$M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 cap M_2 = 0$ implies M is...
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I saw the two properties mentioned by the post on
Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A cong B$.
With intuition I think if $M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 cap M_2 = 0$ implies M is isomorphic to $M_1 oplus M_2$, but I did not find this in books I could find. Is this true?
modules direct-sum
asked Nov 16 at 1:11
Eric Curtis
163
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up vote
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favorite
I saw the two properties mentioned by the post on
Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A cong B$.
With intuition I think if $M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 cap M_2 = 0$ implies M is isomorphic to $M_1 oplus M_2$, but I did not find this in books I could find. Is this true?
modules direct-sum
asked Nov 16 at 1:11
Eric Curtis
163
1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I saw the two properties mentioned by the post on
Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A cong B$.
With intuition I think if $M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 cap M_2 = 0$ implies M is isomorphic to $M_1 oplus M_2$, but I did not find this in books I could find. Is this true?
modules direct-sum
asked Nov 16 at 1:11
Eric Curtis
163
I saw the two properties mentioned by the post on
Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A cong B$.
With intuition I think if $M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 cap M_2 = 0$ implies M is isomorphic to $M_1 oplus M_2$, but I did not find this in books I could find. Is this true?
modules direct-sum
modules direct-sum
asked Nov 16 at 1:11
Eric Curtis
163
asked Nov 16 at 1:11
Eric Curtis
163
asked Nov 16 at 1:11
Eric Curtis
163
asked Nov 16 at 1:11
Eric Curtis
163
asked Nov 16 at 1:11
Eric Curtis
163
163
1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09
|
show 1 more comment
1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09
1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09
|
show 1 more comment
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1) Have you tried proving it? 2) Have you tried to mimic the proof for vector spaces and see if it works in this case?
– dyf
Nov 16 at 1:33
You probably didn't find this specific property in the books because most books prove (or leave as an exercise) a much more general property to which this is a special case. The mentioned property is true. To prove it yourself, I suggest you start with building the most natural homomorphism $phi: M_1oplus M_2to M_1+M_2$ you can think of...
– Hamed
Nov 16 at 1:36
I see. Just being not confident. Thank you for your help!
– Eric Curtis
Nov 16 at 1:39
Also, @Hamed could you tell me some books providing the more general property? Thanks!
– Eric Curtis
Nov 16 at 1:44
If you look at any commutative algebra book and look for "split exact sequence", that's a (first order) generalization of this statement. See for example Atiyah and MacDonald (but again, any other book also has this).
– Hamed
Nov 16 at 2:09