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NBtarheel_33 · 2013-11-25, 09:08

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!

2013-11-25, 09:08	#3
NBtarheel_33 "Nathan" Jul 2008 Maryland, USA 5×223 Posts	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