Can a ring have both finite and infinite residue field

Question

Let $R$ be a commutative ring with unity. Assume $R$ is integral (i.e. it has no zerodivisors, i.e. it is reduced and $\operatorname{Spec} R$ is irreducible) and noetherian. Let $\mathfrak{m}, \mathfrak{n} \subseteq R$ be two maximal ideals. Can it happen that $R/\mathfrak{m}$ is a finite field, and $R/\mathfrak{n}$ is an infinite field?

Some observations:

• If $R$ is a finitely generated algebra over field $k$, then by the weak Nullstellensatz all residue fields are finite extensions of $k$, hence they are either all finite or all infinite, depending on whether $k$ is finite or infinite (integrality is not really needed here).
• If $R$ is a finitely generated algebra over $\mathbb{Z}$, then it is a well-known lemma, that all residue fields of $R$ are finite (integrality is not really needed here).
• If we drop the integrality condition, then there are obvious examples: take any two fields $K, L$, where $K$ is finite, and $L$ is infinite, and consider $R = K \times L$.
• If we drop the condition that $R$ is noetherian, examples exist. Take $R = \mathbb{Z}[x_1, x_2, \ldots]$. It is easy to see that both $\mathbb{F}_p$ and $\mathbb{Q}$ are quotients of $R$.
• Note that the characteristic of a residue field does not need to remain constant across all maximal ideals: take $R = \mathbb{Z}[x]$, then any finite field is a quotient of $R$.
• The question is trivial for local rings.

In case one can find an example, I would also like to know what additional conditions on $R$ would imply that residue fields are either all finite or all infinite.

Edit Thanks for the answers. I've found a related thread on MathOverflow: https://mathoverflow.net/questions/176117/existence-of-a-ring-with-specified-residue-fields . It seems that the linked paper of Heitmann implies existence of a PID with residue fields $\{ \mathbb{Q} \} \cup \{ \mathbb{F}_p | p \in \mathbb{P}\}$.

Answer 1

Take $R = \mathbb{Z}_{(p)}[x]$, where $\mathbb{Z}_{(p)} = \{ \frac{a}{b} \in \mathbb{Q} : p \nmid b \}$ is the localization away from $(p)$. By construction $R$ has a quotient map $R \to \mathbb{F}_p[x]$ and hence finite residue fields. On the other hand, there is also a quotient map

$$R \ni f(x) \mapsto f \left( \frac{1}{p} \right) \in \mathbb{Q}$$

giving $R$ an infinite residue field. This example is very similar to Mohan's in the comments, a polynomial ring over a local ring. I wanted to give an example which is clearly as small and well-behaved as possible (in particular a subring of $\mathbb{Q}[x]$ and a localization of $\mathbb{Z}[x]$) while just barely not being either a f.g. algebra over a field or a f.g. algebra over $\mathbb{Z}$.

Edit: both mine and Mohan's examples have the feature that the finite residue fields all have the same characteristic. Localizing away from a finite set of primes produces an example with any finite set of finite characteristics. I don't know if infinitely many characteristics are possible.