determine $m$ such that $\langle m \rangle = \langle 28 \rangle \cap \langle 35 \rangle$

Question

I need to find a value for $m$, thought of finding the greatest common divisor between the two numbers, but $\langle \rangle$ represents an ideal, don't know if that's the way to solve this problem.

I should also mention that I already know that the intersection of two ideals is also an ideal.

Help would be appreciated.

Answer 1

You want $m \in \langle 28 \rangle \cap \langle 35 \rangle$. So $m=28k$ and $m=35l$ for some $k,l \in \mathbb{Z}$. Thus $28k=35l \implies 4k=5l$. But $\gcd(4,5)=1$, therefore $4 | l$ and $5 | k$. This implies $$m=140j \qquad \text{ for some } j \in \mathbb{Z}.$$ But $m$ is a generator so least positive such value is $m=140$.

Answer 2

You want to find the least common multiple of $28$ and $35$

That is $140$ so you ideal is$$ <140> = <28> \cap <35> $$