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