Let $I=\left \langle a \right \rangle$ and $J = \left \langle b \right \rangle$ be ideals in a ring A. Show that $I\cdot J = \{xy\mid x\in I,y\in J\} $ is an ideal in A and $I\cdot J = \left \langle ab \right \rangle$.
I can show that the product rule for ideals hold for $I\cdot J$ (i.e. $xij \in I\cdot J $ where $x \in A$ and $ij \in I\cdot J $). But i'm stuck trying to show that $ij - i'j'$ is in $I\cdot J$. I'm thinking that this isn't valid for all cases, but I can't find a counterexample (I tried to use ideals in $\mathbb{Z}$ for that but it didn't work).
This is true since $I$ and $J$ are principal ideals. If $i, i' \in (a)$ and $j, j' \in (b)$, write $i = ra, i' = r'a, j = sb, j' = s'b$. Then $ij - i'j' = (ra)(sb) - (r'a)(s'b) = (rs - r's')ab \in (ab)$.
More generally, if $I = (a)$ is principal and $J = (b_1, \ldots, b_n)$, then $I \cdot J = (ab_1, \ldots, ab_n)$.