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While trying to prove one property of commutative rings with units I can't prove one fact without assuming existence of infinitely many different prime ideals or elements. I tried to test if it was the neccesary assumptions, but I failed, since I don't know any "toy"-examples of such rings.

I know only one example of this kind ($\mathbf{Q} [x]/ (x^2) $ ) but it's not a domain.


Are there any infinite domains with finite number of prime ideals?

If no, then are there any infinite domains with finite (but nontrivial) number of prime elements?

I am interested in noncommutative examples as well. Sorry if this question is too elementary.

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Well, infinite fields do have a finite number of prime ideals, but I suppose that’s not what you want. – Emil Jeřábek Jun 14 '12 at 14:00
Discrete valuation rings, such as $F[[x]]$, $\mathbb Z_p$, or $\mathbb Z_{(p)}$, are a slightly better example. – Emil Jeřábek Jun 14 '12 at 14:22
up vote 5 down vote accepted

For the commutative case. Take the set of all rational numbers whose denominators are coprime with a fixed integer $n$. Then the only prime ideals are generated by the prime divisors of $n$.

More generally, take any PID and take its ring of fractions wrt all the elements coprime with some fixed element; you get a desired ring.

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For a noncommutative example you could take the first Weyl algebra $A_1(k)$ with generated over a field $k$ by elements $x$ and $y$ subject to the relation $xy-yx-1$. This is a Noetherian domain which is simple, that is the only two-sided ideals are 0 and the ring itself.

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