652: For Unique Factorization Domain and Finite Subset, iff Greatest Common Divisors of Each Pair Subset of Subset Are Unit Associates, Least Common Multiples of Subset Are Associates of Multiple of Elements of Subset

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description/proof of that for unique factorization domain and finite subset, iff greatest common divisors of each pair subset of subset are unit associates, least common multiples of subset are associates of multiple of elements of subset

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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 unique factorization domain.
• The reader knows a definition of greatest common divisors of subset of commutative ring.
• 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 knows a definition of representatives set of quotient set.
• The reader admits the proposition that for any unique factorization domain, the method of getting the greatest common divisors of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works.
• The reader admits the proposition that for any unique factorization domain, the method of getting the least common multiples of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works.

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Target Context

• The reader will have a description and a proof of the proposition that for any unique factorization domain and any finite subset with more than 1 elements, if and only if the greatest common divisors of each pair subset of the subset are the unit associates, the least common multiples of the subset are the associates of the multiple of the elements of the subset.

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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:
$R$: $\in \{\text{ the unique factorization domains }\}$
$U$: $=\{\text{ the units of }R\}$
$I$: $=\{\text{ the irreducible elements of }R\}$
$R/Asc$: $=\text{ the quotient set by the associates equivalence relation }$
$f$: $:R/Asc\to R$ such that $\mathrm{\forall }p\in R/Asc,f(p)\in p$
$\overline{R/Asc-f}$: $=\text{ the representatives set of }R/Asc\text{ by }f$
$S$: $=\{p_1,...,p_n\}$, $\in \{\text{ the finite subsets of }R\}$, with more than 1 elements
$gcd(S)$:
$lcm(S)$:
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Statements:
$\mathrm{\forall }{p}_j,p_k\in S\text{ such that }{p}_j\ne p_k(gcd(\{p_j,p_k\})=Asc(1))$
$⟺$
$lcm(S)=Asc(p_1...p_n)$
//

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2: Natural Language Description

For any unique factorization domain, $R$, the set of the units of $R$, $U$, the set of the irreducible elements of $R$, $I$, the quotient set by the associates equivalence relation, $R/Asc$, any map, $f:R/Asc\to R$, such that for each $p\in R/Asc$, $f(p)\in p$, the representatives set of $R/Asc$ by $f$, $\overline{R/Asc-f}$, and any finite subset, $S=\{p_1,...,p_n\}\subseteq R$, with more than 1 elements, if and only if $\mathrm{\forall }{p}_j,p_k\in S\text{ such that }{p}_j\ne p_k(gcd(\{p_j,p_k\})=Asc(1))$, $lcm(S)=Asc(p_1...p_n)$.

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3: Note

We are not talking about $gcd(S)=Asc(1)$.

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4: Proof

As this proof is based on the proposition that for any unique factorization domain, the method of getting the greatest common divisors of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works and the proposition that for any unique factorization domain, the method of getting the least common multiples of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works, those propositions are supposed to be understood.

Let $I^{\prime }=\{i_1,...,i_l\}$ be the set of some irreducible elements cited in those propositions: it is the same in the both propositions. The point is that the factorizations of the elements of $S$ with respect to $I^{\prime }$ are unique, because we have chosen the fixed representatives of the associates equivalence classes.

Let us suppose that $\mathrm{\forall }{p}_1,p_2\in S(gcd(\{p_1,p_2\})=Asc(1))$.

While $p_j=u_j{i}_1^{c_{j,1}}...i_l^{c_{j,l}}$, if $0<c_{j,k}$, $c_{s,k}=0$ for each $s\ne j$, because otherwise, $i_k$ would be a common divisor of $\{p_j,p_s\}$ and $1$ would not be any greatest common divisor of $\{p_j,p_s\}$, because $1=q{i}_k$ would not hold, because that would mean that $i_k$ was a unit.

$p_1...p_n=u_1{i}_1^{c_{1,1}}...i_l^{c_{1,l}}...u_n{i}_1^{c_{n,1}}...i_l^{c_{n,l}}=u_1...u_n{i}_1^{c_{1,1}}...i_1^{c_{n,1}}...i_l^{c_{1,l}}...i_l^{c_{n,l}}=u_1...u_n{i}_1^{c_{1,1}+...+c_{n,1}}...i_l^{c_{1,l}+...+c_{n,l}}$, but $c_{1,k}+...+c_{n,k}=max(\{c_{s,k}|s\in \{1,...,n\}\})=M_k$ because only 1 term is nonzero, and so, $=u_1...u_n{i}_1^{M_1}...i_l^{M_l}=u_1...u_{n}m$.

So, $m=(u_1...u_n{)}^{-1}{p}_1...p_n$, and $lcm(S)=Asc(m)=Asc((u_1...u_n{)}^{-1}{p}_1...p_n)=Asc(p_1...p_n)$.

Let us suppose that $lcm(S)=Asc(p_1...p_n)$.

$lcm(S)=Asc(m)$.

So, $Asc(m)=Asc(p_1...p_n)$, which implies that $m=u{p}_1...p_n$ for a $u\in U$.

$i_1^{M_1}...i_l^{M_l}=u{u}_1...u_n{i}_1^{c_{1,1}+...+c_{n,1}}...i_l^{c_{1,l}+...+c_{n,l}}$, which implies that $M_k=max(\{c_{s,k}|s\in \{1,...,n\}\})=c_{1,k}+...+c_{n,k}$, because the factorizations are unique. That means that only 1 of the terms is nonzero, because while each term is non-negative, if 2 terms were positive, the sum would be larger than the maximum.

So, for each pair, $p_j,p_s\in S$, for each $k\in \{1,...,l\}$, $m_k:=min(\{c_{j,k},c_{s,k}\})=0$. So, $d:=i_1^{m_1}...i_l^{m_l}=i_1^0...i_l^0=1$.

So, $gcd(\{p_j,p_s\})=Asc(1)$.

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References

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