# Is there an integral domain with a lot of residue fields of the same characteristic?

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

• every nonzero prime ideal $Q$ is maximal, and
• for every prime power $q\equiv 3 \bmod 8$, there is a maximal ideal $Q$ of $R$ such that $R/Q$ is a field of size $q$?

I am looking for a ring where "$q\equiv 3 \bmod 8$" describes finite fields, rather than just finite local rings like in the ring $\mathbb{Z}$. The only examples I can think of that satisfy the first condition have a finite number of residue fields of each characteristic. The only examples I can think of that satisfy the second condition have a ton of non-maximal non-proper prime ideals with all sorts of bizarre fields associated to them.

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Have you seen mathoverflow.net/questions/42430/… ? – Qiaochu Yuan Dec 20 '10 at 17:50
In particular, take a look at answer mathoverflow.net/questions/42430/… It's not directly relevant because of the condition that there only be finitely many residue fields for each p, but it seems similar. – David Speyer Dec 20 '10 at 18:51
Thanks, I had not. The first answer does not directly address the problem since there are infinitely many fields of the same characteristic, but it does take care of (finitely many fields per each) infinitely many characteristics all at once. Maybe it means finding such a domain R is pretty hard. – Jack Schmidt Dec 20 '10 at 18:54
To be honest, I just assumed R was some silly universal ring covered in every algebraic number theory class and that people would just say "oh you mean the completion of the adele ring" or "oh you want the ring of all algebraic integers" or something. If the ring is too horrible or unknown, then I guess it isn't too useful to me. – Jack Schmidt Dec 20 '10 at 18:57