Strange isomorphism: $R/(A + B) cong (R/B)/barA$.
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$R$ is a ring and $A, B$ are ideals of $R.$
I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?
abstract-algebra
edited Aug 20 at 9:28
Bernard
111k635103
asked Aug 20 at 9:19
伽罗瓦
934615
add a comment |Â
up vote
0
down vote
favorite
$R$ is a ring and $A, B$ are ideals of $R.$
I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?
abstract-algebra
edited Aug 20 at 9:28
Bernard
111k635103
asked Aug 20 at 9:19
伽罗瓦
934615
Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
1
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$R$ is a ring and $A, B$ are ideals of $R.$
I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?
abstract-algebra
edited Aug 20 at 9:28
Bernard
111k635103
asked Aug 20 at 9:19
伽罗瓦
934615
$R$ is a ring and $A, B$ are ideals of $R.$
I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?
abstract-algebra
edited Aug 20 at 9:28
Bernard
111k635103
asked Aug 20 at 9:19
伽罗瓦
934615
edited Aug 20 at 9:28
Bernard
111k635103
edited Aug 20 at 9:28
Bernard
111k635103
edited Aug 20 at 9:28
Bernard
111k635103
111k635103
asked Aug 20 at 9:19
伽罗瓦
934615
asked Aug 20 at 9:19
伽罗瓦
934615
asked Aug 20 at 9:19
伽罗瓦
934615
934615
Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
1
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29
add a comment |Â
Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
1
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29
Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
1
1
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$
answered Aug 20 at 9:32
JayTuma
1,333118
add a comment |Â
up vote
1
down vote
We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
answered Aug 20 at 9:34
伽罗瓦
934615
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$
answered Aug 20 at 9:32
JayTuma
1,333118
add a comment |Â
up vote
1
down vote
accepted
This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$
answered Aug 20 at 9:32
JayTuma
1,333118
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$
answered Aug 20 at 9:32
JayTuma
1,333118
This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$
answered Aug 20 at 9:32
JayTuma
1,333118
edited Aug 20 at 9:39
answered Aug 20 at 9:32
JayTuma
1,333118
answered Aug 20 at 9:32
JayTuma
1,333118
answered Aug 20 at 9:32
JayTuma
1,333118
1,333118
add a comment |Â
add a comment |Â
up vote
1
down vote
We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
answered Aug 20 at 9:34
伽罗瓦
934615
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
add a comment |Â
up vote
1
down vote
We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
answered Aug 20 at 9:34
伽罗瓦
934615
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
answered Aug 20 at 9:34
伽罗瓦
934615
We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
answered Aug 20 at 9:34
伽罗瓦
934615
answered Aug 20 at 9:34
伽罗瓦
934615
answered Aug 20 at 9:34
伽罗瓦
934615
answered Aug 20 at 9:34
伽罗瓦
934615
934615
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
add a comment |Â
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
nice thing here is that you actually proved the second ismorphism theorem :)
– JayTuma
Aug 20 at 9:37
add a comment |Â
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Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25
Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25
1
It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29