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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!

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    $\begingroup$ Great job writing up the question! If only all posters mae the effort to write out their thoughts and make questions so self-contained! I remember crossing this point a long time ago in algebra :) $\endgroup$
    – rschwieb
    Jan 21, 2014 at 18:14
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    $\begingroup$ Haha thankyou! I think a lot of the time people's problems are actually to do with the fact that they haven't properly got an idea of what they're trying to prove, and sometimes just writing it down is all that's needed to see what was wrong. Unfortunately not the case this time though! $\endgroup$
    – Lammey
    Jan 21, 2014 at 18:38

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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.)

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  • $\begingroup$ Ah ok, yeah I see now that the mistake I made was in trying to prove it step by step, so I didn't use some of the required information! If the radicals are all equal then $y^n\in P_{i_0} \Rightarrow y^n \in P_i$ for all $i$, which is all I needed! Thankyou very much for the fast and informative reply! $\endgroup$
    – Lammey
    Jan 21, 2014 at 18:36

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