ring. why?

(i meant this is true in general, i.e. if is a ring and is a Noetherian subring of and is finitely generated -module, then is also a Noetherian ring.)

2) the direct product of finitely many Noetherian rings is always a Noetherian ring. why?

(never mind this part! and are not isomorphic "as rings".)

3) and is Noetherian by the Hilbert bases theorem.

well, is a free -module and every free module is faithfully flat.faithfully flat -algebra:

Let a exact sequence of -moduls. Then

is exact and we have

is exact is exact, this means that is a faithfully flat -algebra.

correct.Now the other part:

There is a one-to-onecorrespondencebetween the ideals in and the ideals in with .

is a principal ideal domain, this means, that for every ideal in there exists an element with .

Let .

.

There is only a finite number of divisors of a. It follows, that there is only a finite number of ideals in with , thus there is only a finite number of ideals in the ring and so is a artinian ring.

also correct.It follows: Every prime ideal in is a maximal ideal in ,

why?hence every prime ideal in is a maximal ideal in .