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Old 2013-11-25, 04:48   #1
jinydu
 
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Default Discrete Ordered Rings?

Are there any examples other than \mathbb{Z}?

(Alright, alright. I know of one other example: \prod_U\mathbb{Z}, i.e. ultrapowers of [tex]\mathbb{Z}[tex]. But this is not helpful for the problem I've got.)

Last fiddled with by jinydu on 2013-11-25 at 04:52
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Old 2013-11-25, 09:01   #2
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Never mind, found an example: The ring of polynomials (in one variable) with natural number coefficients.
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Old 2013-11-25, 09:08   #3
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This paper, entitled (appropriately enough) "Discrete Ordered Rings", might be of some help.

In particular, look at Theorem 11.1 on page 135. It states that if R is an ordered ring with unity and if a is an element of R, then the order in R extends to the ring of polynomials S = R[x] / <(x - a)^2>. Moreover, if R is discrete, then so is S. (The proof follows in the paper, and explains how the ordering works.)

So, given that the integers give you a discrete ordered ring, it seems as though you could just pick your favorite integer (I like 8) and then form the polynomial ring Z[x] / <(x - 8)^2>, and that would then be yet another example of a discrete ordered ring by the above theorem.

Hopefully I have understood this correctly, and this helps you out!

Last fiddled with by NBtarheel_33 on 2013-11-25 at 09:11
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