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This question already has an answer here:

Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$. Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.

I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true.

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marked as duplicate by BCLC, Jendrik Stelzner, Namaste abstract-algebra Aug 31 '18 at 15:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Yes, the $\mid$ symbol here is just a separator in a set definition. You could replace it with a colon character for clarity. $\endgroup$ – Thomas Andrews Oct 20 '15 at 16:51
  • $\begingroup$ Why did you put the "equals" sign outside of MathJax in $a\mathbb Z \cap b\mathbb Z = \operatorname{lcm}(a,b)\mathbb Z$ in both of the places where you wrote that? There seem to be a fair number of people on m.s.e. who follow that incorrect usage. Is there some source instructing people to do it that way? ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 20 '15 at 16:59
  • $\begingroup$ It doesn't say $\text{“}mk\mid m\text{''}$; it says $\text{“}\{\cdots \mid \cdots\cdots\}\text{''}$, where the vertical bar does not mean "divides", but rather means what it means in the context of expressions like $\text{“}\{\cdots \mid \cdots\cdots\}\text{''}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 20 '15 at 17:03
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Hint:

Step 1: Can you prove that $\operatorname{lcm}(a,b)\in a\mathbb{Z}\cap b\mathbb{Z}$?

Step 2: Can you prove that if $a$ and $b$ both divide $c$, then $c\in a\mathbb{Z}\cap b\mathbb{Z}$?

Step 3: How do steps 1 and 2, together, imply your result?

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First of all, to clear up your confusion:

$$m\mathbb Z=\{mk\mid k\in\mathbb Z\}$$

does not mean that $mk$ divides $k$. The vertical line can be read as "so that" or "where". It means that the set $m\mathbb Z$ is the set of numbers in the form of $mk$, where $k$ is any element of $\mathbb Z$.

That said, to prove that $a\mathbb Z\cap b\mathbb Z=\operatorname{lcm}(a,b)\mathbb Z$, you need to prove that

  1. if $x\in a\mathbb Z\cap b\mathbb Z$, then $x\in \operatorname{lcm}(a,b)\mathbb Z$.
  2. if $x\in\operatorname{lcm}(a,b)\mathbb Z$, then $x\in a\mathbb Z\cap b\mathbb Z$.
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    $\begingroup$ I changed what appeared to be an obvious typo, which said $m\mathbb Z\{mk\mid k\in\mathbb Z\}$, to $m\mathbb Z = \{mk\mid k\in\mathbb Z\}$, which seemed obviously intended. You changed it back. How does $m\mathbb Z\{mk\mid k\in\mathbb Z\}$ with no $\text{“}=\text{''}$ make sense? ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 20 '15 at 17:13
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    $\begingroup$ @MichaelHardy Your change also changed the nice enumeration back into the inline enumeration, which is why I had to re-edit it. In doing so, I accidentally re-introduced a mistake that you correctly pointed out. Thanks for the edit. $\endgroup$ – 5xum Oct 20 '15 at 17:15
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    $\begingroup$ I don't see how my edit could have affected the enumeration. I did nothing with the enumeration. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 20 '15 at 17:17
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    $\begingroup$ @MichaelHardy That's because while you were editing the mistakes in the top part, I edited the bottom one to improve enumeration. So first, my edit came in and fixed the enumeration, then your edit of the original post fixed everything else, but reverted the enumeration. Then, I wanted to re-fix the enumeration but ended up ruining everything else. $\endgroup$ – 5xum Oct 20 '15 at 17:21
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I give another demonstration you might find useful, which allows you to reach the result without double implication, using only the definition of $lcm(a,b)$.


Definition: $d \in \Bbb Z$ is defined least common multiple of $a$ and $b$, in symbols $d := lcm(a,b)$ if:

  1. $\exists m,n \in \Bbb Z$ such that $am = d = bn$, otherwise written as $a|d$ and $b|d$
  2. If $c \in \Bbb Z$ is such that $a|c$ and $b|c$, then $d|c$

So we can write $$a\Bbb Z \cap b\Bbb Z =_{(1)} \left \{ x \in \Bbb Z : a|x, b|x \right \} =_{(2)} \left \{ x \in \Bbb Z : lcm(a,b)|x \right \} =_{(3)} lcm(a,b)\Bbb Z$$

Where:

  • $=_{(1)}, =_{(3)}$ follows from the definition of $a\Bbb Z \cap b\Bbb Z$ and $lcm(a,b)\Bbb Z$
  • $=_{(2)}$ follows from the point $2$ of the definition of $lcm(a,b)$
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Well firstly, how do you define $\text{lcm}(a,b)$? I'll define it by the converse of a proposition in Artin Algebra (Prop 2.3.8)

enter image description here

Namely, we'll prove the converse of Prop 2.3.8 where $m:=\text{lcm}(a,b)$ is defined by the integer s.t.

(a) $m$ is divisible by both $a$ and $b$

(b) If $n$ is divisible by $a$ and $b$, then $n$ is divisible by $m$.

Pf: $(\subseteq)$

Let $n \in \mathbb Z m$. Then there is an integer $n_m$ s.t. $n_m = \frac n m$. Observe that $n_m = \frac{n_a}{m_a} = \frac{n_b}{m_b}$ where we define $n_a := \frac n a, m_a := \frac m a, m_b := \frac m b, n_b := \frac n b$. Observe that $m_a, m_b$ are integers by assumption (a) while we want to show that $n_a, n_b$ are integers because showing such is equivalent to showing $n \in \mathbb Za \cap \mathbb Zb$.

Now, $n_m m_a = n_a$ is a product of integers and hence an integer. The same is true for $n_m m_b = n_b$. Therefore, $n_a, n_b$ are integers and thus, $n \in \mathbb Za \cap \mathbb Zb$

$(\supseteq)$

This one is easier. Let $n \in \mathbb Za \cap \mathbb Zb$. Then $n_a, n_b$ as defined earlier are integers, i.e. $n$ is divisible by both $a$ and $by$. By assumption (b), $n$ is divisible by $m$.

QED

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