2
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ @Randall- Yeah sorry I was confusing $(4)\cup (6)$ with $(4)+(6)$ $\endgroup$
    – user67803
    Commented Jul 30, 2018 at 17:33

1 Answer 1

2
$\begingroup$

$(14)$ 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 $(4)\mathbf{+}(6)=(2)$.

$\endgroup$
1
  • 1
    $\begingroup$ Ah yes. I was confusing $(4)\cup (6)$ with $(4)+(6)$. Thanks! $\endgroup$
    – user67803
    Commented Jul 30, 2018 at 17:29

You must log in to answer this question.