The question is

Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$. Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$

I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?

  • $\begingroup$ You seem to have one of your inequalities backwards. $\endgroup$ – Daniel McLaury Jul 20 '13 at 5:51
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    $\begingroup$ Have you not seen the theorem that says that $(a,b)$ can be written as $ar+bs$ for some integers $r$, $s$? $\endgroup$ – Gerry Myerson Jul 20 '13 at 5:58
  • $\begingroup$ Daniel: I'm not sure I understood what do you mean. @GerryMyerson: so it'll be enough to take $k\in(a,b)\mathbb Z$ and say $k=\alpha a+\beta b$ which is in $a\mathbb Z +b\mathbb Z$? Thanks. Another question is how can I procceed to $a\mathbb Z\cap b\mathbb Z$? $\endgroup$ – user65985 Jul 20 '13 at 6:15
  • $\begingroup$ I don't know --- I'm not sure why you were given that hint. I'd do it instead by showing that if $x$ is a multiple of $a$ and a multiple of $b$ then it's a multiple of the least common multiple of $a$ and $b$. $\endgroup$ – Gerry Myerson Jul 20 '13 at 6:25
  • $\begingroup$ Possible duplicate of Prove that $a\mathbb Z \cap b\mathbb Z = \operatorname{lcm}(a,b)\mathbb Z$. $\endgroup$ – BCLC Aug 31 '18 at 13:49


$$a\mid b\iff a\Bbb Z\supset b\Bbb Z$$

P First, suppose $b=ak$. Pick $x\in b\Bbb Z$. Then $x=by=a(ky)\in a\Bbb Z$. Conversely, if $a\Bbb Z\geq b\Bbb Z$ we have $b\in a\Bbb Z$ so $b=ak$, $a\mid b$.

$$a\Bbb Z\cap b\Bbb Z=\ell\Bbb Z\;,\; \ell=\lcm(a,b)$$

P Since $a\mid \ell$, $a\Bbb Z \supset \ell \Bbb Z$. Similarily, $b\Bbb Z \supset \ell \Bbb Z$, so $a\Bbb Z\cap b\Bbb Z\supset \ell \Bbb Z$. Now pick $x\in a\Bbb Z\cap b\Bbb Z$. Then $x=ak=bj$ for some $k,j$. Thus $a,b\mid x$. So $x$ is a common multiple, whence $ \ell\mid x$, that is $\ell =xm$ for some $m$ and $x\in\ell\Bbb Z$

$$a\Bbb Z+b\Bbb Z=d\Bbb Z\;,\;d=\gcd(a,b)$$

P Pick $y\in a\Bbb Z+b\Bbb Z$. Then $y=am+bn$. But then $d\mid y$, since $d\mid a,b$, and thus $y=kd\in d\Bbb Z$. Now pick $y\in d\Bbb Z$. Then $y=dk$ for some $k$. Bezout tells us we can write $d=an+bm$ so $y=adn+bdm\in a\Bbb Z+b\Bbb Z$.


An idea:

Denoting by $\,p_n\,$ a prime that divides an integer $\,n\,$ , we have

$$m\in a\Bbb Z\cap b\Bbb Z\implies m=ar=bs\;,\;r,s\in\Bbb Z\implies \begin{cases}p_a\mid bs\;\forall\,p_a\\{}\\p_b\mid ar\;\;\forall\,p_b\end{cases}\;\implies$$

$$\implies m=[a,b]t\;,\;t\in\Bbb Z\implies m\in [a,b]\Bbb Z$$

since we know that $\;[a,b]=\frac{ab}{(a,b)}\;$ (and thus every prime common to both $\,a,b\,$ appears both in $\,ar\,$ and in $\;bs\;$).

The other direction is trivial


Well firstly, how do you define $[a,b]$ aka $\text{lcm}(a,b)$? 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, $n \in \mathbb Za \cap \mathbb Zb$


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 $m$.



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