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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.

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  • $\begingroup$ Let $F$ be a field of cardinality greater than the one of $\mathbb{C}$, consider $R=F[X]$. $\endgroup$
    – Aphelli
    Aug 19, 2019 at 13:57
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    $\begingroup$ @Mindlack there are more than countably many prime ideals there. $\endgroup$
    – user693936
    Aug 19, 2019 at 13:59
  • $\begingroup$ 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 $>c$". $\endgroup$
    – YCor
    Aug 21, 2019 at 22:27
  • $\begingroup$ @YCor I would not necessarily call it complicated. $\endgroup$
    – user693936
    Aug 21, 2019 at 22:36

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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 $n$.

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    $\begingroup$ 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 $\mathbb R$.) $\endgroup$ Aug 19, 2019 at 15:44
  • $\begingroup$ @MishaLavrov maybe we can, didn't think about it too much. Don't forget to take algebraic closure though. $\endgroup$
    – user693936
    Aug 19, 2019 at 15:47

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