Mixing
![The average value of a list, in chunks of 100 items [duplicate]](https://lh5.googleusercontent.com/-1CItGZeGD5s/AAAAAAAAAAI/AAAAAAAAAAA/ABtNlbDlnaNFz--BBSKX9AM7lwlEnLDtSQ/s72-c-mo/photo.jpg?sz=32)
The radical of the product of two ideals is the intersection of the radicals of the ideals
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Let $R$ be a Noetherian ring (not sure if "Noetherianness" is needed). I'm trying to show that $r(IJ)=r(I)cap r(J)$ ($r$ for radical).
One direction looks easy. If $xin r(IJ)$, then for some $k$, $x^kin IJ$. Since $IJsubset Icap J$, $x^kin Icap J$, so $xin r(I)cap r(J)$.
The other direction is not clear to me. Let $xin r(I)cap r(J)$. Then $x^nin I, x^min J$. So $x^{max(n,m)}in Icap J$. If $I+J=R$, then $Icap Jsubset IJ$, and I would be able to conclude that $xin r(IJ)$. But it need not be the casse that $I+J=R$. How do I proceed then?
abstract-algebra ring-theory ideals
asked Jan 26 at 22:11
user437309user437309
766314
$endgroup$
add a comment |
$begingroup$
Let $R$ be a Noetherian ring (not sure if "Noetherianness" is needed). I'm trying to show that $r(IJ)=r(I)cap r(J)$ ($r$ for radical).
One direction looks easy. If $xin r(IJ)$, then for some $k$, $x^kin IJ$. Since $IJsubset Icap J$, $x^kin Icap J$, so $xin r(I)cap r(J)$.
The other direction is not clear to me. Let $xin r(I)cap r(J)$. Then $x^nin I, x^min J$. So $x^{max(n,m)}in Icap J$. If $I+J=R$, then $Icap Jsubset IJ$, and I would be able to conclude that $xin r(IJ)$. But it need not be the casse that $I+J=R$. How do I proceed then?
abstract-algebra ring-theory ideals
asked Jan 26 at 22:11
user437309user437309
766314
$endgroup$
3
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19
add a comment |
$begingroup$
Let $R$ be a Noetherian ring (not sure if "Noetherianness" is needed). I'm trying to show that $r(IJ)=r(I)cap r(J)$ ($r$ for radical).
One direction looks easy. If $xin r(IJ)$, then for some $k$, $x^kin IJ$. Since $IJsubset Icap J$, $x^kin Icap J$, so $xin r(I)cap r(J)$.
The other direction is not clear to me. Let $xin r(I)cap r(J)$. Then $x^nin I, x^min J$. So $x^{max(n,m)}in Icap J$. If $I+J=R$, then $Icap Jsubset IJ$, and I would be able to conclude that $xin r(IJ)$. But it need not be the casse that $I+J=R$. How do I proceed then?
abstract-algebra ring-theory ideals
asked Jan 26 at 22:11
user437309user437309
766314
$endgroup$
Let $R$ be a Noetherian ring (not sure if "Noetherianness" is needed). I'm trying to show that $r(IJ)=r(I)cap r(J)$ ($r$ for radical).
One direction looks easy. If $xin r(IJ)$, then for some $k$, $x^kin IJ$. Since $IJsubset Icap J$, $x^kin Icap J$, so $xin r(I)cap r(J)$.
The other direction is not clear to me. Let $xin r(I)cap r(J)$. Then $x^nin I, x^min J$. So $x^{max(n,m)}in Icap J$. If $I+J=R$, then $Icap Jsubset IJ$, and I would be able to conclude that $xin r(IJ)$. But it need not be the casse that $I+J=R$. How do I proceed then?
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Jan 26 at 22:11
user437309user437309
766314
asked Jan 26 at 22:11
user437309user437309
766314
asked Jan 26 at 22:11
user437309user437309
766314
asked Jan 26 at 22:11
user437309user437309
766314
asked Jan 26 at 22:11
user437309user437309
766314
766314
3
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19
add a comment |
3
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19
3
3
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19
add a comment |
1 Answer
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Hint: What can you say about $x^mtimes x^n$?
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
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add a comment |
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$begingroup$
Hint: What can you say about $x^mtimes x^n$?
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
$endgroup$
add a comment |
$begingroup$
Hint: What can you say about $x^mtimes x^n$?
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
$endgroup$
add a comment |
$begingroup$
Hint: What can you say about $x^mtimes x^n$?
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
$endgroup$
Hint: What can you say about $x^mtimes x^n$?
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
answered Jan 26 at 22:17
ScientificaScientifica
6,82941335
6,82941335
add a comment |
add a comment |
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3
$begingroup$
If $x^nin I$ and $x^min J$, then $x^{n+m}in IJ$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:14
$begingroup$
@LordSharktheUnknown Oh well apparently I didn't bother to consider $x^nx^m=x^{n+m}$...
$endgroup$
– user437309
Jan 26 at 22:19