If $g = gcd(a,b)$ prove (a,b)=(g). Furthermore, if $k = lcm(a,b)$ prove that $(a)\cap(b) = (k)$

Question

• $a,b \in \mathbb{Z}$.

• $(a,b),\hspace{0.4mm}(g)$ and $(k)$ are principle ideals

I'm new to this kind of problems, so I don't even know how to start it. Some help would be appreciated.

Answer 1

You want to prove an arbitrary $n\in\Bbb Z$ is of the form $ax+by$ iff a multiple of $g$, and divisible by both $a,\,b$ iff divisible by $k$.

Let $S$ denote the set of positive integers in $(a,\,b)$, which is clearly nonempty as it contains $|a|$. Clearly, its elements are all multiples of $g$. Let $d$ denote $S$'s minimal element, so $S$ is precisely the positive elements of $d$ (it includes them all, and any other element's remainder on division by $d$ would be an element of $S$ too, smaller than $d$ and thus contradicting its definition). Thus $d$ divides both $a$ and $b$, so $d|g$. Hence $d=g$.

For the second result let $T$ denote the set of positive common multiples of $a,\,b$, so $k$ is the minimal element of $T$. By the same remainder-contradicts-minimality logic as above, $T$ is precisely the multiples of $k$.