# Ideals whose union is an ideal [duplicate]

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

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.

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