1. ## Interception is lcm

Hi!

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

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

Thanks

2. ## Re: Interception is lcm

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

3. ## Re: Interception is lcm

Ok

- Definition

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

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

- Definition

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

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

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

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

4. ## Re: Interception is lcm

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

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

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

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

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

Thanks