Existence of a ring with specified residue fields 
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 $R$:

*

*$\operatorname{Spec}(R)$ should be connected (otherwise take $R = k_1 \times \ldots \times k_n$)


*$R$ 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 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.)
 A: 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, 
$$|F_r| \leq |F_{r-1}|^{\aleph_0} \leq (|F_{r-2}|^{\aleph_0})^{\aleph_0} \leq \cdot \leq ( \cdots ((|F_0|^{\aleph_0})^{\aleph_0}) \cdots )^{\aleph_0} = |F_0|^{\aleph_0 \times \cdots \times \aleph_0} = |F_0|^{\aleph_0}$$
A: Embed every $k_i$ in a huge field $K$, and consider the subring of $k_1[t]\times \ldots \times k_n[t]$ consisting of $n$-uples $(P_1,\ldots ,P_n)$ with $P_1(0)=\ldots =P_n(0)$. Take for $\mathfrak{m}_i$ the ideal of $n$-uples with $P_i(1)=0$.
Edit : As pointed out in the comments, this construction works only in equal characteristic, and doesn't give a noetherian ring in general.
A: The answer to your question is basically yes. R. Heitmann proved in his thesis that under mild conditions there exists a PID $R$ with specified residue fields. The conditions will always hold if your collection of residue fields is finite, and each of them are countable.
See: http://www.projecteuclid.org/euclid.dmj/1077310578
