
> 
> Given a finite set of fields $k\_1, \ldots, k\_n$, is there a (commutative with $1$) ring $R$ with (maximal) ideals $m\_i$ such that $R/m\_i \cong k\_i$?
> 
> 
> 


To prevent things from being too easy, I require two conditions on $R$:


1. $\operatorname{Spec}(R)$ should be connected (otherwise take $R = k\_1 \times \ldots \times k\_n$)
2. $R$ should be Noetherian (otherwise take $R$ to be a polynomial ring over $\mathbb{Z}$ in sufficiently many variables)


I would be happy with the case $n = 2$ (although I don't currently see how to get the general case from this). However, I do insist that the fields be arbitrary - it is known that any finite collection of *countable* fields is the set of residue fields of a PID (see [this article](https://doi.org/10.1215/S0012-7094-74-04162-3) by Heitmann - first page only).


(I've included the algebraic-geometry tag in hopes for some geometric insight. If however someone feels that this is sufficiently non-geometric, feel free to edit the tags.)



Existence of a ring with specified residue fields
This answer basically fills in details in Oliver Benoist's comments: If $K$ and $L$ are fields with $|L| > |K|^{\aleph\_0}$, then $L$ and $K$ cannot be residue fields of $R$, with $\mathrm{Spec}(R)$ Noetherian and connected.


Proof: Suppose otherwise. Then there is a sequence $K = F\_0$, $F\_1$, $F\_2$, ..., $F\_r = L$ of residue fields of $R$, and of irreducible components $\mathrm{Spec}(R\_1)$, $\mathrm{Spec}(R\_2)$, ..., $\mathrm{Spec}(R\_r)$ so that $F\_i$ and $F\_{i+1}$ are quotients of $R\_{i+1}$. By the [previous answer](https://mathoverflow.net/a/162218/297), 
$$|F\_r| \leq |F\_{r-1}|^{\aleph\_0} \leq (|F\_{r-2}|^{\aleph\_0})^{\aleph\_0} \leq \cdot \leq ( \cdots ((|F\_0|^{\aleph\_0})^{\aleph\_0}) \cdots )^{\aleph\_0} = |F\_0|^{\aleph\_0 \times \cdots \times \aleph\_0} = |F\_0|^{\aleph\_0}$$


