649: For Unique Factorization Domain, if Multiple of Elements Is Divisible by Irreducible Element, at Least 1 Constituent Is Divisible by Irreducible Element

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description/proof of that for unique factorization domain, if multiple of elements is divisible by irreducible element, at least 1 constituent is divisible by irreducible element

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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: Proof

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Starting Context

• The reader knows a definition of unique factorization domain.

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Target Context

• The reader will have a description and a proof of the proposition that for any unique factorization domain, if the multiple of any elements is divisible by any irreducible element, at least 1 constituent is divisible by the irreducible element.

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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:
$R$: $\in \{\text{ the unique factorization domains }\}$
$I$: $=\{\text{ the irreducible elements of }R\}$
$p_1...p_n$: $p_j\in R$
$i$: $\in I$
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Statements:
$i|p_1...p_n$
$⟹$
$\mathrm{\exists }{p}_k\in \{p_1,...,p_n\}(i|p_k)$
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2: Natural Language Description

For any unique factorization domain, $R$, the set of the irreducible elements of $R$, $I$, any $p_1...p_n$ where $p_j\in R$, and any $i\in I$, if $i|p_1...p_n$, there is at least 1 $p_k$ such that $i|p_k$.

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3: Proof

Let $U$ be the set of the units of $R$.

$p_1...p_n=qi$ for a $q\in R$.

$p_j=u_j{i}_{j,1}...i_{j,l_j}$ where $u_j\in U$ and $i_{j,k}\in I$.

$q=u{i}_1...i_l$ where $u\in U$ and $i_j\in I$.

So, $u_1{i}_{1,1}...i_{1,l_1}...u_n{i}_{n,1}...i_{n,l_n}=(u_1...u_n)i_{1,1}...i_{1,l_1}...i_{n,1}...i_{n,l_n}=u{i}_1...i_{l}i$ where $u_1...u_n\in U$.

As the factorizations are unique, $i=u^{\prime }{i}_{j,k}$ for a $i_{j,k}$ and a $u^{\prime }\in U$. Then, $i|p_j$, because $p_j=u_j{i}_{j,1}...i_{j,l_j}=u_j{i}_{j,1}...i_{j,k}...i_{j,l_j}=u_j{i}_{j,1}...u^{\prime -1}i...i_{j,l_j}=u_j{u}^{\prime -1}{i}_{j,1}...\widehat{i_{j,k}}...i_{j,l_j}i$ where $u_j{u}^{\prime -1}{i}_{j,1}...\widehat{i_{j,k}}...i_{j,l_j}\in R$.

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References

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