Why do biadditive, balenced maps define a homomorphism of bimodules?
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While studying Morita's theorem, I came across the following statement which really confuses me:
Let A, B be rings and $_A M_B$,$_B N_A$ be bimodules. A biadditive, $B$-balanced map $f:Mtimes N rightarrow A$ defines a homomorphism of bimodules $tilde{f}:Motimes_BN rightarrow A$.
Now I understand that, by the universal property of the tensor product, the map $f$ defines a homomorphism of abelian groups. I also know that the abelian group $Motimes_BN$ can be equiped with a left and right action, turning it into an $(A,A)$-bimodule. What I don't get is why the induced morphism $tilde{f}$ is actually a homomorphism of bimodules. Specifically, the biadditive and $B$-balanced properties of $f$ don't seem to be enough for me to be able to prove that
- $tilde{f}(a(motimes n))=atilde{f}(motimes n)$
- $tilde{f}((motimes n)a)=tilde{f}(motimes n)a$
Moreover, wouldn't that imply that the map $f$ has the same properties on the first and second component?
abstract-algebra tensor-products
asked Jan 8 at 17:21
user102845user102845
183
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add a comment |
$begingroup$
While studying Morita's theorem, I came across the following statement which really confuses me:
Let A, B be rings and $_A M_B$,$_B N_A$ be bimodules. A biadditive, $B$-balanced map $f:Mtimes N rightarrow A$ defines a homomorphism of bimodules $tilde{f}:Motimes_BN rightarrow A$.
Now I understand that, by the universal property of the tensor product, the map $f$ defines a homomorphism of abelian groups. I also know that the abelian group $Motimes_BN$ can be equiped with a left and right action, turning it into an $(A,A)$-bimodule. What I don't get is why the induced morphism $tilde{f}$ is actually a homomorphism of bimodules. Specifically, the biadditive and $B$-balanced properties of $f$ don't seem to be enough for me to be able to prove that
- $tilde{f}(a(motimes n))=atilde{f}(motimes n)$
- $tilde{f}((motimes n)a)=tilde{f}(motimes n)a$
Moreover, wouldn't that imply that the map $f$ has the same properties on the first and second component?
abstract-algebra tensor-products
asked Jan 8 at 17:21
user102845user102845
183
$endgroup$
$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14
add a comment |
$begingroup$
While studying Morita's theorem, I came across the following statement which really confuses me:
Let A, B be rings and $_A M_B$,$_B N_A$ be bimodules. A biadditive, $B$-balanced map $f:Mtimes N rightarrow A$ defines a homomorphism of bimodules $tilde{f}:Motimes_BN rightarrow A$.
Now I understand that, by the universal property of the tensor product, the map $f$ defines a homomorphism of abelian groups. I also know that the abelian group $Motimes_BN$ can be equiped with a left and right action, turning it into an $(A,A)$-bimodule. What I don't get is why the induced morphism $tilde{f}$ is actually a homomorphism of bimodules. Specifically, the biadditive and $B$-balanced properties of $f$ don't seem to be enough for me to be able to prove that
- $tilde{f}(a(motimes n))=atilde{f}(motimes n)$
- $tilde{f}((motimes n)a)=tilde{f}(motimes n)a$
Moreover, wouldn't that imply that the map $f$ has the same properties on the first and second component?
abstract-algebra tensor-products
asked Jan 8 at 17:21
user102845user102845
183
$endgroup$
While studying Morita's theorem, I came across the following statement which really confuses me:
Let A, B be rings and $_A M_B$,$_B N_A$ be bimodules. A biadditive, $B$-balanced map $f:Mtimes N rightarrow A$ defines a homomorphism of bimodules $tilde{f}:Motimes_BN rightarrow A$.
Now I understand that, by the universal property of the tensor product, the map $f$ defines a homomorphism of abelian groups. I also know that the abelian group $Motimes_BN$ can be equiped with a left and right action, turning it into an $(A,A)$-bimodule. What I don't get is why the induced morphism $tilde{f}$ is actually a homomorphism of bimodules. Specifically, the biadditive and $B$-balanced properties of $f$ don't seem to be enough for me to be able to prove that
- $tilde{f}(a(motimes n))=atilde{f}(motimes n)$
- $tilde{f}((motimes n)a)=tilde{f}(motimes n)a$
Moreover, wouldn't that imply that the map $f$ has the same properties on the first and second component?
abstract-algebra tensor-products
abstract-algebra tensor-products
asked Jan 8 at 17:21
user102845user102845
183
asked Jan 8 at 17:21
user102845user102845
183
edited Jan 9 at 14:12
user102845
asked Jan 8 at 17:21
user102845user102845
183
asked Jan 8 at 17:21
user102845user102845
183
asked Jan 8 at 17:21
user102845user102845
183
183
$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14
add a comment |
$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14
$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14
add a comment |
1 Answer
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One needs $f(am,n)=af(m,n)$ and $f(m,na)=f(m,n)a$. Then those imply
$tilde f(amotimes n)=a tilde f(motimes n)$ and
$tilde f(motimes na)=tilde f(motimes n)a$ by the definition of $tilde f$.
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
One needs $f(am,n)=af(m,n)$ and $f(m,na)=f(m,n)a$. Then those imply
$tilde f(amotimes n)=a tilde f(motimes n)$ and
$tilde f(motimes na)=tilde f(motimes n)a$ by the definition of $tilde f$.
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
add a comment |
$begingroup$
One needs $f(am,n)=af(m,n)$ and $f(m,na)=f(m,n)a$. Then those imply
$tilde f(amotimes n)=a tilde f(motimes n)$ and
$tilde f(motimes na)=tilde f(motimes n)a$ by the definition of $tilde f$.
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
add a comment |
$begingroup$
One needs $f(am,n)=af(m,n)$ and $f(m,na)=f(m,n)a$. Then those imply
$tilde f(amotimes n)=a tilde f(motimes n)$ and
$tilde f(motimes na)=tilde f(motimes n)a$ by the definition of $tilde f$.
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
One needs $f(am,n)=af(m,n)$ and $f(m,na)=f(m,n)a$. Then those imply
$tilde f(amotimes n)=a tilde f(motimes n)$ and
$tilde f(motimes na)=tilde f(motimes n)a$ by the definition of $tilde f$.
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 8 at 17:26
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
add a comment |
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
$begingroup$
Thanks! That I can understand! I've been following the book "Methods of Representation Theory" by Curtis & Reiner and they seem to state that the biaddictive and balanced properties are enough. Maybe I just interpreted it wrong...
$endgroup$
– user102845
Jan 8 at 17:46
add a comment |
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$begingroup$
Just a comment. The word you're looking for is biadditive. Biaddictive is not a word, but it sounds pretty funny.
$endgroup$
– jgon
Jan 9 at 1:26
$begingroup$
Whoops... I didn't even noticed that I was compulsively writing it wrong (not just in here). Thanks!
$endgroup$
– user102845
Jan 9 at 14:14