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