Serre gives first the definition of a discrete valuation ring as a "principal ideal domain that has a unique non-zero prime ideal $m(A)$."

Next, he says: Let $A$ be a commutative ring. In order that A be a discrete valuation ring, it is necessary and sufficient that it be a Noetherian local ring, and that its maximal ideal be generated by a non-nilpotent element.

I do not understand why the unique non-zero prime ideal in a principal ideal domain is necessarily generated by a non-nilpotent element.

I started looking at the nilradical of a ring -- for a commutative ring, it is the intersection of all prime ideals. In this case we only have one prime ideal so the nilradical of our discrete valuation ring would be exactly the unique non-zero prime ideal given by the definition of a DVR. The nilradical is made up only by nilpotent elements, though, so surely it is generated by a nilpotent element, if we are in a principal ideal domain. To me, this would suggest that the maximal ideal of a DVR (by the first definition) is its nilradical, and it would be generated by a nilpotent element.

Clearly, I have made some mistake. Can somebody help me understand where I've gone wrong?

  • $\begingroup$ The nilradical is the intersection of ALL the prime ideals. Here, $(0)$ is also a prime ideal. $\endgroup$ – Prithviraj Chowdhury Jul 6 '16 at 18:52
  • $\begingroup$ How many nilpotent elements does a domain have? $\endgroup$ – carmichael561 Jul 6 '16 at 18:52
  • $\begingroup$ OH! I see. I just wasn't thinking about the conditions guaranteed by a principal ideal domain. Thank you. I knew it was something stupid like this. Thank you thank you. $\endgroup$ – ctesta01 Jul 6 '16 at 18:56
  • $\begingroup$ Actually, the way better question would have been, why this condition is also sufficient. $\endgroup$ – MooS Jul 6 '16 at 18:58
  • $\begingroup$ yes, definitely the better question, but I'm reading Serre's proof to answer that question. The necessary condition was skipped over as 'obvious' by Serre. It is obvious, I just wasn't looking at it the right way at all. $\endgroup$ – ctesta01 Jul 6 '16 at 19:00

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