# Constructing a local non-Noetherian domain with spectrum $\{(0), \mathfrak{m} \}$.

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?

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 '18 at 19:43

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

• Ok give me an example verifying the existence of a non discrete valuation ring of height 1 – Prince M Nov 15 '18 at 1:34
• @Prince: I construct exactly such a thing in my answer. – Qiaochu Yuan Nov 15 '18 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 '18 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 '18 at 2:17