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Does a commutative ring satisfying the following two properties exist?

  1. All ideals are finitely generated;

  2. There are prime ideals with arbitrarily large (finite) minimal generating sets.

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This seems contradictory to me. A ring with the property that all ideals are finitely generated but the ring has ideals with arbitrarily large number of generators? Do I misunderstand something? – anon271828 Jan 13 '13 at 21:08
@anon271828: How is that contradictory? – Chris Eagle Jan 13 '13 at 21:18
@ChrisEagle: I don't see how an ideal can have finitely many generators AND infinitely many... – anon271828 Jan 13 '13 at 22:08
@anon271828 It looks like ast's phrasing misled you a bit. What Ast is asking about is: "is there a Noetherian ring such that for every $n\in \Bbb N$, there exists an ideal whose generating sets all have more than $n$ elements?" It is not a question about multiple generating sets for a single ideal. – rschwieb Jan 14 '13 at 14:05
up vote 10 down vote accepted

The example of Nagata of a noetherian ring with infinite Krull dimension satisfies your requirements: all ideals are finitely generated and prime ideals must have arbitrarily large number of generators, otherwise their heights would be bounded, contradiction.

Edit. After posting this answer I wondered if a noetherian ring containing ideals with arbitrary large minimal generating sets has necessarily infinite Krull dimension. The answer is NO and this paper provides examples of prime ideals of height $2$ in the power series ring $K[[X,Y,Z]]$ having at least $n$ generators for all $n\ge 1$.

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Is Nagata's example what you would call a ‘non-catenary ring’? – Haskell Curry Jan 13 '13 at 21:28
Thank you for your help! – ast Jan 14 '13 at 13:33

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