I am given that $R$ is a commutative ring. I have to show that the intersection of finitely many P-Primary ideals is $P$-Primary. To do this I first need to show that the intersection of finitely many primary ideals is primary.
The definition I am given of a primary ideal is:
"An ideal Q of a commutative ring R is primary when $Q\neq R$ and for all $x,y\in R, xy \in Q$ implies $x\in Q$ or $y^n\in Q$ for some $n>0$. An ideal $Q$ is $P$-primary when $Q$ is primary and Rad($Q$)=$P$."
Let $X=\cap_{i=1}^nP_i$ where $P_i$ are primary ideals. Suppose $xy \in X.$ Then $xy \in P_i$ for all $i$. Then for each $P_i$, the following two statements are true (the second by swapping xy, as R is commutative)
$x\in P_i$ or $\exists m_i$ such that $y^m_i \in P_i$.
$y\in P_i$ or $\exists n_i$ such that $x^n_i \in P_i$.
But the weakest result I need for my proof is either that there exists a power of $y$ in ALL $P_i$ or there exists a power of $x$ in all $P_i$. The above two statements combined don't give me that as far as I can see. I'm at a loss. Any help would be greatly appreciated!
Best Answer
Unfortunately, this is impossible! If the intersection of a pair of primary ideals was primary, then the intersection of any two prime ideals would be prime. But we know this is not the case. It's a misstep to just ignore the added condition that both ideals be $P$ primary.
In the integers, for example, $(2)$ and $(3)$ are primary (actually prime!) ideals, but their intersection $(6)$ is not a primary ideal.
You will need to make full use that they are $P$-primary. The previous example I gave is not a contradiction to the $P$-primary statement since the $P$ primary ideals in $\Bbb Z$ form a chain.
So starting over again with your notation, add in additionally that $Rad(P_1)=Rad(P_2)=P$. Notice that right away you have $a,b\in P=Rad(P_1)=Rad(P_2)$, which is critical. (Solve this and do $n$-ary intersections with induction.)