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Is there an isomorphism between these two finitely generated modules?
Clash Royale CLAN TAG#URR8PPP
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Let $A$ be a ring and $M$ a finitely generated module over $A.$ Then $M = Aw_1 + ldots + Aw_s.$ When exactly is there an isomorphism between $M/Aw_s$ and $Aw_1 + ldots + Aw_s - 1.$ Intuitively, to me, these two should be isomorphic. I guess the most natural way to define a map would be to send $a_1w_1 + ldots + a_s - 1w_s - 1 + Aw_s mapsto a_1w_1 + ldots + a_s - 1w_s - 1.$ However, I d0n't think this is even well defined. When exactly can we form an isomorphism between these two modules? My guess is when $w_1, ldots, w_s - 1$ is linearly independent...
abstract-algebra
asked Jul 30 at 19:40
伽罗瓦
781615
add a comment |Â
up vote
1
down vote
favorite
Let $A$ be a ring and $M$ a finitely generated module over $A.$ Then $M = Aw_1 + ldots + Aw_s.$ When exactly is there an isomorphism between $M/Aw_s$ and $Aw_1 + ldots + Aw_s - 1.$ Intuitively, to me, these two should be isomorphic. I guess the most natural way to define a map would be to send $a_1w_1 + ldots + a_s - 1w_s - 1 + Aw_s mapsto a_1w_1 + ldots + a_s - 1w_s - 1.$ However, I d0n't think this is even well defined. When exactly can we form an isomorphism between these two modules? My guess is when $w_1, ldots, w_s - 1$ is linearly independent...
abstract-algebra
asked Jul 30 at 19:40
伽罗瓦
781615
You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be a ring and $M$ a finitely generated module over $A.$ Then $M = Aw_1 + ldots + Aw_s.$ When exactly is there an isomorphism between $M/Aw_s$ and $Aw_1 + ldots + Aw_s - 1.$ Intuitively, to me, these two should be isomorphic. I guess the most natural way to define a map would be to send $a_1w_1 + ldots + a_s - 1w_s - 1 + Aw_s mapsto a_1w_1 + ldots + a_s - 1w_s - 1.$ However, I d0n't think this is even well defined. When exactly can we form an isomorphism between these two modules? My guess is when $w_1, ldots, w_s - 1$ is linearly independent...
abstract-algebra
asked Jul 30 at 19:40
伽罗瓦
781615
Let $A$ be a ring and $M$ a finitely generated module over $A.$ Then $M = Aw_1 + ldots + Aw_s.$ When exactly is there an isomorphism between $M/Aw_s$ and $Aw_1 + ldots + Aw_s - 1.$ Intuitively, to me, these two should be isomorphic. I guess the most natural way to define a map would be to send $a_1w_1 + ldots + a_s - 1w_s - 1 + Aw_s mapsto a_1w_1 + ldots + a_s - 1w_s - 1.$ However, I d0n't think this is even well defined. When exactly can we form an isomorphism between these two modules? My guess is when $w_1, ldots, w_s - 1$ is linearly independent...
abstract-algebra
asked Jul 30 at 19:40
伽罗瓦
781615
asked Jul 30 at 19:40
伽罗瓦
781615
asked Jul 30 at 19:40
伽罗瓦
781615
asked Jul 30 at 19:40
伽罗瓦
781615
781615
You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03
add a comment |Â
You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03
You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
In any case, there exists a canonical morphism $A w_1 + cdots + A w_s-1 hookrightarrow M to M / A w_s$ in the other direction. Moreover, it is easy to see that this map is surjective in general. However, the kernel of this map is exactly $(A w_1 + cdots + A w_s-1) cap A w_s$ which is not zero in general; but if it is zero, then yes, the two modules are isomorphic.
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In any case, there exists a canonical morphism $A w_1 + cdots + A w_s-1 hookrightarrow M to M / A w_s$ in the other direction. Moreover, it is easy to see that this map is surjective in general. However, the kernel of this map is exactly $(A w_1 + cdots + A w_s-1) cap A w_s$ which is not zero in general; but if it is zero, then yes, the two modules are isomorphic.
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
add a comment |Â
up vote
3
down vote
accepted
In any case, there exists a canonical morphism $A w_1 + cdots + A w_s-1 hookrightarrow M to M / A w_s$ in the other direction. Moreover, it is easy to see that this map is surjective in general. However, the kernel of this map is exactly $(A w_1 + cdots + A w_s-1) cap A w_s$ which is not zero in general; but if it is zero, then yes, the two modules are isomorphic.
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In any case, there exists a canonical morphism $A w_1 + cdots + A w_s-1 hookrightarrow M to M / A w_s$ in the other direction. Moreover, it is easy to see that this map is surjective in general. However, the kernel of this map is exactly $(A w_1 + cdots + A w_s-1) cap A w_s$ which is not zero in general; but if it is zero, then yes, the two modules are isomorphic.
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
In any case, there exists a canonical morphism $A w_1 + cdots + A w_s-1 hookrightarrow M to M / A w_s$ in the other direction. Moreover, it is easy to see that this map is surjective in general. However, the kernel of this map is exactly $(A w_1 + cdots + A w_s-1) cap A w_s$ which is not zero in general; but if it is zero, then yes, the two modules are isomorphic.
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
answered Jul 30 at 19:58
Daniel Schepler
6,6681513
6,6681513
add a comment |Â
add a comment |Â
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You have this isomorphism when $Aw_s cap sum_i=1^s-1 Aw_i = 0$ (which is sort of saying $w_s$ is independent from the other $w_i$'s); otherwise, the map is not well-defined.
– JJC94
Jul 30 at 19:55
I mean the map you defined above, btw.
– JJC94
Jul 30 at 20:03