$I$ and $J$ are coprime ideals iff $x to (x + I, x + J)$ is surjective.
$begingroup$
I'm stuck on this exercise and any help would be well appreciated:
Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.
Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.
Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.
Again, any help, pointers or references are much appreciated.
Thanks in advance.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 27 '14 at 17:56
user26857
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
$endgroup$
add a comment |
$begingroup$
I'm stuck on this exercise and any help would be well appreciated:
Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.
Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.
Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.
Again, any help, pointers or references are much appreciated.
Thanks in advance.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 27 '14 at 17:56
user26857
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
$endgroup$
add a comment |
$begingroup$
I'm stuck on this exercise and any help would be well appreciated:
Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.
Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.
Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.
Again, any help, pointers or references are much appreciated.
Thanks in advance.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 27 '14 at 17:56
user26857
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
$endgroup$
I'm stuck on this exercise and any help would be well appreciated:
Let $R$ be a commutative ring with ideals $I,J$. Show that $R=I+J$ if and only if $phi(x)= (x + I, x + J)$ is surjective from $R$ to $R/I times R/J$.
Assuming surjectivity I got as far as to realize that I need to show $Rsubset I+J$ since $I+Jsubset R$ always holds (as they are both subrings of R). Now I'm not sure how to proceed.
Also, I don't even know where to begin in the direction $R=I+J$ implies $phi$ is onto.
Again, any help, pointers or references are much appreciated.
Thanks in advance.
abstract-algebra ring-theory commutative-algebra ideals
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 27 '14 at 17:56
user26857
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
edited Dec 27 '14 at 17:56
user26857
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
edited Dec 27 '14 at 17:56
user26857
39.4k124183
edited Dec 27 '14 at 17:56
user26857
39.4k124183
edited Dec 27 '14 at 17:56
user26857
39.4k124183
39.4k124183
asked Dec 15 '14 at 11:02
WinstonWinston
374313
asked Dec 15 '14 at 11:02
WinstonWinston
374313
asked Dec 15 '14 at 11:02
WinstonWinston
374313
374313
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.
Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
$endgroup$
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.
Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
$endgroup$
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
add a comment |
$begingroup$
If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.
Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
$endgroup$
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
add a comment |
$begingroup$
If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.
Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
$endgroup$
If $R=I+J$ then for $a_iin R$ we get $x_iin I$ and $y_iin J$ such that $a_i=x_i+y_i$ ($i=1,2$), so $(a_1bmod I,a_2bmod J)=phi(y_1+x_2)$.
Conversely, if $phi$ is surjective and $ain R$ then $(abmod I,0bmod J)=phi(x)$, and thus $a=(a-x)+xin I+J$.
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
edited Dec 15 '14 at 12:13
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
answered Dec 15 '14 at 11:39
user26857user26857
39.4k124183
39.4k124183
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
add a comment |
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
3
3
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
$begingroup$
You know this better than I do, so hope you don't mind me asking. Is there a textbook on Commutative algebra that does not assume that rings have a $1$? The once I have perused do (IIRC) make this assumption in the preface or in the intro chapter. After all, modules over rings are a necessary tool / subject, and those require the unity. I realize that the OP called ideals subrings, which may have prompted you to include that assumption. But even so?
$endgroup$
– Jyrki Lahtonen
Dec 15 '14 at 11:47
add a comment |
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