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Question About Tensor Products over Different Base Rings

1

$begingroup$

I can't prove the following remark which appears in Matsumura's 'Commutative Ring Theory':

If $B$ is an $A$-algebra and $M$ and $N$ are $B$-modules, $Motimes_BN$ is the quotient of $Motimes_AN$ by the submodule generated by elements of the form $bxotimes_Ay-xotimes_Aby$.

I think I was able to do it using the universal property (in a completely straightforward fashion), but I am suspicious of this as I couldn't do it by proving directly that $ker(Motimes_ANrightarrow Motimes_BN)$ is generated by these elements.

abstract-algebra ring-theory commutative-algebra tensor-products

share|cite|improve this question

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  One can prove directly that the kernel is what is given by directly using the definition of tensor product. We have $M otimes_A N$ is the free $A$-module on $M times N$ with some relations $R$ whereas $M otimes_B N$ is the free $B$-module on $M times N$ with some relations $Q$. One shows that one can get all relations $Q$ by using the relations from $R$ and the relations $(bx otimes_A y - x otimes_A by$. This is not very precise but I hope it leads you in the right direction.
  $endgroup$
  – equin
  Mar 25 at 6:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, this works, although I was hoping to avoid a particular construction.
  $endgroup$
  – Jehu314
  Mar 25 at 6:55
  

  
  
  
  

add a comment | 

1

$begingroup$

I can't prove the following remark which appears in Matsumura's 'Commutative Ring Theory':

If $B$ is an $A$-algebra and $M$ and $N$ are $B$-modules, $Motimes_BN$ is the quotient of $Motimes_AN$ by the submodule generated by elements of the form $bxotimes_Ay-xotimes_Aby$.

I think I was able to do it using the universal property (in a completely straightforward fashion), but I am suspicious of this as I couldn't do it by proving directly that $ker(Motimes_ANrightarrow Motimes_BN)$ is generated by these elements.

abstract-algebra ring-theory commutative-algebra tensor-products

share|cite|improve this question

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  One can prove directly that the kernel is what is given by directly using the definition of tensor product. We have $M otimes_A N$ is the free $A$-module on $M times N$ with some relations $R$ whereas $M otimes_B N$ is the free $B$-module on $M times N$ with some relations $Q$. One shows that one can get all relations $Q$ by using the relations from $R$ and the relations $(bx otimes_A y - x otimes_A by$. This is not very precise but I hope it leads you in the right direction.
  $endgroup$
  – equin
  Mar 25 at 6:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, this works, although I was hoping to avoid a particular construction.
  $endgroup$
  – Jehu314
  Mar 25 at 6:55
  

  
  
  
  

add a comment | 

$begingroup$

I can't prove the following remark which appears in Matsumura's 'Commutative Ring Theory':

If $B$ is an $A$-algebra and $M$ and $N$ are $B$-modules, $Motimes_BN$ is the quotient of $Motimes_AN$ by the submodule generated by elements of the form $bxotimes_Ay-xotimes_Aby$.

I think I was able to do it using the universal property (in a completely straightforward fashion), but I am suspicious of this as I couldn't do it by proving directly that $ker(Motimes_ANrightarrow Motimes_BN)$ is generated by these elements.

abstract-algebra ring-theory commutative-algebra tensor-products

share|cite|improve this question

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

$endgroup$

share|cite|improve this question

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

share|cite|improve this question

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

edited Mar 27 at 8:20

Eric Wofsey

193k14221352

asked Mar 25 at 6:37

Jehu314Jehu314

1549

asked Mar 25 at 6:37

Jehu314Jehu314

1549

1 Answer
1

active

oldest

votes

4

$begingroup$

Using the universal property for $otimes_A$ one gets a map
$$ Motimes_ANto Motimes_BN, quad motimes_A nmapsto motimes_B n. $$
This sends the submodule $L$ generated by all $mbotimes_A n-motimes_A bn$ to zero, so we get the induced map
$$ (Motimes_AN)/Lto Motimes_BN, quad motimes_An+L mapsto motimes_Bn. $$

We can construct the inverse map using the universal property for $otimes_B$. For, the map $$ Mtimes Nto (Motimes_AN)/L, quad (m,n)mapsto motimes_A n+L $$
is $B$-bilinear, so induces the map
$$ Motimes_BNto (Motimes_AN)/L, quad motimes_Bn mapsto motimes_An+L. $$
These two maps are clearly mutually inverse, as required.

share|cite|improve this answer

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

$endgroup$

add a comment | 

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4

$begingroup$

Using the universal property for $otimes_A$ one gets a map
$$ Motimes_ANto Motimes_BN, quad motimes_A nmapsto motimes_B n. $$
This sends the submodule $L$ generated by all $mbotimes_A n-motimes_A bn$ to zero, so we get the induced map
$$ (Motimes_AN)/Lto Motimes_BN, quad motimes_An+L mapsto motimes_Bn. $$

We can construct the inverse map using the universal property for $otimes_B$. For, the map $$ Mtimes Nto (Motimes_AN)/L, quad (m,n)mapsto motimes_A n+L $$
is $B$-bilinear, so induces the map
$$ Motimes_BNto (Motimes_AN)/L, quad motimes_Bn mapsto motimes_An+L. $$
These two maps are clearly mutually inverse, as required.

share|cite|improve this answer

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

$endgroup$

add a comment | 

4

$begingroup$

Using the universal property for $otimes_A$ one gets a map
$$ Motimes_ANto Motimes_BN, quad motimes_A nmapsto motimes_B n. $$
This sends the submodule $L$ generated by all $mbotimes_A n-motimes_A bn$ to zero, so we get the induced map
$$ (Motimes_AN)/Lto Motimes_BN, quad motimes_An+L mapsto motimes_Bn. $$

We can construct the inverse map using the universal property for $otimes_B$. For, the map $$ Mtimes Nto (Motimes_AN)/L, quad (m,n)mapsto motimes_A n+L $$
is $B$-bilinear, so induces the map
$$ Motimes_BNto (Motimes_AN)/L, quad motimes_Bn mapsto motimes_An+L. $$
These two maps are clearly mutually inverse, as required.

share|cite|improve this answer

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

$endgroup$

add a comment | 

$begingroup$

Using the universal property for $otimes_A$ one gets a map
$$ Motimes_ANto Motimes_BN, quad motimes_A nmapsto motimes_B n. $$
This sends the submodule $L$ generated by all $mbotimes_A n-motimes_A bn$ to zero, so we get the induced map
$$ (Motimes_AN)/Lto Motimes_BN, quad motimes_An+L mapsto motimes_Bn. $$

We can construct the inverse map using the universal property for $otimes_B$. For, the map $$ Mtimes Nto (Motimes_AN)/L, quad (m,n)mapsto motimes_A n+L $$
is $B$-bilinear, so induces the map
$$ Motimes_BNto (Motimes_AN)/L, quad motimes_Bn mapsto motimes_An+L. $$
These two maps are clearly mutually inverse, as required.

share|cite|improve this answer

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

$endgroup$

share|cite|improve this answer

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

edited Mar 25 at 20:34

darij grinberg

11.5k33168

edited Mar 25 at 20:34

darij grinberg

11.5k33168

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125

answered Mar 25 at 8:58

Andrew HuberyAndrew Hubery

56125