642: Least Common Multiples of Subset of Commutative Ring

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definition of least common multiples of subset of commutative ring

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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

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Starting Context

• The reader knows a definition of ring.

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Target Context

• The reader will have a definition of least common multiples of subset of commutative ring.

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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:
$R$: $\in \{\text{ the rings }\}$
$S$: $\subseteq R$
$S^{\prime }$: $=\{m^{\prime }\in R|\mathrm{\forall }p\in S(\mathrm{\exists }q\in R(m=qp))\}$
$\ast lcm(S)$: $=\{m\in S^{\prime }|\mathrm{\forall }{m}^{\prime }\in S^{\prime }(\mathrm{\exists }q\in R(m^{\prime }=qm))\}$
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Conditions:
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$S^{\prime }$ is called "set of common multiples of $S$".

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2: Natural Language Description

For any ring, $R$, any subset of $R$, $S$, and the set of the common multiples of $S$, $S^{\prime }:=\{m^{\prime }\in R|\mathrm{\forall }p\in S(\mathrm{\exists }q\in R(m=qp))\}$, $lcm(S):=\{m\in S^{\prime }|\mathrm{\forall }{m}^{\prime }\in S^{\prime }(\mathrm{\exists }q\in R(m^{\prime }=qm))\}$

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3: Note

This definition does not require $R$ to have any order: "least" is not according to any order.

$S=\mathrm{\varnothing }$ is not excluded, although it is not particularly assumed to be useful.

When $S=\mathrm{\varnothing }$, vacuously, $S^{\prime }=R$, and $1\in lcm(S)$, because $d^{\prime }=d^{\prime }1$ for each $d^{\prime }\in S^{\prime }$; if $0\in lcm(S)$, $lcm(S)=\{0\}$, because $d^{\prime }=q0=0$, which does not contradict $1\in lcm(S)$, because that means that $1=0$, which means that $R=\{0=1\}$; what others $lcm(S)$ has depends on $R$: when $d$ is a unit, $d^{\prime }=d^{\prime }{d}^{-1}d$, so, $d\in lcm(S)$; but otherwise, whether there is a $q\in R$ such that $d^{\prime }=qd$ for each $d^{\prime }\in R^{\prime }$ is not clear.

Hereafter, let us suppose that $S\ne \mathrm{\varnothing }$.

Always $0\in S^{\prime }$: $0=0p$ for each $p\in S$.

But when $0\in lcm(S)$, $lcm(S)=\{0\}$: $m^{\prime }=0=q0$ for each $m^{\prime }\in S^{\prime }$, so, $S^{\prime }=\{0\}$.

When $S$ is finite, $\prod _{p_j\in S}{p}_j\in S^{\prime }$: $\mathrm{\forall }{p}_k\in S(\prod _{p_j\in S}{p}_j=(\prod _{p_j\in S\setminus \{p_k\}}{p}_j)pk)$.

$lcm(S)$ may be empty or have multiple elements.

If $0\in S$, $S^{\prime }=\{0\}$: $0=q0$ for each $q\in R$. The reverse is not necessarily true: as a counterexample, let $R=\mathbb{Z}/(6\mathbb{Z})$ and $S=\{[2],[3]\}$, then, $S^{\prime }=\{[0]\}$: the multiples of $[2]$ are $[2\ast 0]=[0],[2\ast 1]=[2],[2\ast 2]=[4],[2\ast 3]=[6]=[0],[2\ast 4]=[8]=[2],[2\ast 5]=[10]=[4]$ and the multiples of $[3]$ are $[3\ast 0]=[0],[3\ast 1]=[3],[3\ast 2]=[6]=[0],[3\ast 3]=[9]=[3],[3\ast 4]=[12]=[0],[3\ast 5]=[15]=[3]$.

If and only if $S^{\prime }=\{0\}$, $lcm(S)=\{0\}$: if $S^{\prime }=\{0\}$, $0=00$, so, $0\in lcm(S)$, while $lcm(S)$ is a subset of $S^{\prime }$; if $lcm(S)=\{0\}$, $\{0\}\subseteq S^{\prime }$, but as $0=q0$ for each $q\in R$, only $0$ can be in $S^{\prime }$.

This definition does not exactly specialize to 'least common multiple of subset of integers': the least common multiples of $\{2,3\}$ of $\mathbb{Z}$ by this definition are $\{-6,6\}$ ($-6=(-3)2$; $-6=(-2)3$; $6=(-1)(-6)$), while the least common multiple of $\{2,3\}$ by 'least common multiple of subset of integers' is $6$.

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References

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