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Does someone have an example of three ideals such that no one is contained in another, yet their union is an ideal?

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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.

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  • $\begingroup$ One can also consider $C_2\times C_2$ as a ring with zero multiplication. $\endgroup$ – user26857 May 25 '20 at 7:42

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