# Ordered Domains

• Apr 22nd 2008, 03:49 PM
hercules
Ordered Domains
Prove or disprove: If E is a subring of an ordered integral domain D, and E is also an integral domain, then E is ordered.

Trying to satisfy the ordered integral domain definiton:
since E is a ring ...it is closed under addition..then it's positive elements certainly are.
also E's nonzero elements are closed under multiplication because it is an integral domain.

help ...i don't know how to complete and write a proper proof for this.
• Apr 22nd 2008, 07:02 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
Prove or disprove: If E is a subring of an ordered integral domain D, and E is also an integral domain, then E is ordered.

Since $\displaystyle D$ is ordered it means there is a subset $\displaystyle P$ so that:
• Exactly one holds: $\displaystyle a\in P,0\in P,-a\in P$ for any $\displaystyle a\in D$
• If $\displaystyle a,b\in P$ then $\displaystyle a+b\in P$
• If $\displaystyle a,b\in P$ then $\displaystyle ab\in P$

Note this same definition applies to $\displaystyle E$ as well because it is a subset.
• Apr 23rd 2008, 06:09 AM
hercules
Quote:

Originally Posted by ThePerfectHacker
Since $\displaystyle D$ is ordered it means there is a subset $\displaystyle P$ so that:
• Exactly one holds: $\displaystyle a\in P,0\in P,-a\in P$ for any $\displaystyle a\in D$
• If $\displaystyle a,b\in P$ then $\displaystyle a+b\in P$
• If $\displaystyle a,b\in P$ then $\displaystyle ab\in P$
Note this same definition applies to $\displaystyle E$ as well because it is a subset.

What you posted is exactly what i had in mind.
i thought i needed to show also that E has positive elements inside. Identity is in E+ but what else is guaranteed. Or am i thinking needlessly and overlooking an easy definition.
• Apr 23rd 2008, 09:35 AM
ThePerfectHacker
Quote:

Originally Posted by hercules
What you posted is exactly what i had in mind.
i thought i needed to show also that E has positive elements inside. Identity is in E+ but what else is guaranteed. Or am i thinking needlessly and overlooking an easy definition.

We need to show there is a subset $\displaystyle P'$ of $\displaystyle E$ having those conditions. We cannot simply choose $\displaystyle P' = P$ because it is not necessarily a subset. But if we choose $\displaystyle P' = P\cap E$ then $\displaystyle P'\subseteq E$ and the three conditions will follow.