Ideals whose union is an ideal Does someone have an example of three ideals such that no one is contained in another, yet their union is an ideal?
 A: I have no better idea than to built on the example of the three proper subgroups of $C_2\times C_2$: let $M$ be a module over the commutative ring $R$. Then one can define a ring structure on the product $R\times M$ as follows:


*

*$(r,m)+(s,n):=(r+s,m+n)$,

*$(r,m)\cdot(s,n):=(rs,rn+sm)$.


For every submodule $N$ of $M$ the subset $0\times N$ is an ideal of this ring.
Now consider $C_2\times C_2$ as a $\mathbb{Z}$-module and look at the ring $\mathbb{Z}\times (C_2\times C_2)$ with the operations just described. The three subgroups then become ideals in that ring and their union is an ideal as well.
Remark: a technical result in commutative algebra says that given ideals $p_1,\ldots ,p_n$, $n\geq 3$, such that $p_1,\ldots ,p_{n-2}$ are prime ideals and an ideal $I$ with the property
$I\subseteq\bigcup\limits_{k=1}^n p_k$
then $I\subseteq p_k$ for some $k$.
This result yields that in any example of three ideals pairwise not contained in each other and such that their union is an ideal none of the three can be a prime ideal.
