Need Help with the definition of product of two ideals I know this are not ideals but for the purpose of being clear imagine $I=\{1,2,3\}$ $J=\{4,5,6\}$ where I and J are ideals
$IJ = \{ a_1b_1 + a_2b_2 + ...+a_nb_n \mid a_i\in I b_i\in J \space i=1,2...,n \space \land \space n\in \mathbb{N}^*\}$
Then $IJ= \{1*4,1*4+2*5,1*4+2*5+3*6\} =\{4,14,32\}$
Would this be right or Am I missing something about the definition?
 A: In the definition of the product of two ideals, we're not saying take the first element of one ideal and multiply it by the first element of the next ideal, and so on. What the definition is saying is that you need to allow for arbitrary sums of any product of an element of $I$ with an element of $J$. Maybe a definition you would be more comfortable with is to define $IJ$ to be the ideal generated by the set $\{ab:a\in I\, b\in J\}$.
As an example, let's show that the ideal of even integers $I$, and the ideal of integers divisible by $3$ which we'll call $J$, has as its product the ideal of all integers divisible by $6$ which we'll call $K$.
Every multiple of $6$ can be written as $6z$ for some integer $z$, which is the same as $2\cdot 3\cdot z$ so that $6z$ is a multiple of $2$ times a multiple of $3$. Thus every multiple of $6$ is an element of $IJ$, so that $K\subseteq IJ$.
Conversely, every element of $IJ$ has the form $\sum_{i=1}^{n}a_ib_i$ where $a_i$ is in $I$ and $b_i \in J$.
But then every summand is divisible by $6$, so that the sum is divisible by $6$. This means that every element of $IJ$ is a multiple of $6$ so that $IJ\subseteq K$.
Hope this helps!
