Residue fields of a principal ideal domain

Question

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.

Answer 1

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.