# When do two ideals have trivial intersection?

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?

• 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. – RMWGNE96 May 23 at 20:02
• In $\mathbb{Z}$, for example, two ideals have trivial intersection if and only if at least one of them is the zero ideal. – RMWGNE96 May 23 at 20:05
• $\mathbb{Z}$ is a PID, though, so that might not be such a good example. – Nicolas May 23 at 20:10
• 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 $R/I^2$. – egreg May 23 at 20:44

## 1 Answer

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 $$y$$.