description/proof of that for integral domain, if principal ideal by element is also by another element, elements are associates with each other, and principal ideal is by any associate
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of integral domain.
- The reader knows a definition of principal ideal of ring.
- The reader knows a definition of associates of element of commutative ring.
- The reader admits the proposition that the cancellation rule holds on any integral domain.
Target Context
- The reader will have a description and a proof of the proposition that for any integral domain, if any principal ideal by any element is also by any another element, the elements are associates with each other, and the principal ideal is by any associate.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the integral domains }\}\)
\(p\): \(\in R\)
\(p R\): \(= \text{ the principal ideal by } p\)
//
Statements:
(
\(\exists p' \in R (p R = p' R)\)
\(\implies\)
\(p' \in Asc (p) \land p \in Asc (p')\)
)
\(\land\)
\(\forall p' \in Asc (p) (p R = p' R)\)
//
2: Natural Language Description
For any integral domain, \(R\), any \(p \in R\), and the principal ideal, \(p R\), if there is a \(p' \in R\) such that \(p R = p' R\), \(p' \in Asc (p)\) and \(p \in Asc (p')\). And for each \(p' \in Asc (p)\), \(p R = p' R\).
3: Proof
\(p = p 1 \in p' R\), so, \(p = p' r'\) for an \(r' \in R\). \(p' = p' 1 \in p R\), so, \(p' = p r\) for an \(r \in R\).
\(p = p r r'\). By the proposition that the cancellation rule holds on any integral domain, \(1 = r r'\). As \(R\) is a commutative ring, \(1 = r' r\).
So, \(r\) and \(r'\) are units.
For each \(p' \in Asc (p)\), \(p' = u p\) with a unit, \(u\).
\(p R = p' R\), because for each \(p'' \in p R\), \(p'' = p r = u^{-1} p' r = p' u^{-1} r \in p' R\); for each \(p'' \in p' R\), \(p'' = p' r = u p r = p u r \in p R\).
