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Finding a basis of $S otimes_R N$ with $N$ a free $R$-module

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0

$begingroup$

Let $S$ and $R$ be commutative rings with $f: R to S$ a ring morphism. Suppose that $N$ is free $R$-module with basis $B$. I want to understand why $N otimes_R S$ is a free $S$-module where the action is given by $s'cdot (notimes s) = n otimes s's$.

It's easy to see that $B$ generates $N otimes_R S$, but I'm having trouble showing directly that $B$ is linearly independent over $S$. Perhaps there is an easy way to do this via the universal property of tensor products, but I'm unsure. Or perhaps I'm looking at the wrong set altogether.

Any help is appreciated.

abstract-algebra ring-theory tensor-products free-modules

share|cite|improve this question

asked yesterday

CuriousKid7CuriousKid7

1,693717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The basis deduced from $B$ is $;eotimes 1_S,;ein B$.
  $endgroup$
  – Bernard
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bernard Yes, exactly.
  $endgroup$
  – CuriousKid7
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can prove it's a basis by a relevant isomorphism.
  $endgroup$
  – Bernard
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $S$ and $R$ be commutative rings with $f: R to S$ a ring morphism. Suppose that $N$ is free $R$-module with basis $B$. I want to understand why $N otimes_R S$ is a free $S$-module where the action is given by $s'cdot (notimes s) = n otimes s's$.

It's easy to see that $B$ generates $N otimes_R S$, but I'm having trouble showing directly that $B$ is linearly independent over $S$. Perhaps there is an easy way to do this via the universal property of tensor products, but I'm unsure. Or perhaps I'm looking at the wrong set altogether.

Any help is appreciated.

abstract-algebra ring-theory tensor-products free-modules

share|cite|improve this question

asked yesterday

CuriousKid7CuriousKid7

1,693717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The basis deduced from $B$ is $;eotimes 1_S,;ein B$.
  $endgroup$
  – Bernard
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bernard Yes, exactly.
  $endgroup$
  – CuriousKid7
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can prove it's a basis by a relevant isomorphism.
  $endgroup$
  – Bernard
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let $S$ and $R$ be commutative rings with $f: R to S$ a ring morphism. Suppose that $N$ is free $R$-module with basis $B$. I want to understand why $N otimes_R S$ is a free $S$-module where the action is given by $s'cdot (notimes s) = n otimes s's$.

It's easy to see that $B$ generates $N otimes_R S$, but I'm having trouble showing directly that $B$ is linearly independent over $S$. Perhaps there is an easy way to do this via the universal property of tensor products, but I'm unsure. Or perhaps I'm looking at the wrong set altogether.

Any help is appreciated.

abstract-algebra ring-theory tensor-products free-modules

share|cite|improve this question

asked yesterday

CuriousKid7CuriousKid7

1,693717

$endgroup$

share|cite|improve this question

asked yesterday

CuriousKid7CuriousKid7

1,693717

share|cite|improve this question

asked yesterday

CuriousKid7CuriousKid7

1,693717

asked yesterday

CuriousKid7CuriousKid7

1,693717

asked yesterday

CuriousKid7CuriousKid7

1,693717

1 Answer
1

active

oldest

votes

1

$begingroup$

Here are a couple possible approaches: pick your favorite.

• Show that $bigoplus_b in B S$ satisfies the universal property of $N otimes_R S$, using the $R$-module homomorphism $N to bigoplus_b in B S$, $sum_b in B lambda_b b mapsto (lambda_b)_b in B$.


• Use the universal property of $N otimes_R S$ to conclude that for any $R$-linear functional $phi : N to R$, there exists a unique $S$-linear functional $tildephi : N otimes_R S to S$ such that $tildephi(x otimes lambda) = lambda cdot phi(x)$ for $lambda in S$, $x in N$. Now, if you have a basis $B$ for $N$, and $sum_bin B b otimes lambda_b = 0$, then apply this with the dual elements of $N^* = operatornameHom_R(N, R)$ to conclude that $lambda_b = 0$ for all $b$.

share|cite|improve this answer

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the first approach, to show that $bigoplus_b S$ is a universal object, don't we need a bilinear map $N times S to bigoplus_b S$?
  $endgroup$
  – CuriousKid7
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I was thinking of $- otimes_R S$ as being the scalar extension functor, and as such as being adjoint to the scalar restriction functor. (Which is in fact equivalent to the definition as a tensor product.) The corresponding $R$-bilinear map $N times S to bigoplus_bin B S$ would be $(sum_bin B lambda_b b, mu) mapsto (mu lambda_b)_bin B$ (where $lambda_b in R$, $mu in S$).
  $endgroup$
  – Daniel Schepler
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In case you might not be familiar with adjoint functors: what I mean is that I view $N otimes_R S$ as the $S$-module with an $R$-module homomorphism $N to N otimes_R S$, such that for any $S$-module $M$ and an $R$-module homomorphism $N to M$, there is a unique $S$-module homomorphism $N otimes_R S to M$ such that the composition $N to N otimes_R S to M$ results in the given map $N to M$. (This $S$-module is then unique up to unique isomorphism.)
  $endgroup$
  – Daniel Schepler
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm actually confused now. What do you mean by $lambda cdot phi(x)$?
  $endgroup$
  – CuriousKid7
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means that $f : R to S$ makes $S$ into an $R$-algebra, so you can multiply $lambda in S$ and $phi(x) in R$ to get an element of $S$ (technically, you would get $lambda f(phi(x))$).
  $endgroup$
  – Daniel Schepler
  14 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

Here are a couple possible approaches: pick your favorite.

• Show that $bigoplus_b in B S$ satisfies the universal property of $N otimes_R S$, using the $R$-module homomorphism $N to bigoplus_b in B S$, $sum_b in B lambda_b b mapsto (lambda_b)_b in B$.


• Use the universal property of $N otimes_R S$ to conclude that for any $R$-linear functional $phi : N to R$, there exists a unique $S$-linear functional $tildephi : N otimes_R S to S$ such that $tildephi(x otimes lambda) = lambda cdot phi(x)$ for $lambda in S$, $x in N$. Now, if you have a basis $B$ for $N$, and $sum_bin B b otimes lambda_b = 0$, then apply this with the dual elements of $N^* = operatornameHom_R(N, R)$ to conclude that $lambda_b = 0$ for all $b$.

share|cite|improve this answer

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the first approach, to show that $bigoplus_b S$ is a universal object, don't we need a bilinear map $N times S to bigoplus_b S$?
  $endgroup$
  – CuriousKid7
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I was thinking of $- otimes_R S$ as being the scalar extension functor, and as such as being adjoint to the scalar restriction functor. (Which is in fact equivalent to the definition as a tensor product.) The corresponding $R$-bilinear map $N times S to bigoplus_bin B S$ would be $(sum_bin B lambda_b b, mu) mapsto (mu lambda_b)_bin B$ (where $lambda_b in R$, $mu in S$).
  $endgroup$
  – Daniel Schepler
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In case you might not be familiar with adjoint functors: what I mean is that I view $N otimes_R S$ as the $S$-module with an $R$-module homomorphism $N to N otimes_R S$, such that for any $S$-module $M$ and an $R$-module homomorphism $N to M$, there is a unique $S$-module homomorphism $N otimes_R S to M$ such that the composition $N to N otimes_R S to M$ results in the given map $N to M$. (This $S$-module is then unique up to unique isomorphism.)
  $endgroup$
  – Daniel Schepler
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm actually confused now. What do you mean by $lambda cdot phi(x)$?
  $endgroup$
  – CuriousKid7
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means that $f : R to S$ makes $S$ into an $R$-algebra, so you can multiply $lambda in S$ and $phi(x) in R$ to get an element of $S$ (technically, you would get $lambda f(phi(x))$).
  $endgroup$
  – Daniel Schepler
  14 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Here are a couple possible approaches: pick your favorite.

• Show that $bigoplus_b in B S$ satisfies the universal property of $N otimes_R S$, using the $R$-module homomorphism $N to bigoplus_b in B S$, $sum_b in B lambda_b b mapsto (lambda_b)_b in B$.


• Use the universal property of $N otimes_R S$ to conclude that for any $R$-linear functional $phi : N to R$, there exists a unique $S$-linear functional $tildephi : N otimes_R S to S$ such that $tildephi(x otimes lambda) = lambda cdot phi(x)$ for $lambda in S$, $x in N$. Now, if you have a basis $B$ for $N$, and $sum_bin B b otimes lambda_b = 0$, then apply this with the dual elements of $N^* = operatornameHom_R(N, R)$ to conclude that $lambda_b = 0$ for all $b$.

share|cite|improve this answer

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the first approach, to show that $bigoplus_b S$ is a universal object, don't we need a bilinear map $N times S to bigoplus_b S$?
  $endgroup$
  – CuriousKid7
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I was thinking of $- otimes_R S$ as being the scalar extension functor, and as such as being adjoint to the scalar restriction functor. (Which is in fact equivalent to the definition as a tensor product.) The corresponding $R$-bilinear map $N times S to bigoplus_bin B S$ would be $(sum_bin B lambda_b b, mu) mapsto (mu lambda_b)_bin B$ (where $lambda_b in R$, $mu in S$).
  $endgroup$
  – Daniel Schepler
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In case you might not be familiar with adjoint functors: what I mean is that I view $N otimes_R S$ as the $S$-module with an $R$-module homomorphism $N to N otimes_R S$, such that for any $S$-module $M$ and an $R$-module homomorphism $N to M$, there is a unique $S$-module homomorphism $N otimes_R S to M$ such that the composition $N to N otimes_R S to M$ results in the given map $N to M$. (This $S$-module is then unique up to unique isomorphism.)
  $endgroup$
  – Daniel Schepler
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm actually confused now. What do you mean by $lambda cdot phi(x)$?
  $endgroup$
  – CuriousKid7
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means that $f : R to S$ makes $S$ into an $R$-algebra, so you can multiply $lambda in S$ and $phi(x) in R$ to get an element of $S$ (technically, you would get $lambda f(phi(x))$).
  $endgroup$
  – Daniel Schepler
  14 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Here are a couple possible approaches: pick your favorite.

• Show that $bigoplus_b in B S$ satisfies the universal property of $N otimes_R S$, using the $R$-module homomorphism $N to bigoplus_b in B S$, $sum_b in B lambda_b b mapsto (lambda_b)_b in B$.


• Use the universal property of $N otimes_R S$ to conclude that for any $R$-linear functional $phi : N to R$, there exists a unique $S$-linear functional $tildephi : N otimes_R S to S$ such that $tildephi(x otimes lambda) = lambda cdot phi(x)$ for $lambda in S$, $x in N$. Now, if you have a basis $B$ for $N$, and $sum_bin B b otimes lambda_b = 0$, then apply this with the dual elements of $N^* = operatornameHom_R(N, R)$ to conclude that $lambda_b = 0$ for all $b$.

share|cite|improve this answer

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

$endgroup$

share|cite|improve this answer

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721

answered yesterday

Daniel ScheplerDaniel Schepler

9,0641721