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I came across the following question

Let $P_1, P_2$, and $P_3$ be prime ideals of $R$. If an ideal $I$ satisfies $I \subset P_1\cup P_2 \cup P_3$, then prove $I \subset P_i$ for some $i$.

I proved it by doing many different cases and found out afterward that this a special case of the prime avoidance lemma (it is very long, looks correct to me). But, I am just curious, my solution never used the fact that we were working in ideals. In fact, it never used the fact that a ring is needed as I never multiplied anything.

So, I am wondering if this result holds for abelian groups? I assume not as I have not seen a result like this anywhere.

NOTE: After writing up my solution, I found a mistake. Comments below give a counterexample.

Thank You

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  • $\begingroup$ Klein four group is a union of three proper subgroups. $\endgroup$
    – user26857
    Commented Jul 4, 2019 at 5:52

1 Answer 1

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This is false for abelian groups, and even for ideals that are not required to be prime. For instance, let $R=\mathbb{F}_2[x,y]/(x^2,xy,y^2)$, let $P_1=(x)$, let $P_2=(x+y)$, and $P_3=(y)$. Then $I=(x,y)$ is equal to the union $P_1\cup P_2\cup P_3$, but is not contained in any of them. (If you have trouble verifying these claims, note that $R$ has only $8$ elements; its non-unit elements are just $0,x,y,$ and $x+y$.)

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  • $\begingroup$ That is weird, but I thought my proof worked for 3 abelian groups. Did I do something wrong? $\endgroup$
    – Mike
    Commented Jul 2, 2019 at 20:18
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    $\begingroup$ How are we supposed to know that without you showing us your proof? $\endgroup$
    – Con
    Commented Jul 2, 2019 at 20:22
  • $\begingroup$ @ThorWittich Right, let me write it up now $\endgroup$
    – Mike
    Commented Jul 2, 2019 at 20:23
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    $\begingroup$ The result does work for 2 abelian groups (and the usual proof of prime avoidance indeed does not use multiplication in that case). But for more than 2 a more complicated argument is needed that does not work without prime ideals. $\endgroup$ Commented Jul 2, 2019 at 20:24
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    $\begingroup$ You should rather thank @EricWofsey. I only pointed out that we were not able to see your proof. $\endgroup$
    – Con
    Commented Jul 2, 2019 at 20:51

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