Let $A'$ be an ideal in ring $A$ and a$\in A$( commutative ring with 1).
Let $ P_1 , P_2,…,P_r$ be prime ideals in $A$ such that $a+ A'\subseteq \bigcup P_1 \bigcup P_2 … \bigcup P_r$.
(a) Then prove that the ideal $<a,A'> = Aa+A' \subseteq P_i$ for some $i\in \{ 1,…,r\}.$
(b) Prime avoidence lemma: Let $P_1 ,…,P_r$ be ideals in A such that almost two of them are not prime. If $A' \subseteq P_1 \bigcup P_2,…\bigcup P_r$ , then prove that $A' \subseteq P_i$ for some $i\in \{1,…,r\}$.
For (a) I thought of assuming that that $<a,A'>$ is not subset of $P_i$ for any $i\in \{1,…,r\}$ in hope of of getting a contradiction. Some property of primes ideals is to be used but I am unable to think of it.
(b) In case every ideal is a prime ideal then I can use the (a) with $a=0$, but I am unable to think about the case when one ideal is not prime or two are not primes and will appreciate hints.
Can you please help?
Best Answer
Induct on $r$. If $r=1$, then done.
Let $r=2$, then $A'\subset P_1\cup P_2.$ We can assume $A'\nsubseteq P_1$ and $A'\nsubseteq P_2$. Let $x_1\in A'$ and $x_1\notin P_1$. Similarly $x_2\in A'$ and $x_2\notin P_2$. If $x_1\notin P_2$, we get a contradiction that $A'\subset P_1\cup P_2$. So we have $x_1\in P_2$. Similarly $x_2\in P_1.$ Then $x_1+x_2\in A'$, but $x_1+x_2\notin P_1\cup P_2$.
Next let $r\geq 3$. Now assume the statement for $r-1$. That is for any choice of $r-1$ ideals out of which atmost two of them are not prime ideals, then the result holds.
If $A'$ is contained in any of the $r-1$ many ideals out of $P_1,\dots, P_r$, then we are done by induction. So assume otherwise. We can safely assume that $P_r$ is a prime ideal.
Let $x\in A'$ such that $x\notin P_j$ for all $j=1,\cdots r-1$. Now if $x\notin P_r$ , then $x\notin \cup_{j=1}^r P_j$, a contradiction. So we can assume that $x\in P_r$.
Note $A'P_1\cdots P_{r-1}\nsubseteq P_r$, for otherwise either of the factors is contained in $P_r$, in that case we are done by induction. Let $y\in A'P_1\cdots P_{r-1}$ and $\notin P_r$. Now $x+y\in A'$, such that $x+y\notin P_i$ for all $i.$
So we are done.