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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?

Anuj
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1 Answers1

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Here are the comments I think will be most helpful:

  1. 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.

  2. 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.

  3. 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.)

rschwieb
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  • So my first part is correct and I need to find an example to prove the 2nd part, right? – Anuj Nov 01 '18 at 19:06
  • The reason i didnt took up the case that you have mentioned is it contradicts the problem statement because union of 2 ideals is an ideal iff one is contained in the other. – Anuj Nov 01 '18 at 19:08
  • Let's say if I had to prove the second part as well (not just providing an example), then what is wrong with my answer (of second part) – Anuj Nov 01 '18 at 19:10
  • @Anuj OK, well if you have some fact already that eliminated the case I mention, I could not have known you were using it. Frankly I think it is begging the question to apply that Lemma, and it would be better to be explain the critical case in detail. The cases you covered are trivial in comparison. – rschwieb Nov 01 '18 at 21:15
  • @Anuj The second question is “ show there exists such a ring” the only answer to such a question is an example. There is no proof to be conducted beyond finding the example. You are writing out the hypotheses you want to have in the example, but there is nothing to deduce from them besides necessary conditions. That line of thought does not produce an example. Said another way, you are engaging in a line of thought that is not worth doing, considering the ease of finding an example. – rschwieb Nov 01 '18 at 21:19
  • @Anuj Did you also confirm that apparently your question is about rings without identity? – rschwieb Nov 01 '18 at 21:20