Talk:Hilbert's Basis Theorem
Do we need to specify "a Noetherian ring with identity" since by definition a N.R is already a "commutative ring with identity"? --prime mover 05:37, 10 September 2010 (UTC)
- Well as defined on this site, a ring automatically has identity, since the multiplicative part of the ring forms a group in the definition. The trick, of course, is that not all texts define ring this way; in many texts they call a ring what this site calls a semiring. Secondly, the existence of an ideal in an algebraic structure does not necessarily imply the existence of an identity in that structure (I think - does it?) So although my class notes specify that the ring have an identity, I'm not sure that it's necessary. Something to think on. J D Bowen 06:00, 10 September 2010 (UTC)
- "as defined on this site, a ring automatically has identity" - nope, go check ring. The additive part does form a group, but the multiplicative part may only be a semigroup. This is important.
- $\left\{{\ldots, -4, -2, 0, 2, 4, \ldots}\right\}$ is a ring, but has no multiplicative identity. Can't stop to discuss it at the moment, I have to go to work, I'll be back to look at it again this evening. --prime mover 07:00, 10 September 2010 (UTC)
On the subject of the statement, "an extension of transcendence degree 1" would be more precise.
--Linus44 18:53, 6 February 2011 (UTC)
Redone proof
It dosen't hurt to say "commutative with 1" as a reminder, since most people will know the Noetherian condition, but exactly what rings it applies to isn't so memorable, so I left that in.
Otherwise it was a bit vague, I added a proof taken from Eisenbud - Commutative Algebra; in my opinion, the neatest proof. --Linus44 15:55, 5 February 2011 (UTC)
There might be a case to add a (more or less trivial inductive) extension that any polynomial ring in finitely many indeterminates will be Noetherian; this is important in eg. algebraic geometry. --Lord_Farin 12:10, 28 April 2012 (EDT)