Product of two ideals doesn't equal the intersection The product of two ideals is defined as the set of all finite sums $\sum  f_i  g_i$, with $f_i$ an element of $I$, and $g_i$ an element of $J$. I'm trying to think of an example in which $IJ$ does not equal $I \cap J$.
I'm thinking of letting $I = 2\mathbb{Z}$, and $J = \mathbb{Z}$, and $I\cap J = 2\mathbb{Z}$? Can someone point out anything faulty about this logic of working with an even ideal and then an odd ideal?
Thanks in advance.

Maybe it is helpful for you to realise what really happens for ideals in the integers. You probably know that any ideal in $\mathbb Z$ is of the form $(a)$ for $a\in \mathbb Z$, i.e. is generated by one element. The elements in $(a)$ are all integers which are divisible by $a$.
If we are given two ideals $(a)$ and $(b)$, their intersection consists of those numbers which are divisible by $a$ and divisible by $b$. Their product consists of all numbers which are divisible by the product $ab$.
If $a$ and $b$ are coprime they are the same. E.g. all numbers which are divisible by $2$ and $3$ are also divisible by $6$, and vice versa. If they are not coprime the situation changes. If a number is divisible by $4$ and $2$, then it is not necessarily divisible by $8$. 
Another way of saying that two integers $a$, $b$ are coprime is that there exist $x,y$, such that $xa+by=1$ (cf. Euclidean algorithm). In the language of ideals this translates to $(a)+(b)=\mathbb Z$ and the circle closes.
