658: For Commutative Ring, if Each Elements Pair Has Greatest Common Divisor, Each Finite Subset Has Greatest Common Divisor, Which Can Be Gotten Sequentially

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description/proof of that for commutative ring, if each elements pair has greatest common divisor, each finite subset has greatest common divisor, which can be gotten sequentially

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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 ring.
• The reader knows a definition of greatest common divisors of subset of commutative ring.
• The reader admits 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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Target Context

• The reader will have a description and a proof of the proposition that for any commutative ring, if each elements pair has a greatest common divisor, each finite subset has a greatest common divisor, which can be gotten sequentially.

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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:
$R$: $\in \{\text{ the commutative rings }\}$
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Statements:
$\mathrm{\forall }\{p_1,p_2\}\subseteq R(gcd(\{p_1,p_2\})\ne \mathrm{\varnothing })$
$⟹$
(
$\mathrm{\forall }S=\{p_1,...,p_n\}\in \{\text{ the finite subsets of R }\}(gcd(S)\ne \mathrm{\varnothing })$
$\wedge$
$gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})\subseteq gcd(\{p_1,...,p_n\})$
)
//

$gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})\subseteq gcd(\{p_1,...,p_n\})$ means that if we take a greatest common divisor of $\{p_{n-1},p_n\}$ as $d_{n-1}$, then, take a greatest common divisor of $\{p_{n-2},d_{n-1}\}$ as $d_{n-2}$, then, ..., and so on, then a greatest common divisor of $\{p_1,d_2\}$ is a common divisor of $S$.

This proposition does not claim $gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})=gcd(\{p_1,...,p_n\})$, but when $R$ is an integral domain, it holds, because while there is a $d\in gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})$, for each $d^{\prime }\in gcd(\{p_1,...,p_n\})$, $d^{\prime }=ud$, by 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, and $d^{\prime }\in gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})$, by 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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2: Natural Language Description

For any commutative ring, $R$, if for each $\{p_1,p_2\}\subseteq R$, $gcd(\{p_1,p_2\})\ne \mathrm{\varnothing })$, for each finite subset, $S=\{p_1,...,p_n\}\subseteq R$, $gcd(S)\ne \mathrm{\varnothing }$, and $gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})\subseteq gcd(\{p_1,...,p_n\})$.

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3: Note

This proposition does not suppose that $R$ is a greatest common divisors domain ($R$ can be a non-integral-domain), but when $R$ is an integral domain, $R$ is a greatest common divisors domain, and this proposition holds for any greatest common divisors domain.

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4: Proof

Let us prove it inductively with respect to $n$.

For $n=2$, it obviously holds.

Let us suppose that it holds through $n=n^{\prime }$.

Let us think of $S=\{p_1,...,p_{n^{\prime }+1}\}$.

There is a $d\in gcd(\{p_2,...,p_{n^{\prime }+1}\})$. $p_j=q_{j}d$ for each $j\in \{2,...,n^{\prime }+1\}$.

There is a $d^{\prime }\in gcd(\{g_1,d\})$. $p_1=q_1^{\prime }{d}^{\prime }$ and $d=q^{\prime }{d}^{\prime }$.

$p_j=q_j{q}^{\prime }{d}^{\prime }$ for each $j\in \{2,...,n^{\prime }+1\}$. So, $d^{\prime }$ is a common divisor of $S=\{p_1,...,p_{n^{\prime }+1}\}$.

Let $d^{″}$ be any common divisor of $S$. $p_1=q_1^{″}{d}^{″}$ and $p_j=q_j^{″}{d}^{″}$ for each $j\in \{2,...,n^{\prime }+1\}$. $d^{″}$ is a common divisor of $\{p_2,...,p_{n^{\prime }+1}\}$. So, $d=q^{″}{d}^{″}$.

So, $d^{″}$ is a common divisor of $\{p_1,d\}$. So, $d^{\prime }=q^{‴}{d}^{″}$.

So, $d^{\prime }$ is a greatest common divisor of $S$.

So, by the induction principle, each $S$ for each $n$ has a greatest common divisor.

We have seen that $gcd(\{p_1,gcd(\{p_2,...,p_n\})\})\subseteq gcd(\{p_1,...,p_n\})$.

As $gcd(\{p_2,gcd(\{p_3,...,p_n\})\})\subseteq gcd(\{p_2,...,p_n\})$ and so on, $gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...,p_n\})\})\})\subseteq gcd(\{p_1,...,p_n\})$ and so on, and after all, $gcd(\{p_1,gcd(\{p_2,gcd(\{p_3,...gcd(\{p_{n-1},p_n\})...\})\})\})\subseteq gcd(\{p_1,...,p_n\})$.

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References

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