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We know examples of non Noetherian Prüfer domains, which do not contain any irreducible elements.

On the other hand, a Dedekind domain (not being a field) always contains irreducible elements since it is Noetherian and therefore atomic.

Now my question is if there are Dedekind domains which do not contain prime elements. Equivalently, one could ask after a Dedekind domain without principal maximal ideals.

I can neither find a proof that every Dedekind domain has a principal maximal ideal nor a counterexample of one that has no such ideal.

Every help will be appreciated! Thanks in advance!

user26857
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Daniel W.
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  • Is it possible to take a Dedekind domain that isn't a UFD and localize at a nonprincipal maximal ideal to get an example? i'm rather ignorant about commutative algebra, so I don't know if the number of generators is preserved or not. – rschwieb Apr 16 '19 at 16:00
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    Looks like no: the localization would be a local Dedekind ring, a discrete evaluation ring, hence a PID. – rschwieb Apr 16 '19 at 16:08

1 Answers1

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Let $R=\mathbb C[X,Y]/(Y^2-X^3-X)$. This is a Dedekind domain, and its non-zero prime ideals are of the form $(x-a,y-b)$ with $b^2=a^3+a$.

I let you as an exercise to prove that these are not principal. (Hint. Show that $x-a$ and $y-b$ are irreducible in $R$.)

user26857
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  • $R=\mathbb C[X,Y]/(Y^2-X^3+X)$ works as well. – user26857 Apr 16 '19 at 22:58
  • Wonderful example. Thank you! – Daniel W. Apr 17 '19 at 09:02
  • It should actually work for $R = \mathbb{C}[X,Y]/(f)$, for any $f \in \mathbb{C}[X,Y]$ such that the degree of $f$ in $x$ is bigger than 1 or the degree of $f$ in $y$ is bigger than 1. Am I wrong? – Daniel W. Apr 17 '19 at 09:06
  • f irreducible, of course. – Daniel W. Apr 17 '19 at 09:22
  • In such generality you don't know if the ring is integrally closed. For example, it is not if $f=x^2-y^3$. – user26857 Apr 17 '19 at 12:47
  • Cool example! I thought it must be possible. I'd like to add this to DaRT and if you created an account there I'd be more than happy to associate your name with having suggested it. – rschwieb Apr 17 '19 at 13:35
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    I realize this must be totally trivial to you but what are good ways to reason it is Dedekind? The closest I got was to prove it is $1$-dimensional and integrally closed, but that isn't easy for me either. Or perhaps you have a different geometric argument to suggest. Thanks – rschwieb Apr 17 '19 at 13:43
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    If I take a closer look, I also cannot see that it is integrally closed. – Daniel W. Apr 17 '19 at 13:53
  • One can show that this ring is regular by using the Jacobian criterion. (This is Exercise 6.2, page 46, from Hartshorne, Algebraic Geometry.) There is also an elementary approach, and I think I used it many times on this site; see here. See also Matsumura, Commutative Ring Theory, Example 4, page 65 for a general result. – user26857 Apr 17 '19 at 14:13
  • Btw, to me the hardest part of this example is to prove the irreducibility of $x-a$ and $y-b$. – user26857 Apr 17 '19 at 14:17
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    Thanks for the pointers. – rschwieb Apr 17 '19 at 15:34