TITLE: Stuggling to understand ideal powers
QUESTION [7 upvotes]: In my current algebraic number theory course we have defined the multiplication of 2 ideals as the smallest ideal containing all products of elements of both, 
[i.e: let I and J be ideals of a ring R, then
IJ:={$\sum_1^k a_ib_i : a_i\in I , b_i\in J$} where k depends on R ]
we haven't however, formally defined ideal powers, by which I mean ideals of the form $I^x$ where x is an integer. 
However, using the first, informal definition, it seems to me that all products of elements from I will just again form I
i.e. using the formal definition, and taking x = 2 I find that 
$I^2$ = II = {$\sum_i^k a_ib_i : a_i\in I , b_i\in I$} = {$\sum_1^k c_i : c_i\in I$} = {$d : d\in I$} = I
and clearly extending this definition to higher powers results in the same thing?
Thanks for any clarification!

REPLY [2 votes]: The third equality sign in your chain of equalities is, in fact, only $\subseteq$. 
There is no reason to expect that every sum of elements in $I$ can be expressed as a sum of products of two elements of $I$. 
To pick an analogy from group theory and sidestep the added complexity of the sums needed for products of ideals: for an abelian group $G$ (say, for example, the multiplicative group of positive rationals) the subset $H=\{x^2\mid x\in G\}$ turns out to be a subgroup of $G$. But, of course, not every rational is the square of a rational, so $H$ is a proper subgroup of $G$.
It's simple to give examples of ideals to convince you of the inequality too, and the one given by zarathustra in the comments is hard to improve upon:  $I=2\mathbb Z\lhd \mathbb Z$ is an ideal, and every product of two elements in $I$ is divisible by $4$, so $I^2\neq I$.