# Residue fields of a principal ideal domain

Let $$R$$ be a principal ideal domain with exactly two maximal ideals, say $$m$$ and $$m'$$. Is it possible that $$R/m \cong \mathbb F_2$$ and $$R/m'\cong \mathbb Q$$? If not, then can we at least get an example where $$R/m$$ is finite but $$\mathbb Q \subseteq R/m'$$? I know examples where the second case happens when $$R$$ is a unique factorization domain but not a principal ideal domain.

• What about the intersection of two DVRs having the desired residue fields? Commented Sep 14, 2020 at 7:33
• If there is any such example, then it'll be of the form which you mentioned. But how to find such DVRs? Commented Sep 14, 2020 at 9:28
• In Heitmann, Raymond C. PID’s with specified residue fields it is proved that for any finite collection of countable (including finite) fields there is a countable PID having this collection as its residue fields. Unfortunately, I didn't find a complete online version of this paper.
– Ulli
Commented Sep 14, 2020 at 10:55
• Thank you very much. Commented Sep 14, 2020 at 14:04

It suffices to find discrete valuation rings $$O_1$$ and $$O_2$$ with the required residue fields and common fraction field $$K$$; then $$R:=O_1\cap O_2$$ is an example of the desired type.
Consider the rational function field $$K:=\mathbb{Q}(x)$$ in one indeterminate over the rationals. The valuation ring $$O_1$$ of the $$x$$-adic valuation is discrete and its residue field equals $$\mathbb{Q}$$.
Next consider the completion $$\mathbb{Q}_2$$ of $$\mathbb{Q}$$ with respect to the $$2$$-adic valuation with its natural valuation $$v$$: it is discrete and its residue field equals $$\mathbb{F}_2$$. $$\mathbb{Q}_2$$ contains elements that are transcendental over $$\mathbb{Q}$$; let $$y$$ be such a $$2$$-adic number. A field embedding $$\phi:\mathbb{Q}(x)\rightarrow\mathbb{Q}_2$$ can then be defined via $$\phi(x):=y$$. This yields a discrete valuation $$w:=v\circ\phi$$. By construction the residue field of $$w$$ equals $$\mathbb{F}_2$$ so that one can chose $$O_2$$ to be the valuation ring of $$w$$.
The approach also covers the case of more than two prime ideals with finite extensions of $$\mathbb{Q}$$ or finite fields as residue fields. By replacing $$\mathbb{Q}$$ with other fields one can include even more types of residue fields.