description/proof of that for integral domain, if least common multiples of subset exist, they are associates of a least common multiple
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: Note
- 4: Proof
Starting Context
- The reader knows a definition of integral domain.
- The reader knows a definition of least common multiples of subset of commutative 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 and any subset, if the least common multiples of the subset exist, they are the associates of a least common multiple.
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:
\(R\): \(\in \{\text{ the integral domains }\}\)
\(S\): \(\subseteq R\)
\(lcm (S)\): \(= \text{ the set of the least common multiples of } S\)
//
Statements:
\(\exists m \in lcm (S)\)
\(\implies\)
\(lcm (S) = Asc (m)\).
//
2: Natural Language Description
For any integral domain, \(R\), and any subset, \(S \subseteq R\), if there is an element, \(m \in lcm (S)\), \(lcm (S) = Asc (m)\).
3: Note
This proposition is not claiming that such an \(m\) always exists.
4: Proof
Let us suppose that there is an \(m \in lcm (S)\).
When \(0 \in lcm (S)\), \(lcm (S) = \{0\}\): refer to Note for the definition of least common multiples of subset of commutative ring. Then, \(m = 0\), and \(lcm (S) = \{u 0 \vert u \in \{\text{ the units in } R\}\} = \{0\}\).
Let us suppose that \(0 \notin lcm (S)\) hereafter.
\(m \neq 0\).
Let \(S'\) be the set of the common multiples.
Let us prove that \(u m \in S'\).
When \(S = \emptyset\), \(S' = R\): refer to Note for the definition of least common multiples of subset of commutative ring, and so, \(u m \in S'\).
When \(S \neq \emptyset\), for each \(p \in S\), there is a \(q \in R\) such that \(m = q p\), so, \(u m = u q p\) where \(u q \in R\). So, \(u m \in S'\).
Let us prove that for each unit, \(u\), \(u m \in lcm (S)\).
For each \(m' \in S'\), \(m' = q m\) for a \(q \in R\), so, \(m' = q u^{-1} u m\) where \(q u^{-1} \in R\).
Let us prove that each \(m' \in lcm (S)\) is \(m' = u m\) for a unit, \(u\).
\(m = q' m'\) and \(m' = q m\) for some \(q, q' \in R\). \(m = m 1 = q' q m = m q' q\). By the cancellation rule (\(m \neq 0\)), \(1 = q' q\). So, \(q\) is a unit, \(u\), and \(m' = u m\).
