Let $I$ be an ideal of a commutative ring $R$ with $1$ such that $\sqrt{I}=I_1\cdots I_n$ where $I_i's$ are pairwise comaximal ideal of $R$. Are there ideals $J_1,...,J_n$ of $R$ such that $V(I_i)=V(J_i)$ for each $i=1,...n$ and $I=J_1\cdots J_n$, where $V(A):=\{P\in Spec(R): A\subseteq P\}$ for every $A\subseteq R$. (If this statment is not true in general under what conditions it is true?)
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2$\begingroup$ Define $J_i=\{x\in R|(I_1\cdots I_{i-1}\cdot I_{i+1}\cdots I_n)^m\cdot x\subset I, \text{for some}\, m\}$. I will leave you to check that $I=\prod J_i$. $\endgroup$– MohanCommented Jul 26, 2018 at 16:33
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$\begingroup$ Can you explain why $V(I_i)=V(J_i)$? $\endgroup$– user127045Commented Jul 27, 2018 at 13:58
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2 Answers
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Ok, so I am understanding comaximal to mean that $I_i + I_j = R$ for all $i \neq j$. Given this, then I believe the answer should be yes, this is always possible.
Heuristically, the point is that the $V(I_i)$ are disjoint in Spec $R$. Thus, for a given $i$, we can isolate $I$ from all the other $I_j$ by localization to construct $J_i$.
Via a geometric construction, I believe we can do the following. For each $i$, let $U_i$ denote an open neighborhood of $V(I_i)$ that does not intersect any $V(I_j)$. Let $W_i$ denote the complement of $V(I_i)$. Then $U_i$ and $W_i$ cover Spec $R$. Set $J_i$ to be the ideal sheaf which agrees with $I$ on $U_i$ and agrees with $\mathcal{O}$ elsewhere (ie, we are gluing two sheaves along $U_i \cap W_i$). This should do the trick I believe.
(Hopefully I didn't secretly use Noetherian somewhere...)
answered Jul 26, 2018 at 16:48
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$\begingroup$ Thanks for your answer, can you explain the algebraic constraction a littel more( I know no sheaves theory). $\endgroup$ Commented Jul 26, 2018 at 18:25
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$\begingroup$ I thinke, the localization must be by $S:=R-\cap_{P\in \cup_{j\not= i}V(I_j)}P$. But $S$ dose not work, am I right? $\endgroup$ Commented Jul 26, 2018 at 18:30
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I will explain my comment for the case $n=2$ (I have not carefully checked the other cases, but I do not see any problems except notational difficulties).
So, assume $\sqrt{I}=I_1I_2$ as above. Pick $x\in I_1, y\in I_2$ such that $x+y=1$. Define $J_1=\{a\in R\,|\, y^ma\in I\,\text{ for some }\, m\}$ and similarly $J_2 = \{a\in R\,|\, x^na\in I\,\text{ for some }\, n\}$. Since $xy\in \sqrt{I}$, we have $x^Ny^N\in I$ for some $N \ge 1$ and thus, $x^N\in J_1, y^N\in J_2$. In particular the ideals $J_1$ and $J_2$ are comaximal, i.e., $J_1 + J_2 = R$.
Next, we show that $\sqrt{J_1} = \sqrt{I_1}$ (the proof of $\sqrt{J_2} = \sqrt{I_2}$ is identical). If $a\in J_1$, we have $y^ma\in I$. We can write $px+qy^m=1$ for some $p,q\in R$ and thus $a=pxa+qy^ma$. One has $pxa\in I_1$, since $x \in I_1$ and also $qy^ma\in I\subset I_1$. So, we get $a\in I_1$. We have established $J_1 \subseteq I_1$, and hence $\sqrt{J_1} \subseteq \sqrt{I_1}$. Conversely, let $b\in I_1$. Then $by\in I_1I_2$ and thus $b^my^m\in I$ for some $m$ and then, $b^m\in J_1$. Thus $I_1 \subseteq \sqrt{J_1}$, which entails $\sqrt{I_1} \subseteq \sqrt{J_1}$
Finally, we show that $J_1J_2=I$. Let $a\in J_1, b\in J_2$. Then $x^mab\in I$ and $y^nab\in I$ and thus $ab\in I$, since $Rx^m + Ry^n = R$. Conversely, let $a\in I$. Write $px^N+qy^N=1$ for $p,q\in R$. Then, $a=px^Na+qy^Na$. Since $px^Na\in I$, we have $pa\in J_2$. But, $x^N\in J_1$, so $px^Na\in J_1J_2$ and similarly, $qy^Na\in J_1J_2$ and so $a\in J_1J_2$ finishing the proof.
