645: For Integral Domain, if Greatest Common Divisors of Subset Exist, They Are Associates of a Greatest Common Divisor

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description/proof of that for integral domain, if greatest common divisors of subset exist, they are associates of a greatest common divisor

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Topics

About: ring

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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

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the integral domains }\}$
$S$: $\subseteq R$
$gcd(S)$: $=\text{ the set of the greatest common divisors of }S$
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Statements:
$\mathrm{\exists }d\in gcd(S)$
$⟹$
$gcd(S)=Asc(d)$
//

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2: Natural Language Description

For any integral domain, $R$, and any subset, $S\subseteq R$, if there is an element, $d\in gcd(S)$, $gcd(S)=Asc(d)$.

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3: Note

This proposition is not claiming that such a $d$ always exists.

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4: Proof

Let us suppose that there is a $d\in gcd(S)$.

When $S=\mathrm{\varnothing }$ 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 $d=0$. $gcd(S)=\{0\}=\{u0|u\in \{\text{ the units in }R\}\}$.

Let us suppose that $S\ne \mathrm{\varnothing }$ and $S\ne \{0\}$ hereafter.

$0\notin gcd(S)$: refer to Note for the definition of greatest common divisors of subset of commutative ring. So, $d\ne 0$.

Let us prove that for each unit, $u$, $ud\in gcd(S)$.

For each $p\in S$, there is a $q\in R$ such that $p=qd$, so, $p=q{u}^{-1}ud$ where $q{u}^{-1}\in R$. So, $ud$ is a common divisor.

Let $S^{\prime }$ be the set of the common divisors of $S$. For each $d^{\prime }\in S^{\prime }$, $d=q^{\prime }{d}^{\prime }$ for a $q^{\prime }\in R$, so, $ud=u{q}^{\prime }{d}^{\prime }$ where $u{q}^{\prime }\in R$. So, $ud$ is a greatest common divisor.

Let us prove that each $d^{\prime }\in gcd(S)$ is $d^{\prime }=ud$ for a unit, $u$.

$d=q^{\prime }{d}^{\prime }$ and $d^{\prime }=qd$ for some $q,q^{\prime }\in R$. $d=d1=q^{\prime }qd=d{q}^{\prime }q$. By the cancellation rule ($d\ne 0$), $1=q^{\prime }q$. So, $q$ is a unit, $u$, and $d^{\prime }=ud$.

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References

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