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[parent] positivity in ordered ring (Theorem)
Theorem 1   If $ (R,\,\leq)$ is an ordered ring, then it contains a subset $ R_+$ having the following properties:

Proof. We take $ R_+ = \{r\in R:\,\, 0 < r\} = \{r\in R:\,\, 0\leq r\, \wedge \,0 \neq r\}$. Let $ a,\,b \in R_+$. Then $ 0 < a$, $ 0 < b$, and therefore we have $ 0 < a\!+\!0 < a\!+\!b$, i.e. $ a\!+\!b \in R_+$. If $ R$ has no zero-divisors, then also $ ab \neq 0$ and $ 0 = a0 < ab$, i.e. $ ab\in R_+$. Let $ r$ be an arbitrary non-zero element of $ R$. Then we must have either $ 0 < r$ or $ r < 0$ (not both) because $ R$ is totally ordered. The latter alternative gives that $ 0 = -r\!+\!r < -r\!+\!0 = -r$. The both cases mean that either $ r\in R_+$ or $ -r\in R_+$.



"positivity in ordered ring" is owned by pahio.
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See Also: positive cone, topic entry on real numbers


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ordered integral domain with well-ordered positive elements (Theorem) by Wkbj79
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Cross-references: totally ordered, proof, ring multiplication, zero divisors, ring, ring addition, subset, contains, ordered ring
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This is version 9 of positivity in ordered ring, born on 2004-10-26, modified 2006-10-10.
Object id is 6424, canonical name is PositivityInOrderedRing.
Accessed 1294 times total.

Classification:
AMS MSC: 13J25 (Commutative rings and algebras :: Topological rings and modules :: Ordered rings)
 12J15 (Field theory and polynomials :: Topological fields :: Ordered fields)
 06F25 (Order, lattices, ordered algebraic structures :: Ordered structures :: Ordered rings, algebras, modules)
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