$R$ is commutative ring with unity, prove $I,Jtrianglelefteq R wedge I+J=R Rightarrow IJ=Icap J$ [duplicate]
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For a ring R, and ideals $A$, $B$, then $AB=A cap B$ if $A + B = R$
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Let $R$ a commutative ring with unity, and let $I,Jtrianglelefteq R$ ideals such that $I+J=R$. Prove that $IJ=Icap J$.
Recall $IJ={sum_{k=1}^n i_kj_k:i_kin I,j_kin J, kin mathbb{N} }$.
I have shown that $IJsubseteq Icap J$. The rest of the solution does not use the commutativity of $R$ so my question is where is the mistake:
Let $ain Icap J$. $a=acdot 1in IJ$ thus $Icap J=IJ$.
ring-theory ideals
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
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marked as duplicate by Arnaud D., rschwieb ring-theory
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Dec 12 '18 at 14:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
For a ring R, and ideals $A$, $B$, then $AB=A cap B$ if $A + B = R$
1 answer
Let $R$ a commutative ring with unity, and let $I,Jtrianglelefteq R$ ideals such that $I+J=R$. Prove that $IJ=Icap J$.
Recall $IJ={sum_{k=1}^n i_kj_k:i_kin I,j_kin J, kin mathbb{N} }$.
I have shown that $IJsubseteq Icap J$. The rest of the solution does not use the commutativity of $R$ so my question is where is the mistake:
Let $ain Icap J$. $a=acdot 1in IJ$ thus $Icap J=IJ$.
ring-theory ideals
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
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marked as duplicate by Arnaud D., rschwieb ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
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Dec 12 '18 at 14:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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You have assumed $1in Jimplies J=R$.
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– Yadati Kiran
Dec 12 '18 at 12:42
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@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
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– J. Doe
Dec 12 '18 at 13:17
1
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@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
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– Yadati Kiran
Dec 12 '18 at 13:24
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Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
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– rschwieb
Dec 12 '18 at 14:04
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I meant, "please stop posting as more than one account," that is.
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– rschwieb
Dec 12 '18 at 14:10
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show 1 more comment
$begingroup$
This question already has an answer here:
For a ring R, and ideals $A$, $B$, then $AB=A cap B$ if $A + B = R$
1 answer
Let $R$ a commutative ring with unity, and let $I,Jtrianglelefteq R$ ideals such that $I+J=R$. Prove that $IJ=Icap J$.
Recall $IJ={sum_{k=1}^n i_kj_k:i_kin I,j_kin J, kin mathbb{N} }$.
I have shown that $IJsubseteq Icap J$. The rest of the solution does not use the commutativity of $R$ so my question is where is the mistake:
Let $ain Icap J$. $a=acdot 1in IJ$ thus $Icap J=IJ$.
ring-theory ideals
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
$endgroup$
This question already has an answer here:
For a ring R, and ideals $A$, $B$, then $AB=A cap B$ if $A + B = R$
1 answer
Let $R$ a commutative ring with unity, and let $I,Jtrianglelefteq R$ ideals such that $I+J=R$. Prove that $IJ=Icap J$.
Recall $IJ={sum_{k=1}^n i_kj_k:i_kin I,j_kin J, kin mathbb{N} }$.
I have shown that $IJsubseteq Icap J$. The rest of the solution does not use the commutativity of $R$ so my question is where is the mistake:
Let $ain Icap J$. $a=acdot 1in IJ$ thus $Icap J=IJ$.
This question already has an answer here:
For a ring R, and ideals $A$, $B$, then $AB=A cap B$ if $A + B = R$
1 answer
ring-theory ideals
ring-theory ideals
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
asked Dec 12 '18 at 12:40
J. DoeJ. Doe
15911
15911
marked as duplicate by Arnaud D., rschwieb ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
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Dec 12 '18 at 14:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud D., rschwieb ring-theory
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Dec 12 '18 at 14:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You have assumed $1in Jimplies J=R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:42
$begingroup$
@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
$endgroup$
– J. Doe
Dec 12 '18 at 13:17
1
$begingroup$
@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 13:24
$begingroup$
Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
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– rschwieb
Dec 12 '18 at 14:04
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I meant, "please stop posting as more than one account," that is.
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– rschwieb
Dec 12 '18 at 14:10
|
show 1 more comment
1
$begingroup$
You have assumed $1in Jimplies J=R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:42
$begingroup$
@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
$endgroup$
– J. Doe
Dec 12 '18 at 13:17
1
$begingroup$
@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 13:24
$begingroup$
Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
$endgroup$
– rschwieb
Dec 12 '18 at 14:04
$begingroup$
I meant, "please stop posting as more than one account," that is.
$endgroup$
– rschwieb
Dec 12 '18 at 14:10
1
1
$begingroup$
You have assumed $1in Jimplies J=R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:42
$begingroup$
You have assumed $1in Jimplies J=R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:42
$begingroup$
@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
$endgroup$
– J. Doe
Dec 12 '18 at 13:17
$begingroup$
@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
$endgroup$
– J. Doe
Dec 12 '18 at 13:17
1
1
$begingroup$
@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 13:24
$begingroup$
@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 13:24
$begingroup$
Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
$endgroup$
– rschwieb
Dec 12 '18 at 14:04
$begingroup$
Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
$endgroup$
– rschwieb
Dec 12 '18 at 14:04
$begingroup$
I meant, "please stop posting as more than one account," that is.
$endgroup$
– rschwieb
Dec 12 '18 at 14:10
$begingroup$
I meant, "please stop posting as more than one account," that is.
$endgroup$
– rschwieb
Dec 12 '18 at 14:10
|
show 1 more comment
1 Answer
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As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $acdot 1in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $acdot 1in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $iin I,jin J$. Now if $ain Icap J$, try to rewrite $a=acdot 1=acdot (i+j)=(acdot i)+(acdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $acdot 1in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $acdot 1in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $iin I,jin J$. Now if $ain Icap J$, try to rewrite $a=acdot 1=acdot (i+j)=(acdot i)+(acdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
$endgroup$
add a comment |
$begingroup$
As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $acdot 1in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $acdot 1in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $iin I,jin J$. Now if $ain Icap J$, try to rewrite $a=acdot 1=acdot (i+j)=(acdot i)+(acdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
$endgroup$
add a comment |
$begingroup$
As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $acdot 1in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $acdot 1in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $iin I,jin J$. Now if $ain Icap J$, try to rewrite $a=acdot 1=acdot (i+j)=(acdot i)+(acdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
$endgroup$
As Yadati Kiran mentioned in their comment, $1$ is not in general an element of $J$. While it is true that $acdot 1in I$ by one of the properties of an ideal (as you mention in your recent comment), how do you deduce that $acdot 1in IJ$? If $1$ were an element of $J$ this would follow, but as Yadati Kiran pointed out, this is not the case in general.
Here is a hint on how to proceed instead: Your intuition to be skeptical of a solution that doesn't use all of the assumptions was right - but you also did not use I+J=R yet. This allows you to write $1=i+j$ with $iin I,jin J$. Now if $ain Icap J$, try to rewrite $a=acdot 1=acdot (i+j)=(acdot i)+(acdot j)$ in a way that shows this is an element of $IJ$ - you will want to use commutativity here.
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
edited Dec 12 '18 at 13:22
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
answered Dec 12 '18 at 13:06
Malte L.Malte L.
462
462
add a comment |
add a comment |
1
$begingroup$
You have assumed $1in Jimplies J=R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:42
$begingroup$
@YadatiKiran Why? One of the properties of an ideal $I$ is $forall ain I forall rin R, ar,rain I$.
$endgroup$
– J. Doe
Dec 12 '18 at 13:17
1
$begingroup$
@J.Doe: Precisely $a=acdot1in I$ not $IJ$ because $1in R$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 13:24
$begingroup$
Interesting, a user named "Jon D" posted a very similar question yesterday. If you are the same person, please stop. One of the suggested duplicates contains an example for noncommutative rings.
$endgroup$
– rschwieb
Dec 12 '18 at 14:04
$begingroup$
I meant, "please stop posting as more than one account," that is.
$endgroup$
– rschwieb
Dec 12 '18 at 14:10