For an ideal $I$, why might not $I$=$I^2$?

Question

Let $I$ be an ideal of a ring $R$. Define $I^2$ to be the set of all finite sums of elements of the form $a_1a_2$ where $a_1,a_2\in I$. Why isn't it necessarily the case that $I=I^2$?

If $a\in I$, then $a=1a\in I^2$, so $I\subseteq I^2$. And any element of $I^2$ must also be in $I$ since $I$ is closed under addition and multiplication, which then means $I^2\subseteq I$, and so it follows that $I=I^2$.

Am I missing something?

Answer 1

Why is $1a\in I^2$? If $I$ is a proper ideal, we must have $1\notin I$.

Answer 2

$I \subseteq I^2$ is not true. Consider $I=2 \Bbb Z$ on $R=\Bbb Z$.

Answer 3

Well, take the polynomial ring $R[x]$ and the ideal $I=\langle x\rangle$. Then $I^2=\langle x^2\rangle$ and both ideals are certainly different as $x\not\in I^2$.