This follows from Hilbert's basis theorem, which is valid for polynomial rings over any noetherian ring. But is there a more elementary proof, knowing that $\mathbb{Z}$ is a PID (even a Euclidean domain)?
Asked
Active
Viewed 2,900 times
1
-
Hilbert's Basis Theorem is elementary, the proof only takes about 5 lines and is self-contained. – Martin Brandenburg Feb 09 '13 at 20:45
-
@MartinBrandenburg Probably, elementary is the wrong word. It's a clever proof, though, and not very constructive, and I wonder if there's a more obvious proof for rings with better structure – user61622 Feb 09 '13 at 20:48
-
Wait two years, then you will find the proof very constructive, obvious and wonder why you haven't found it for yourself ;). – Martin Brandenburg Feb 09 '13 at 20:55
-
There are very constructive proofs of the theorem. – Mariano Suárez-Álvarez Feb 09 '13 at 21:37
1 Answers
3
Using the information you have about the ideals of $\mathbb Z$ and of polynomial rings in one variable over fields (both of which are PIDs), you can —with some work— describe all ideals of the ring, and then check the ACC by hand.
Mariano Suárez-Álvarez
- 135,076
-
1I have never heard of a classification of ideals in $\mathbb{Z}[X]$. Can you state it? – Martin Brandenburg Feb 09 '13 at 20:46
-
1Thanks Mariano - could you also mention (of course not an exact proof) a bit more in detail how you get these generators? – user61622 Feb 09 '13 at 20:48
-
1Ok I've asked here: http://math.stackexchange.com/questions/300170/ideals-of-mathbbzx – Martin Brandenburg Feb 11 '13 at 12:42