1. ## Interception is lcm

Hi!

I have problems with this demostration (I do not know how to start)

If $a,b \in{\mathbb{Z}}$, prove that $(a) \cap{(b)} = m$ where $m = lcm(a,b)$

Thanks

2. ## Re: Interception is lcm

What do $\left(m\right)$ and $\left(n\right)$ mean? Please give the definition.

3. ## Re: Interception is lcm

Ok

- Definition

An ideal in a commutative ring $R$ is a subset $I$ of $R$ such that

$a)$ $0\in{I}$
$b)$ $a,b \in{I} \Rightarrow{a+b \in{I}}$
$c)$ If $a\in{I}$, $r\in{A}$ then $ra \in{I}$

- Definition

If $b_1,b_2,...,b_n$ lie in a commutative ring $R$ , then the set of all linear combinations, denoted by

$(b_1,b_2,...,b_n)$ is an ideal in $R$, called the ideal generated by $b_1,b_2,...,b_n$ . In particular if $n=1$, then

$(b) = \{rb:r \in{R}\}$

consists of all the mulpiples of $b$ ; it is called the principal generated by $b$

4. ## Re: Interception is lcm

since $m$ is a multiple of $a$ and a multiple of $b$

$m\in (a)\cap (b)$

Therefore $(m)\subseteq (a)\cap (b)$

If $x$ is a multiple of $a$ and a multiple of $b$ then $x$ is a multiple of $m$ and therefore an element of $(m)$

Therefore $(a)\cap (b)\subseteq (m)$

Thanks