Embedding $mathbb{Z}$-modules into injective $mathbb{Z}$-modules
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I want to show that any $mathbb{Z}$-module $M$ can be embedded into an injective $mathbb{Z}$-module (I'm in the process of showing this can be done for more general rings, but starting with this case).
My idea is to tensor with $mathbb{Q}$, so define $N=mathbb{Q}otimes M$. This is a $mathbb{Z}$-module via $$n(qotimes m)=qotimes nm=nqotimes m$$ and extending linearly.
It is divisible since $$sum q_im_i=nsum frac{q_i}{n}m_i$$ Now, $mathbb{Z}$ is a PID and for a module over a PID injectivity is equivalent to divisibility, so $N$ is injective, and $M$ embeds via $m rightarrow (0,m)$.
Is this correct?
abstract-algebra modules homological-algebra injective-module
asked Nov 18 at 14:50
probablystuck
27619
add a comment |
up vote
2
down vote
favorite
I want to show that any $mathbb{Z}$-module $M$ can be embedded into an injective $mathbb{Z}$-module (I'm in the process of showing this can be done for more general rings, but starting with this case).
My idea is to tensor with $mathbb{Q}$, so define $N=mathbb{Q}otimes M$. This is a $mathbb{Z}$-module via $$n(qotimes m)=qotimes nm=nqotimes m$$ and extending linearly.
It is divisible since $$sum q_im_i=nsum frac{q_i}{n}m_i$$ Now, $mathbb{Z}$ is a PID and for a module over a PID injectivity is equivalent to divisibility, so $N$ is injective, and $M$ embeds via $m rightarrow (0,m)$.
Is this correct?
abstract-algebra modules homological-algebra injective-module
asked Nov 18 at 14:50
probablystuck
27619
2
The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to show that any $mathbb{Z}$-module $M$ can be embedded into an injective $mathbb{Z}$-module (I'm in the process of showing this can be done for more general rings, but starting with this case).
My idea is to tensor with $mathbb{Q}$, so define $N=mathbb{Q}otimes M$. This is a $mathbb{Z}$-module via $$n(qotimes m)=qotimes nm=nqotimes m$$ and extending linearly.
It is divisible since $$sum q_im_i=nsum frac{q_i}{n}m_i$$ Now, $mathbb{Z}$ is a PID and for a module over a PID injectivity is equivalent to divisibility, so $N$ is injective, and $M$ embeds via $m rightarrow (0,m)$.
Is this correct?
abstract-algebra modules homological-algebra injective-module
asked Nov 18 at 14:50
probablystuck
27619
I want to show that any $mathbb{Z}$-module $M$ can be embedded into an injective $mathbb{Z}$-module (I'm in the process of showing this can be done for more general rings, but starting with this case).
My idea is to tensor with $mathbb{Q}$, so define $N=mathbb{Q}otimes M$. This is a $mathbb{Z}$-module via $$n(qotimes m)=qotimes nm=nqotimes m$$ and extending linearly.
It is divisible since $$sum q_im_i=nsum frac{q_i}{n}m_i$$ Now, $mathbb{Z}$ is a PID and for a module over a PID injectivity is equivalent to divisibility, so $N$ is injective, and $M$ embeds via $m rightarrow (0,m)$.
Is this correct?
abstract-algebra modules homological-algebra injective-module
abstract-algebra modules homological-algebra injective-module
asked Nov 18 at 14:50
probablystuck
27619
asked Nov 18 at 14:50
probablystuck
27619
asked Nov 18 at 14:50
probablystuck
27619
asked Nov 18 at 14:50
probablystuck
27619
asked Nov 18 at 14:50
probablystuck
27619
27619
2
The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56
add a comment |
2
The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56
2
2
The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56
The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56
add a comment |
1 Answer
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No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $mathbb{Z}/nmathbb{Z} otimes mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = mathbb{Z}langle S rangle / R$, where $R subset mathbb{Z}langle S rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = mathbb{Q}langle S rangle / R$ (where you view $R subset mathbb{Z}langle S rangle subset mathbb{Q}langle S rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = mathbb{Z} = mathbb{Z}langle x rangle / 0$, then $I = mathbb{Q}langle x rangle / (0) = mathbb{Q}$. If $G = mathbb{Z}/nmathbb{Z} = mathbb{Z}langle x rangle / (nx)$, then $I = mathbb{Q} langle x rangle / langle nx rangle = mathbb{Q}/nmathbb{Z}$. The result is quite different from $G otimes mathbb{Q}$ (in fact $G otimes mathbb{Q}$ is a quotient of $I$: it's $I / (R otimes mathbb{Q})$).
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $mathbb{Z}/nmathbb{Z} otimes mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = mathbb{Z}langle S rangle / R$, where $R subset mathbb{Z}langle S rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = mathbb{Q}langle S rangle / R$ (where you view $R subset mathbb{Z}langle S rangle subset mathbb{Q}langle S rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = mathbb{Z} = mathbb{Z}langle x rangle / 0$, then $I = mathbb{Q}langle x rangle / (0) = mathbb{Q}$. If $G = mathbb{Z}/nmathbb{Z} = mathbb{Z}langle x rangle / (nx)$, then $I = mathbb{Q} langle x rangle / langle nx rangle = mathbb{Q}/nmathbb{Z}$. The result is quite different from $G otimes mathbb{Q}$ (in fact $G otimes mathbb{Q}$ is a quotient of $I$: it's $I / (R otimes mathbb{Q})$).
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
add a comment |
up vote
2
down vote
accepted
No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $mathbb{Z}/nmathbb{Z} otimes mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = mathbb{Z}langle S rangle / R$, where $R subset mathbb{Z}langle S rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = mathbb{Q}langle S rangle / R$ (where you view $R subset mathbb{Z}langle S rangle subset mathbb{Q}langle S rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = mathbb{Z} = mathbb{Z}langle x rangle / 0$, then $I = mathbb{Q}langle x rangle / (0) = mathbb{Q}$. If $G = mathbb{Z}/nmathbb{Z} = mathbb{Z}langle x rangle / (nx)$, then $I = mathbb{Q} langle x rangle / langle nx rangle = mathbb{Q}/nmathbb{Z}$. The result is quite different from $G otimes mathbb{Q}$ (in fact $G otimes mathbb{Q}$ is a quotient of $I$: it's $I / (R otimes mathbb{Q})$).
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $mathbb{Z}/nmathbb{Z} otimes mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = mathbb{Z}langle S rangle / R$, where $R subset mathbb{Z}langle S rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = mathbb{Q}langle S rangle / R$ (where you view $R subset mathbb{Z}langle S rangle subset mathbb{Q}langle S rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = mathbb{Z} = mathbb{Z}langle x rangle / 0$, then $I = mathbb{Q}langle x rangle / (0) = mathbb{Q}$. If $G = mathbb{Z}/nmathbb{Z} = mathbb{Z}langle x rangle / (nx)$, then $I = mathbb{Q} langle x rangle / langle nx rangle = mathbb{Q}/nmathbb{Z}$. The result is quite different from $G otimes mathbb{Q}$ (in fact $G otimes mathbb{Q}$ is a quotient of $I$: it's $I / (R otimes mathbb{Q})$).
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $mathbb{Z}/nmathbb{Z} otimes mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = mathbb{Z}langle S rangle / R$, where $R subset mathbb{Z}langle S rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = mathbb{Q}langle S rangle / R$ (where you view $R subset mathbb{Z}langle S rangle subset mathbb{Q}langle S rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = mathbb{Z} = mathbb{Z}langle x rangle / 0$, then $I = mathbb{Q}langle x rangle / (0) = mathbb{Q}$. If $G = mathbb{Z}/nmathbb{Z} = mathbb{Z}langle x rangle / (nx)$, then $I = mathbb{Q} langle x rangle / langle nx rangle = mathbb{Q}/nmathbb{Z}$. The result is quite different from $G otimes mathbb{Q}$ (in fact $G otimes mathbb{Q}$ is a quotient of $I$: it's $I / (R otimes mathbb{Q})$).
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
edited Nov 18 at 15:39
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
answered Nov 18 at 15:12
Najib Idrissi
40.6k469136
40.6k469136
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
add a comment |
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
1
1
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
Isn't ${mathbb Q}langle Srangle/R cong {mathbb Z}langle Srangle/Rotimes{mathbb Q}$? I don't see a straightforward construction like this which would e.g. yield ${mathbb Q}/{mathbb Z}$ from ${mathbb Z}/n{mathbb Z}$.
– Hanno
Nov 18 at 15:16
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
@Hanno No, actually $(mathbb{Z}langle S rangle / R) otimes mathbb{Q} = mathbb{Q}langle S rangle / (R otimes mathbb{Q})$. The difference is important. The presentation of $I$ isn't a presentation of $mathbb{Q}$-modules, it's really the quotient of the abelian group $mathbb{Q}langle S rangle$ by the abelian group $R$, the same one that appears in the presentation of $G$. If you apply it to $mathbb{Z}/nmathbb{Z} = mathbb{Z} langle x rangle / (nx)$ then you get $mathbb{Q} / n mathbb{Z} cong mathbb{Q}/mathbb{Z}$.
– Najib Idrissi
Nov 18 at 15:36
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
Ah, I thought you meant you'd take the ${mathbb Q}$-span of $R$. Thanks, the way it's written now it's very clear.
– Hanno
Nov 18 at 15:47
add a comment |
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The problem is that $Mto mathbb{Q}otimes M$ is not necessarily injective. Think of the case $M=mathbb{Z}/nmathbb{Z}$.
– Roland
Nov 18 at 14:56