A DVR necessarily has spectrum $\{ 0, \mathfrak{m} \}$, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum $\{ 0, \mathfrak{m} \}?$

Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $\mathfrak{m} \neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $\mathfrak{m} \neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.

I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.

My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?

up vote 2 down vote accepted

To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^{\mathbb{R}_{\ge 0}}]]$ of formal power series of the form $\sum_{r \in S} c_r x^r$ where $r \in \mathbb{R}_{\ge 0}$ and $S \subseteq \mathbb{R}_{\ge 0}$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.

For $f \in R$ a nonzero power series, write $\nu(f) \in \mathbb{R}_{\ge 0}$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f \in I$ has valuation $\nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g \in R$ can be written as $x^{\nu(g)}$ times a unit, we see that $I$ contains every element of $R$ with valuation $\ge \nu(f)$. This means $I$ is determined by the valuations of its nonzero elements, and the set of all such valuations is an upward-closed subset of $\mathbb{R}_{\ge 0}$ (upward-closed means if $x \in S$ and $y \ge x$ then $y \in S$). There are two infinite families $[r, \infty)$ and $(r, \infty)$ of such subsets, so the ideals of $R$ come in two infinite families, namely

$$I_r = \{ f \in R : \nu(f) \ge r \}, r \in \mathbb{R}_{\ge 0}$$

and

$$J_r = \{ f \in R : \nu(f) > r \}, r \in \mathbb{R}_{\ge 0}$$

(together with the zero ideal, which you can think of as $I_{\infty}$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian. Note that $I_r = (x^r)$ is principal but $J_r$ is not even countably generated.

The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.

  • Great answer, explicit and detailed! – Prince M Nov 15 at 19:43

A non-discrete valuation ring of height $1$ is such an example.

  • 1
    Ok give me an example verifying the existence of a non discrete valuation ring of height 1 – Prince M Nov 15 at 1:34
  • @Prince: I construct exactly such a thing in my answer. – Qiaochu Yuan Nov 15 at 1:57
  • I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short. – Prince M Nov 15 at 2:16
  • Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that! – Prince M Nov 15 at 2:17

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