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Let $R$ be an rng (no unity).

Define $IJ$ as the ideal generated by $\{ab:a\in I, b\in J\}$ for every ideals $I,J$ of $R$.

Let $I=\langle a\rangle , J=\langle b\rangle $ be principal ideals.

What would be an example that $IJ\neq \langle ab\rangle$?

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  • $\begingroup$ Hey is this even true, since $I$ and $J$ are principle ??? $\endgroup$ – brick Nov 14 '14 at 18:48
  • $\begingroup$ @brick I'm not sure whether this is true or not. When $R$ has a unit, this is true, but I'm not really sure about the case $R$ without unity $\endgroup$ – cococomi Nov 14 '14 at 18:49
  • $\begingroup$ The thing you need is the distributive law, which holds no matter if the ring has unity. $\endgroup$ – brick Nov 14 '14 at 18:51
  • $\begingroup$ @brick I have proven that $I(J+K)=IJ+IK$ and $(I+J)K=IK+JK$. How do i prove my question with this? Would you please write your argument as an answer? $\endgroup$ – cococomi Nov 14 '14 at 18:53
  • $\begingroup$ Did you check the answer bellow? I think it's fine... $\endgroup$ – brick Nov 14 '14 at 18:59
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It is even possible that $(a)(a) \neq (a^2)$, even in unital rings.

Take the ring of $2 \times 2$-matrices over any field. Consider the matrix $a = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and define $b = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. Then one computes $a=a \cdot ba$, and this lies in $(a)(a)$. But it doesn't lie in $(a^2)$, since $a^2=0$.

In commutative rings (unital or not), we always have $(a)(b)=(ab)$.

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