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Let $R$ be a ring and $I,J$ two ideals. If $I \cap J=0$ then $ij=0$ for every $i \in I$ and $j \in J$. This happens when $R=A \times B$ and $I=I’ \times \{0\}$ and $J=\{0\} \times J’$ with $I’$ ideal of $A$ and $J’$ ideal of B.

Is this the only case when this happens?

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  • $\begingroup$ Another example is if $R$ is the zero ring, which is in essence already covered by your product example as $R$ is the zero ring if $A$ and $B$ are. $\endgroup$ – RMWGNE96 May 23 at 20:02
  • $\begingroup$ In $\mathbb{Z}$, for example, two ideals have trivial intersection if and only if at least one of them is the zero ideal. $\endgroup$ – RMWGNE96 May 23 at 20:05
  • $\begingroup$ $\mathbb{Z}$ is a PID, though, so that might not be such a good example. $\endgroup$ – Nicolas May 23 at 20:10
  • $\begingroup$ Note that in general $IJ=0$ doesn't imply $I\cap J=0$. Just find an ideal $I$ such that $I^2\subsetneq I$ and consider the ring $R/I^2$. $\endgroup$ – egreg May 23 at 20:44
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No, the ring need not be decomposable into two pieces.

For example, take $F[x,y]$ and localize at the maximal ideal $(x,y)$, then take the quotient by the ideal $(x)\cap (y)$ in the localization.

In the resulting ring $R$, the ideal generated by $x$ and the ideal generated by $y$ are distinct from each other and from $(x,y)$, and they have a trivial intersection because we took the quotient by their intersection.

The ring can't be decomposed into two pieces, though, because it is a local ring with unique maximal ideal generated by $x$ and $y$.

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