I am attempting to show that the product of two ideals $A$ and $B$, $AB$, is also an ideal. I've got the underlying set defined as the following:
$$ AB=\left\{ a_1b_1+a_2b_2+\dots+a_nb_n \mid a_i\in A, b_i \in B, n \in \mathbb{Z}^+ \right\} $$
Good so far. Verifying the $ar, ra \in AB$ requirement is simple enough using associativity. For whatever reason I'm getting hung up on the subtraction closure. I'll define $x, y \in AB$ such that:
$$ \begin{align*} x&= a_1b_1+a_2b_2+\dots+a_nb_n \tag{$a_i \in A, b_i \in B$}\\ y&= c_1d_1+c_2d_2+\dots+c_nd_n \tag{$c_i \in A, d_i \in B$} \end{align*} $$
And then I'll take $x - y$:
$$ \begin{align*} x-y&= a_1b_1+a_2b_2+\dots+a_nb_n-\left(c_1d_1+c_2d_2+\dots+c_nd_n\right) \\ &= a_1b_1+a_2b_2+\dots+a_nb_n-c_1d_1-c_2d_2-\dots-c_nd_n \\ &= a_1b_1-c_1d_1+a_2b_2-c_2d_2+\dots+a_nb_n-c_nd_n \\ &= (a_1b_1-c_1d_1)+(a_2b_2-c_2d_2)+\dots+(a_nb_n-c_nd_n) \end{align*} $$
So that's where I end up. Question is: is that enough? (bonus question: have I screwed something up?). I'm stuck on the idea that I should be able to derive $(a_i-c_i)(b_i-d_i)$ somehow, but I haven't been able to make that happen. So, at the moment I'm not convinced that $(x-y) \in AB$.
My suspicions are that I've either WAY overthought this, or that I've completely forgotten some middle school math rules that will due the trick. Either way, insights will be appreciated. Thanks!
$-y=-\sum c_id_i=\sum(-c_i)d_i\in AB$