646: For Integral Domain, if Least Common Multiples of Subset Exist, They Are Associates of a Least Common Multiple

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description/proof of that for integral domain, if least common multiples of subset exist, they are associates of a least common multiple

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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 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.

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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 least common multiples of the subset exist, they are the associates of a least common multiple.

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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:
$R$: $\in \{\text{ the integral domains }\}$
$S$: $\subseteq R$
$lcm(S)$: $=\text{ the set of the least common multiples of }S$
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Statements:
$\mathrm{\exists }m\in lcm(S)$
$⟹$
$lcm(S)=Asc(m)$.
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2: Natural Language Description

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

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3: Note

This proposition is not claiming that such an $m$ always exists.

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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 $lcm(S)=\{u0|u\in \{\text{ the units in }R\}\}=\{0\}$.

Let us suppose that $0\notin lcm(S)$ hereafter.

$m\ne 0$.

Let $S^{\prime }$ be the set of the common multiples.

Let us prove that $um\in S^{\prime }$.

When $S=\mathrm{\varnothing }$, $S^{\prime }=R$: refer to Note for the definition of least common multiples of subset of commutative ring, and so, $um\in S^{\prime }$.

When $S\ne \mathrm{\varnothing }$, for each $p\in S$, there is a $q\in R$ such that $m=qp$, so, $um=uqp$ where $uq\in R$. So, $um\in S^{\prime }$.

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

For each $m^{\prime }\in S^{\prime }$, $m^{\prime }=qm$ for a $q\in R$, so, $m^{\prime }=q{u}^{-1}um$ where $q{u}^{-1}\in R$.

Let us prove that each $m^{\prime }\in lcm(S)$ is $m^{\prime }=um$ for a unit, $u$.

$m=q^{\prime }{m}^{\prime }$ and $m^{\prime }=qm$ for some $q,q^{\prime }\in R$. $m=m1=q^{\prime }qm=m{q}^{\prime }q$. By the cancellation rule ($m\ne 0$), $1=q^{\prime }q$. So, $q$ is a unit, $u$, and $m^{\prime }=um$.

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References

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