A PID with countably infinitely many prime ideals and no embedding into complex numbers

Does there exist a characteristic $$0$$ principal ideal domain $$R$$ that has countably infinitely many prime ideals and such that there is no injective unital ring homomorphism $$R\rightarrow \mathbb{C}$$?

I am aware of examples of PIDs with countably many prime ideals coming from number theory but they are all subrings of $$\overline{\mathbb{Q}}$$. The PID $$\mathbb{Q}[x]$$ is not a subring of $$\overline{\mathbb{Q}}$$ but it is a subring of $$\mathbb{C}$$ via the homomorphism $$x\rightarrow \pi$$.

There exist fields of cardinality larger than continuum (at least assuming choice, not sure what happens otherwise) so that is a PID that does not embed into $$\mathbb{C}$$ but it does not have infinitely many prime ideals. On the other hand, the ring of univariate polynomials over such a field has more than countably many prime ideals.

• Let $F$ be a field of cardinality greater than the one of $\mathbb{C}$, consider $R=F[X]$. Aug 19, 2019 at 13:57
• @Mindlack there are more than countably many prime ideals there.
– user693936
Aug 19, 2019 at 13:59
• A domain of characteristic zero embeds into $\mathbf{C}$ iff it has cardinal at most continuum. So the last assumption is just a complicated way to say "of cardinal $>c$".
– YCor
Aug 21, 2019 at 22:27
• @YCor I would not necessarily call it complicated.
– user693936
Aug 21, 2019 at 22:36

Take an algebraically closed field $$F$$ of characteristic $$0$$ of cardinality larger than continuum (such a beast exists assuming choice, not sure otherwise). Consider the ring of univariate polynomials $$F[x]$$ and localize it at the multiplicative system of elements that are not multiples of $$(x-n)$$ for any positive integer $$n$$. The resulting PID can not embed in $$\mathbb{C}$$ because of cardinality and its prime ideals are $$(0)$$ and $$(x-n)$$ for positive integers $$n$$.
• To construct $F$, can't we just take the field of rational functions in very many variables? (E.g., index the variables by arbitrary subsets of $\mathbb R$.) Aug 19, 2019 at 15:44