Skip to main content
MathOverflow

Return to Answer

Adds missing radical symbol, plus minor rewording
Source Link
Luc Guyot
Luc Guyot
  • 7.9k
  • 2
  • 30
  • 51

I will explain my comment for the case $n=2$ (haveI 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\}$$J_1=\{a\in R\,|\, y^ma\in I\,\text{ for some }\, m\}$ and similarly $J_2$$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. In particular theythe ideals $J_1$ and $J_2$ are comaximal, i.e., $J_1 + J_2 = R$.

Next, we show that $\sqrt{J_1}=I_1$$\sqrt{J_1} = \sqrt{I_1}$ (and similarly forthe proof of $I_2,J_2$$\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 $x$ is.$x \in I_1$ 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 $x^m, y^n$ are comaximal$Rx^m + Ry^n = R$. 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.

I will explain my comment for the case $n=2$ (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 $J_2$. Since $xy\in \sqrt{I}$, we have $x^Ny^N\in I$ and thus, $x^N\in J_1, y^N\in J_2$, in particular they are comaximal.

Next, we show that $\sqrt{J_1}=I_1$ (and similarly for $I_2,J_2$). 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 $x$ is. $qy^ma\in I\subset I_1$. So, we get $a\in I_1$. Conversely, let $b\in I_1$. Then $by\in I_1I_2$ and thus $b^my^m\in I$ for some $m$ and then, $b^m\in J_1$.

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 $x^m, y^n$ are comaximal. 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.

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.

Source Link
Mohan
Mohan
  • 6.3k
  • 1
  • 23
  • 24

I will explain my comment for the case $n=2$ (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 $J_2$. Since $xy\in \sqrt{I}$, we have $x^Ny^N\in I$ and thus, $x^N\in J_1, y^N\in J_2$, in particular they are comaximal.

Next, we show that $\sqrt{J_1}=I_1$ (and similarly for $I_2,J_2$). 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 $x$ is. $qy^ma\in I\subset I_1$. So, we get $a\in I_1$. Conversely, let $b\in I_1$. Then $by\in I_1I_2$ and thus $b^my^m\in I$ for some $m$ and then, $b^m\in J_1$.

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 $x^m, y^n$ are comaximal. 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.