# A question about the statement of the Prime Avoidance Lemma.

The Prime Avoidance Lemma states the following:

Suppose $$I_1,I_2,\dots,I_n,J$$ are ideals of a ring $$R$$, such that $$J\subset \cup U_i$$. If $$R$$ contains an infinite field, or if at most $$2$$ of the ideals $$I_1,\dots,I_n$$ are not prime, then $$J\subset I_k$$ for some $$k$$.

Consider $$(4)\cup (6)$$ as ideals of the ring $$\Bbb{Z}$$. Although it does not contain in infinite field, the ideals $$(6)$$ and $$(4)$$ satisfy the criterion of at most two ideals being not prime. Hence, the condition of the prime avoidance lemma is satisfied. However, $$(14)\subset (4)\cup (6)$$, and $$(14)$$ is not contained within either of $$(4)$$ or $$(6)$$. Is this not a counter-example? Where am I going wrong?

• @Randall- Yeah sorry I was confusing $(4)\cup (6)$ with $(4)+(6)$
– user67803
Commented Jul 30, 2018 at 17:33

$(14)$ isn't contained in the union $(4) \cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)\mathbf{+}(6)=(2)$.
• Ah yes. I was confusing $(4)\cup (6)$ with $(4)+(6)$. Thanks!