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-one correspondence between 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 .
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 .