Prime Avoidance Lemma for abelian groups?

Question

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

Answer 1

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 $x+y$.)