Ring expressed as union of ideals Show that a ring can't be expressed as union of 2 proper ideals but it is possible to express it as a union of three proper ideals.
My solution
First Part
Let$ R$ be the ring with $ A$ , $B$ & $C$ as proper ideals 
Assuming $ R=A\cup_{}B$
As $R$ is an ideal of itself -> $A\cup_{}B$ is also an Ideal which is possible only if 
(i) $A$ is contained in $B$ 
->$ B = R$ 
-> $B$ is an improper ideal 
-> contradiction
or 
(ii)  $B$ is contained in $A$ .
-> $A = R$ 
-> $A$ is an improper ideal 
-> contradiction
Proved.
Second part 
Let  $ R=A\cup_{}B\cup_{}C $
(i) $C$ is contained in $A\cup_{}B$ 
-> $A\cup_{}B = R$ . This is similar to the above part which results in contradiction
(ii) $A\cup_{}B$ is contained in $C$ .
-> $C = R$ -> $C$ is an improper ideal -> contradiction
I am not able to solve the second part. What is wrong with my approach?
 A: Here are the comments I think will be most helpful:


*

*You haven't finished proving the first part. The case you haven't covered is $A\nsubseteq B$ and $B\nsubseteq A$, which is the only case that really matters.  It is already proven several times on the site why a group is never a union of two proper subgroups.

*What you are doing isn't wrong, but it does suggest you don't understand the task fully. You should be searching for an example not a proof. Your aim is to find an instance of a ring that is a union of three proper ideals.  Doing what you are doing will lead you to necessary conditions that the example must satisfy, but it does not tell you exactly what an example is.

*We must verify the problem statement. Do all your rings have identity? If so, then the problem is bogus. How could a ring with identity ever be a union of any number of proper ideals? By definition, $1\in\cup I_i$ implies $1\in I_j$ for some $j$, but then $I_j$ is not proper after all.
Now, if identity doesn't matter, we may as well look for an abelian group that is a union of three proper subgroups, and then make it into a ring by using the trivial multiplication. (Hint: start simply. The first possible candidate is a group with 4 elements.)
