Is it true that $A otimes_{Bbb Z} B= A otimes _{ R} B$ for every commutative ring $R$?
up vote
1
down vote
favorite
I have been obtaining a weird statement:
Let $R$ be a commutative ring $A,B$ both $R$-modules. Then $A otimes_{Bbb Z} B= A otimes _{ R} B$ as abelian groups.
Proof: We may regard $A otimes _{Bbb Z} B$ as an $R$-module by acting $r(a otimes b) = ra otimes b$. There is thus an $R$-billinear map inducing $$A otimes _R B rightarrow A otimes B$$ where $ a otimes b mapsto a otimes b$. Conversely, there is also a billiner map inducing $$A otimes B rightarrow A otimes _R B$$
where $a otimes b mapsto a otimes b$. They form inverse to one another.
What has gone wrong in the proof?
EDIT: I believe I have found my main problem. So I took the $R$-module structure for $A otimes B$ to be that induced from the map
$$ R times A otimes B rightarrow A otimes B, (r,a otimes b) mapsto ra otimes b $$
But this does not imply $ra otimes b = a otimes rb $. Which is required to induce the map $$A otimes _R B rightarrow A otimes B $$
This is seen in the example given by egreg, that with $Bbb Z$ given the induced qutoient structure of $R= Bbb Z[x]$, in $Bbb Z otimes _{Bbb Z} R cong Bbb Z^{(N)}$
$$ 0 = x 1 otimes 1 not= 1 otimes x $$
Since under the isomorphism, the latter item is $(1) $ in the 1st level of $Bbb Z^{(N)}$.
abstract-algebra ring-theory tensor-products
asked Nov 28 at 8:25
CL.
2,1532822
add a comment |
up vote
1
down vote
favorite
I have been obtaining a weird statement:
Let $R$ be a commutative ring $A,B$ both $R$-modules. Then $A otimes_{Bbb Z} B= A otimes _{ R} B$ as abelian groups.
Proof: We may regard $A otimes _{Bbb Z} B$ as an $R$-module by acting $r(a otimes b) = ra otimes b$. There is thus an $R$-billinear map inducing $$A otimes _R B rightarrow A otimes B$$ where $ a otimes b mapsto a otimes b$. Conversely, there is also a billiner map inducing $$A otimes B rightarrow A otimes _R B$$
where $a otimes b mapsto a otimes b$. They form inverse to one another.
What has gone wrong in the proof?
EDIT: I believe I have found my main problem. So I took the $R$-module structure for $A otimes B$ to be that induced from the map
$$ R times A otimes B rightarrow A otimes B, (r,a otimes b) mapsto ra otimes b $$
But this does not imply $ra otimes b = a otimes rb $. Which is required to induce the map $$A otimes _R B rightarrow A otimes B $$
This is seen in the example given by egreg, that with $Bbb Z$ given the induced qutoient structure of $R= Bbb Z[x]$, in $Bbb Z otimes _{Bbb Z} R cong Bbb Z^{(N)}$
$$ 0 = x 1 otimes 1 not= 1 otimes x $$
Since under the isomorphism, the latter item is $(1) $ in the 1st level of $Bbb Z^{(N)}$.
abstract-algebra ring-theory tensor-products
asked Nov 28 at 8:25
CL.
2,1532822
You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
1
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been obtaining a weird statement:
Let $R$ be a commutative ring $A,B$ both $R$-modules. Then $A otimes_{Bbb Z} B= A otimes _{ R} B$ as abelian groups.
Proof: We may regard $A otimes _{Bbb Z} B$ as an $R$-module by acting $r(a otimes b) = ra otimes b$. There is thus an $R$-billinear map inducing $$A otimes _R B rightarrow A otimes B$$ where $ a otimes b mapsto a otimes b$. Conversely, there is also a billiner map inducing $$A otimes B rightarrow A otimes _R B$$
where $a otimes b mapsto a otimes b$. They form inverse to one another.
What has gone wrong in the proof?
EDIT: I believe I have found my main problem. So I took the $R$-module structure for $A otimes B$ to be that induced from the map
$$ R times A otimes B rightarrow A otimes B, (r,a otimes b) mapsto ra otimes b $$
But this does not imply $ra otimes b = a otimes rb $. Which is required to induce the map $$A otimes _R B rightarrow A otimes B $$
This is seen in the example given by egreg, that with $Bbb Z$ given the induced qutoient structure of $R= Bbb Z[x]$, in $Bbb Z otimes _{Bbb Z} R cong Bbb Z^{(N)}$
$$ 0 = x 1 otimes 1 not= 1 otimes x $$
Since under the isomorphism, the latter item is $(1) $ in the 1st level of $Bbb Z^{(N)}$.
abstract-algebra ring-theory tensor-products
asked Nov 28 at 8:25
CL.
2,1532822
I have been obtaining a weird statement:
Let $R$ be a commutative ring $A,B$ both $R$-modules. Then $A otimes_{Bbb Z} B= A otimes _{ R} B$ as abelian groups.
Proof: We may regard $A otimes _{Bbb Z} B$ as an $R$-module by acting $r(a otimes b) = ra otimes b$. There is thus an $R$-billinear map inducing $$A otimes _R B rightarrow A otimes B$$ where $ a otimes b mapsto a otimes b$. Conversely, there is also a billiner map inducing $$A otimes B rightarrow A otimes _R B$$
where $a otimes b mapsto a otimes b$. They form inverse to one another.
What has gone wrong in the proof?
EDIT: I believe I have found my main problem. So I took the $R$-module structure for $A otimes B$ to be that induced from the map
$$ R times A otimes B rightarrow A otimes B, (r,a otimes b) mapsto ra otimes b $$
But this does not imply $ra otimes b = a otimes rb $. Which is required to induce the map $$A otimes _R B rightarrow A otimes B $$
This is seen in the example given by egreg, that with $Bbb Z$ given the induced qutoient structure of $R= Bbb Z[x]$, in $Bbb Z otimes _{Bbb Z} R cong Bbb Z^{(N)}$
$$ 0 = x 1 otimes 1 not= 1 otimes x $$
Since under the isomorphism, the latter item is $(1) $ in the 1st level of $Bbb Z^{(N)}$.
abstract-algebra ring-theory tensor-products
abstract-algebra ring-theory tensor-products
asked Nov 28 at 8:25
CL.
2,1532822
asked Nov 28 at 8:25
CL.
2,1532822
edited Nov 28 at 10:01
asked Nov 28 at 8:25
CL.
2,1532822
asked Nov 28 at 8:25
CL.
2,1532822
asked Nov 28 at 8:25
CL.
2,1532822
2,1532822
You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
1
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22
add a comment |
You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
1
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22
You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
1
1
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You're doing too many steps implicitly.
In your second step, you want to say that the natural map $Atimes B to A otimes_R B$ is $mathbb{Z}$-bilinear. That's true, and so it gives a map $Aotimes_mathbb{Z} B to Aotimes_R B$.
In your first step, you want to say that the natural map $Atimes B to A otimes_mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a otimes b) = raotimes b$, but that might not equal $aotimes rb$. So there's no universal property to apply that gives us a map $Aotimes_R B to A otimes_mathbb{Z} B$.
answered Nov 28 at 9:31
Slade
24.7k12564
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
add a comment |
up vote
1
down vote
The statement is not true in general.
Consider $R=mathbb{Z}[x]$. If $A$ is any $R$-module, then
$$
Aotimes_R Rcong A
$$
but
$$
Aotimes_{mathbb{Z}}Rcong A^{(mathbb{N})}
$$
(direct sum of a countable number of copies of $A$). Now take $A=mathbb{Z}$, which is an $R$-module because it is $mathbb{Z}[x]/(x)$.
answered Nov 28 at 8:53
egreg
176k1484198
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You're doing too many steps implicitly.
In your second step, you want to say that the natural map $Atimes B to A otimes_R B$ is $mathbb{Z}$-bilinear. That's true, and so it gives a map $Aotimes_mathbb{Z} B to Aotimes_R B$.
In your first step, you want to say that the natural map $Atimes B to A otimes_mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a otimes b) = raotimes b$, but that might not equal $aotimes rb$. So there's no universal property to apply that gives us a map $Aotimes_R B to A otimes_mathbb{Z} B$.
answered Nov 28 at 9:31
Slade
24.7k12564
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
add a comment |
up vote
3
down vote
accepted
You're doing too many steps implicitly.
In your second step, you want to say that the natural map $Atimes B to A otimes_R B$ is $mathbb{Z}$-bilinear. That's true, and so it gives a map $Aotimes_mathbb{Z} B to Aotimes_R B$.
In your first step, you want to say that the natural map $Atimes B to A otimes_mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a otimes b) = raotimes b$, but that might not equal $aotimes rb$. So there's no universal property to apply that gives us a map $Aotimes_R B to A otimes_mathbb{Z} B$.
answered Nov 28 at 9:31
Slade
24.7k12564
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You're doing too many steps implicitly.
In your second step, you want to say that the natural map $Atimes B to A otimes_R B$ is $mathbb{Z}$-bilinear. That's true, and so it gives a map $Aotimes_mathbb{Z} B to Aotimes_R B$.
In your first step, you want to say that the natural map $Atimes B to A otimes_mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a otimes b) = raotimes b$, but that might not equal $aotimes rb$. So there's no universal property to apply that gives us a map $Aotimes_R B to A otimes_mathbb{Z} B$.
answered Nov 28 at 9:31
Slade
24.7k12564
You're doing too many steps implicitly.
In your second step, you want to say that the natural map $Atimes B to A otimes_R B$ is $mathbb{Z}$-bilinear. That's true, and so it gives a map $Aotimes_mathbb{Z} B to Aotimes_R B$.
In your first step, you want to say that the natural map $Atimes B to A otimes_mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a otimes b) = raotimes b$, but that might not equal $aotimes rb$. So there's no universal property to apply that gives us a map $Aotimes_R B to A otimes_mathbb{Z} B$.
answered Nov 28 at 9:31
Slade
24.7k12564
edited Nov 28 at 9:58
answered Nov 28 at 9:31
Slade
24.7k12564
answered Nov 28 at 9:31
Slade
24.7k12564
answered Nov 28 at 9:31
Slade
24.7k12564
24.7k12564
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
add a comment |
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
Thanks Slade. But why do we not have an $R$-module structure on $A otimes _{Bbb Z} B$? It goes as follows: There is a well defined map $$tau_r: A otimes B rightarrow A otimes B$$ $(a otimes b) mapsto ra otimes b$. Thus, there is a well defined map $$R times A otimes B mapsto A otimes B$$ $(r, a otimes b) mapsto tau_r(a otimes b)$. This map gives an $R$-module structure on $A otimes B$. Doesn't this make the map described $R$-bilinear giving $A otimes _R B rightarrow A otimes B$?
– CL.
Nov 28 at 9:37
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
You're right, my answer was a little unclear. The point is that if you define $r(a otimes b) = raotimes b$, you still need to prove that this equals $aotimes rb$ for your map to be $R$-bilinear.
– Slade
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
Thanks Slade, in fact, I have noticed the exact same thing (which I have updated in my post).
– CL.
Nov 28 at 10:01
add a comment |
up vote
1
down vote
The statement is not true in general.
Consider $R=mathbb{Z}[x]$. If $A$ is any $R$-module, then
$$
Aotimes_R Rcong A
$$
but
$$
Aotimes_{mathbb{Z}}Rcong A^{(mathbb{N})}
$$
(direct sum of a countable number of copies of $A$). Now take $A=mathbb{Z}$, which is an $R$-module because it is $mathbb{Z}[x]/(x)$.
answered Nov 28 at 8:53
egreg
176k1484198
add a comment |
up vote
1
down vote
The statement is not true in general.
Consider $R=mathbb{Z}[x]$. If $A$ is any $R$-module, then
$$
Aotimes_R Rcong A
$$
but
$$
Aotimes_{mathbb{Z}}Rcong A^{(mathbb{N})}
$$
(direct sum of a countable number of copies of $A$). Now take $A=mathbb{Z}$, which is an $R$-module because it is $mathbb{Z}[x]/(x)$.
answered Nov 28 at 8:53
egreg
176k1484198
add a comment |
up vote
1
down vote
up vote
1
down vote
The statement is not true in general.
Consider $R=mathbb{Z}[x]$. If $A$ is any $R$-module, then
$$
Aotimes_R Rcong A
$$
but
$$
Aotimes_{mathbb{Z}}Rcong A^{(mathbb{N})}
$$
(direct sum of a countable number of copies of $A$). Now take $A=mathbb{Z}$, which is an $R$-module because it is $mathbb{Z}[x]/(x)$.
answered Nov 28 at 8:53
egreg
176k1484198
The statement is not true in general.
Consider $R=mathbb{Z}[x]$. If $A$ is any $R$-module, then
$$
Aotimes_R Rcong A
$$
but
$$
Aotimes_{mathbb{Z}}Rcong A^{(mathbb{N})}
$$
(direct sum of a countable number of copies of $A$). Now take $A=mathbb{Z}$, which is an $R$-module because it is $mathbb{Z}[x]/(x)$.
answered Nov 28 at 8:53
egreg
176k1484198
edited Nov 28 at 8:59
answered Nov 28 at 8:53
egreg
176k1484198
answered Nov 28 at 8:53
egreg
176k1484198
answered Nov 28 at 8:53
egreg
176k1484198
176k1484198
add a comment |
add a comment |
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You never defined the inverse. And that's what's missing.
– Pedro Tamaroff♦
Nov 28 at 8:40
"Conversely, there is also a bilinear map inducing..." You never define that map in the post.
– Pedro Tamaroff♦
Nov 28 at 9:11
1
The problem with that map, as illustrated by egreg's answer, is it is not well defined.
– Pedro Tamaroff♦
Nov 28 at 9:17
Which I am still confused, why is it not well defined? There is a bilinear map $A times B rightarrow A otimes _R B$, $(a,b) mapsto a otimes b$, then we use Universal Property.
– CL.
Nov 28 at 9:22