Let R be a commutative ring, and I, J denote two ideals in R such that I + J = R. Is it true that IJ = I ∩...












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This question already has an answer here:




  • $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

    2 answers




So far I have that if I + J = R, then 1 ∈ I and/or 1 ∈ J. Then I = R and/or J = R. If both I = J = R, then IJ = I ∩ J = R must be true because we already know that multiplying every element in I or J by every element in R will end up giving us R again. So IJ = I ∩ J = R.



If I = R, but J ≠ R, then IJ = J = I ∩ J because we already know that when all elements of J are multiplied by all elements of R, the result is J again. Therefore, IJ = I ∩ J = J.



Therefore, it is true that if I + J = R, then IJ = I ∩ J.



Does this proof work or am I not able to say that 1 ∈ I or 1 ∈ J?










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Dec 11 '18 at 0:07


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




    $begingroup$
    "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
    $endgroup$
    – darij grinberg
    Dec 10 '18 at 23:51










  • $begingroup$
    How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
    $endgroup$
    – André 3000
    Dec 10 '18 at 23:52












  • $begingroup$
    Try $I cap J(I+J)$
    $endgroup$
    – Andres Mejia
    Dec 10 '18 at 23:54
















0












$begingroup$



This question already has an answer here:




  • $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

    2 answers




So far I have that if I + J = R, then 1 ∈ I and/or 1 ∈ J. Then I = R and/or J = R. If both I = J = R, then IJ = I ∩ J = R must be true because we already know that multiplying every element in I or J by every element in R will end up giving us R again. So IJ = I ∩ J = R.



If I = R, but J ≠ R, then IJ = J = I ∩ J because we already know that when all elements of J are multiplied by all elements of R, the result is J again. Therefore, IJ = I ∩ J = J.



Therefore, it is true that if I + J = R, then IJ = I ∩ J.



Does this proof work or am I not able to say that 1 ∈ I or 1 ∈ J?










share|cite|improve this question









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marked as duplicate by rschwieb abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Dec 11 '18 at 0:07


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




    $begingroup$
    "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
    $endgroup$
    – darij grinberg
    Dec 10 '18 at 23:51










  • $begingroup$
    How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
    $endgroup$
    – André 3000
    Dec 10 '18 at 23:52












  • $begingroup$
    Try $I cap J(I+J)$
    $endgroup$
    – Andres Mejia
    Dec 10 '18 at 23:54














0












0








0





$begingroup$



This question already has an answer here:




  • $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

    2 answers




So far I have that if I + J = R, then 1 ∈ I and/or 1 ∈ J. Then I = R and/or J = R. If both I = J = R, then IJ = I ∩ J = R must be true because we already know that multiplying every element in I or J by every element in R will end up giving us R again. So IJ = I ∩ J = R.



If I = R, but J ≠ R, then IJ = J = I ∩ J because we already know that when all elements of J are multiplied by all elements of R, the result is J again. Therefore, IJ = I ∩ J = J.



Therefore, it is true that if I + J = R, then IJ = I ∩ J.



Does this proof work or am I not able to say that 1 ∈ I or 1 ∈ J?










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

    2 answers




So far I have that if I + J = R, then 1 ∈ I and/or 1 ∈ J. Then I = R and/or J = R. If both I = J = R, then IJ = I ∩ J = R must be true because we already know that multiplying every element in I or J by every element in R will end up giving us R again. So IJ = I ∩ J = R.



If I = R, but J ≠ R, then IJ = J = I ∩ J because we already know that when all elements of J are multiplied by all elements of R, the result is J again. Therefore, IJ = I ∩ J = J.



Therefore, it is true that if I + J = R, then IJ = I ∩ J.



Does this proof work or am I not able to say that 1 ∈ I or 1 ∈ J?





This question already has an answer here:




  • $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=Icap J$

    2 answers








abstract-algebra ring-theory commutative-algebra ideals






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asked Dec 10 '18 at 23:37









Jon D.Jon D.

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marked as duplicate by rschwieb abstract-algebra
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Dec 11 '18 at 0:07


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 rschwieb abstract-algebra
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Dec 11 '18 at 0:07


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




    $begingroup$
    "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
    $endgroup$
    – darij grinberg
    Dec 10 '18 at 23:51










  • $begingroup$
    How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
    $endgroup$
    – André 3000
    Dec 10 '18 at 23:52












  • $begingroup$
    Try $I cap J(I+J)$
    $endgroup$
    – Andres Mejia
    Dec 10 '18 at 23:54














  • 1




    $begingroup$
    "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
    $endgroup$
    – darij grinberg
    Dec 10 '18 at 23:51










  • $begingroup$
    How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
    $endgroup$
    – André 3000
    Dec 10 '18 at 23:52












  • $begingroup$
    Try $I cap J(I+J)$
    $endgroup$
    – Andres Mejia
    Dec 10 '18 at 23:54








1




1




$begingroup$
"if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
$endgroup$
– darij grinberg
Dec 10 '18 at 23:51




$begingroup$
"if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
$endgroup$
– darij grinberg
Dec 10 '18 at 23:51












$begingroup$
How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
$endgroup$
– André 3000
Dec 10 '18 at 23:52






$begingroup$
How does $I+J = R$ imply that $1 in I$ or $1 in J$? This is false: consider for instance the ideals $2 mathbb{Z}$ and $3 mathbb{Z}$ in $mathbb{Z}$. Then $1 = -2 + 3 in 2 mathbb{Z} + 3 mathbb{Z}$, but $1 notin 2mathbb{Z}$ and $1 notin 3 mathbb{Z}$.
$endgroup$
– André 3000
Dec 10 '18 at 23:52














$begingroup$
Try $I cap J(I+J)$
$endgroup$
– Andres Mejia
Dec 10 '18 at 23:54




$begingroup$
Try $I cap J(I+J)$
$endgroup$
– Andres Mejia
Dec 10 '18 at 23:54










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint:



$IJsubset Icap J$ is trivial. For the reverse inclusion:



By hypothesis, there exist $uin I$, $vin J$ such that $u+v =1$. Now let $xin Icap J$. Use that $x=1cdot x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:11










  • $begingroup$
    It is indeed obvious. Check the definition of a product of ideals.
    $endgroup$
    – Bernard
    Dec 11 '18 at 0:27










  • $begingroup$
    I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:45










  • $begingroup$
    Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:54










  • $begingroup$
    Yes. Isn't each of these products in $Icap J$?
    $endgroup$
    – Bernard
    Dec 11 '18 at 1:05


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint:



$IJsubset Icap J$ is trivial. For the reverse inclusion:



By hypothesis, there exist $uin I$, $vin J$ such that $u+v =1$. Now let $xin Icap J$. Use that $x=1cdot x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:11










  • $begingroup$
    It is indeed obvious. Check the definition of a product of ideals.
    $endgroup$
    – Bernard
    Dec 11 '18 at 0:27










  • $begingroup$
    I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:45










  • $begingroup$
    Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:54










  • $begingroup$
    Yes. Isn't each of these products in $Icap J$?
    $endgroup$
    – Bernard
    Dec 11 '18 at 1:05
















0












$begingroup$

Hint:



$IJsubset Icap J$ is trivial. For the reverse inclusion:



By hypothesis, there exist $uin I$, $vin J$ such that $u+v =1$. Now let $xin Icap J$. Use that $x=1cdot x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:11










  • $begingroup$
    It is indeed obvious. Check the definition of a product of ideals.
    $endgroup$
    – Bernard
    Dec 11 '18 at 0:27










  • $begingroup$
    I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:45










  • $begingroup$
    Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:54










  • $begingroup$
    Yes. Isn't each of these products in $Icap J$?
    $endgroup$
    – Bernard
    Dec 11 '18 at 1:05














0












0








0





$begingroup$

Hint:



$IJsubset Icap J$ is trivial. For the reverse inclusion:



By hypothesis, there exist $uin I$, $vin J$ such that $u+v =1$. Now let $xin Icap J$. Use that $x=1cdot x$.






share|cite|improve this answer











$endgroup$



Hint:



$IJsubset Icap J$ is trivial. For the reverse inclusion:



By hypothesis, there exist $uin I$, $vin J$ such that $u+v =1$. Now let $xin Icap J$. Use that $x=1cdot x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 0:27

























answered Dec 10 '18 at 23:54









BernardBernard

120k740113




120k740113












  • $begingroup$
    Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:11










  • $begingroup$
    It is indeed obvious. Check the definition of a product of ideals.
    $endgroup$
    – Bernard
    Dec 11 '18 at 0:27










  • $begingroup$
    I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:45










  • $begingroup$
    Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:54










  • $begingroup$
    Yes. Isn't each of these products in $Icap J$?
    $endgroup$
    – Bernard
    Dec 11 '18 at 1:05


















  • $begingroup$
    Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:11










  • $begingroup$
    It is indeed obvious. Check the definition of a product of ideals.
    $endgroup$
    – Bernard
    Dec 11 '18 at 0:27










  • $begingroup$
    I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:45










  • $begingroup$
    Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
    $endgroup$
    – Jon D.
    Dec 11 '18 at 0:54










  • $begingroup$
    Yes. Isn't each of these products in $Icap J$?
    $endgroup$
    – Bernard
    Dec 11 '18 at 1:05
















$begingroup$
Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
$endgroup$
– Jon D.
Dec 11 '18 at 0:11




$begingroup$
Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals?
$endgroup$
– Jon D.
Dec 11 '18 at 0:11












$begingroup$
It is indeed obvious. Check the definition of a product of ideals.
$endgroup$
– Bernard
Dec 11 '18 at 0:27




$begingroup$
It is indeed obvious. Check the definition of a product of ideals.
$endgroup$
– Bernard
Dec 11 '18 at 0:27












$begingroup$
I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
$endgroup$
– Jon D.
Dec 11 '18 at 0:45




$begingroup$
I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit.
$endgroup$
– Jon D.
Dec 11 '18 at 0:45












$begingroup$
Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
$endgroup$
– Jon D.
Dec 11 '18 at 0:54




$begingroup$
Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made?
$endgroup$
– Jon D.
Dec 11 '18 at 0:54












$begingroup$
Yes. Isn't each of these products in $Icap J$?
$endgroup$
– Bernard
Dec 11 '18 at 1:05




$begingroup$
Yes. Isn't each of these products in $Icap J$?
$endgroup$
– Bernard
Dec 11 '18 at 1:05



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