Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

So,

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.

share|improve this question
3  
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
3  
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
add comment

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.