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I am studying the theorem: Let $R$ be a Noetherian domain of dimension 1. Then every non-zero ideal $I$ of $R$ has a unique expression as a product of primary ideals with distinct radicals.

Let $I=\bigcap_{n=1}^n$$Q_i$ where $Q_i$ are primary. Then $P_i=rad(Q_i)$ is maximal. Moreover $P_i+P_j=R$, so they are comaximal for $i\ne j$. Why does it follow that $Q_i$ are also comaximal for $i$ distinct from $\ne j$? So, why does it follow that $Q_i+Q_j=R$?

Would you help me, please? Thank you in advance.

user26857
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user404634
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    What if you try a little harder before posting such trivial questions? If you are stuck, then can also find good ideas on previous posts like this: https://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring – user26857 Jul 04 '17 at 23:58
  • @user26857 I have tried but I have problems. $R$=$P_i+P_j$=$nilrad (Q_i)+nilrad (Q_j)$. But I don't know how to continue :( – user404634 Jul 05 '17 at 00:01

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Hint:

Let $x\in P_i$, $y\in P_j$ such that $x+y=1$. On the other hans, there exist positive integers $m,n$ such that $x^m\in Q_i$, $y^n\in Q_j$.

What can you say about $(x+y)^{m+n}$?

Bernard
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  • Thank you! You mean: On the other hand, there exists integers $m, n$ such that $x^m$ is in $Q_i$, $x^n$ is in $Q_j$. So, I think $1$=$(x+y)^{m+n}$ is in $Q_i+Q_j$ and hence $Q_i+Q_j$=$R$?? – user404634 Jul 05 '17 at 00:12
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    Oh! yes. I messed up with the notations (it's getting late here…). Thanks for pointing it! – Bernard Jul 05 '17 at 00:14
  • You-re welcome! It's Ok. Also here is too late, but I have to study. One more question, pleaseee... $P_i$ are maximal because every non-zero ideal $I$ in a Noetherian ring accepts a primary decomposition and we always have I is includedto to Q_i=nilrad(P_i) and the latter is included toP_i?? – user404634 Jul 05 '17 at 00:19
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    @User1999: It's much simpler than that: they're maximal because they're non-zero and the ring has Krull dimension 1 by hypothesis. – Bernard Jul 05 '17 at 07:40