If $Acong B$, then $Aotimes Ccong Botimes C$.
up vote
6
down vote
favorite
I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and
$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?
abstract-algebra modules
add a comment |
up vote
6
down vote
favorite
I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and
$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?
abstract-algebra modules
1
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
3
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and
$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?
abstract-algebra modules
I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and
$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?
abstract-algebra modules
abstract-algebra modules
edited Nov 29 at 13:56
asked Nov 22 at 14:48
Xavier Yang
476314
476314
1
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
3
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31
add a comment |
1
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
3
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31
1
1
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
3
3
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$
but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.
So if we take the exact sequence
$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$
where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are
$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$
So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.
add a comment |
up vote
0
down vote
$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
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
6
down vote
accepted
$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$
but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.
So if we take the exact sequence
$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$
where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are
$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$
So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.
add a comment |
up vote
6
down vote
accepted
$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$
but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.
So if we take the exact sequence
$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$
where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are
$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$
So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$
but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.
So if we take the exact sequence
$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$
where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are
$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$
So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.
$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$
but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.
So if we take the exact sequence
$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$
where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are
$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$
So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.
answered Nov 22 at 16:20
Trevor Gunn
14k32046
14k32046
add a comment |
add a comment |
up vote
0
down vote
$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
add a comment |
up vote
0
down vote
$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
add a comment |
up vote
0
down vote
up vote
0
down vote
$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.
$2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.
answered Nov 22 at 15:54
Joel Pereira
47319
47319
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
add a comment |
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
Yes, this is clear.
– Xavier Yang
Nov 22 at 17:16
add a comment |
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1
@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53
Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54
@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56
3
Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01
You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31