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?
abstract-algebra ring-theory commutative-algebra ideals
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
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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.
add a comment |
$begingroup$
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?
abstract-algebra ring-theory commutative-algebra ideals
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
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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.
1
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"if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument.
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– darij grinberg
Dec 10 '18 at 23:51
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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 $1 notin 3 mathbb{Z}$.
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– André 3000
Dec 10 '18 at 23:52
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Try $I cap J(I+J)$
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– Andres Mejia
Dec 10 '18 at 23:54
add a comment |
$begingroup$
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?
abstract-algebra ring-theory commutative-algebra ideals
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
$endgroup$
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
abstract-algebra ring-theory commutative-algebra ideals
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
asked Dec 10 '18 at 23:37
Jon D.Jon D.
164
164
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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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 $x=1cdot x$.
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
$endgroup$
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x=1cdot x$.
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x=1cdot x$.
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x=1cdot x$.
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
$endgroup$
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 $x=1cdot x$.
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
edited Dec 11 '18 at 0:27
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
answered Dec 10 '18 at 23:54
BernardBernard
120k740113
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
add a comment |
$begingroup$
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
add a comment |
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