Suppose ideal $I = (m)$ and ideal $J = (n)$. Prove $IJ = (mn)$.

Question

$I$ and $J$ are two ideals of integers. $IJ = \{a_1b_1 + a_2b_2 + ... + a_nb_n\}$ in $\Bbb Z$

$a_1....a_n$ in $I$, $b_1....b_n$ in $J$

Suppose $I = (m)$ and $J = (n)$. Prove $IJ = (mn)$.

I don't know where to start this proof.

Answer 1

To prove an equality like $IJ = (mn)$, it is often useful to prove two different inclusions: $IJ \subseteq (mn)$ and $(mn) \subseteq IJ$.

Let's look at $(mn) \subseteq IJ$: any element in $(mn)$ is of the form $amn$, where $a$ is some integer. Can you write this element as the product of an element in $I$ and an element in $J$?

Try using a similar approach for the other inclusion. Good luck!

Answer 2

Note: I have changed the subscript of the last term in the set definition from $n$ to $k$ to avoid the ambiguity Gerry Myerson pointed out.

Now for the hint: check that the g.c.d of $\sum_{i = 1}^k a_i b_i$ for any choice of $a_i \in I, b_i \in J$ and $k \geq 1$ is $mn$. Now prove the inclusion in either direction as Harry suggested.