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Is there a commutative integral domain $R$ in which:

  • every nonzero prime ideal $Q$ is maximal, and
  • there are maximal ideals $Q$ with $R/Q$ of sizes $3$, $11$, and $27$?

This doesn't happen with number rings of Galois extensions of $\mathbb{Q}$, as far as I know, since what happens to $3$ seriously happens to $3$. You cannot have residue fields of sizes both $3$ and $27$. Now $(\mathbb{Z}/3\mathbb{Z})[X]$ has residue fields of sizes $3$ and $27$, but not $11$.

Edit: I'll separate the next into its own question.

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The answer to your first question is yes. Indeed, for any $d \geq 4$, you can construct infinitely many degree $d$ algebraic number fields $K/\mathbb{Q}$ such that the ring of integers $\mathbb{Z}_K$ has this property.

There is a basic philosophy here that you can build a global object that has prescribed local behavior at any finite set of primes. In this particular case, Theorem 23 of these notes does what you want: you want to take as a $\mathbb{Q}_3$ algebra $L \oplus \mathbb{Q}_3 \oplus A'$ where $L/\mathbb{Q}_3$ is an unramified cubic extension (and $A'$ is any separable $\mathbb{Q}_p$-algebra of dimension $d-4$), and similarly you want to choose your $\mathbb{Q}_{11}$-algebra to have a $\mathbb{Q}_{11}$ factor.

You are right that these conditions cannot be realized by a global Galois extension, since then the local invariants of ramification index and inertial degree would have to agree for each of the factors.

I am still parsing the rest of your question...

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    Thanks! Second part wants infinitely many primes to work out (didn't realize that was a separate step) and to have there be a very simple way to detect the primes that work, similar to "3 mod 8". I might be using language only suitable for principal ideals, but I don't know how to phrase it in general.2010-12-19
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    Ok, Q[x]/(x^4-x-11) works for the first question. Obviously doesn't work for the second question, since there are infinitely many fields of characteristic 3 and size congruent to 3 mod 8, but (3) only splits so many ways in a finite extension.2010-12-19
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    I inserted a middle question to make the last question (my actual concern) make more sense. 1. Can we get close? 2. Can we get them all? 3. Can we get them all in some nice regular way that sounds like "3 mod 8"?2010-12-19