Constructing injection into injective group
$begingroup$
$newcommand{ZZ}{mathbb{Z}}$
$newcommand{QQ}{mathbb{Q}}$
$newcommand{Hom}{mathrm{Hom}}$
Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which should be an injective object (that is a divisible abelian group) and provide an injection $0 to M hookrightarrow I(M)$.
I formed
$$
I(M) = Hom_{ZZ}(F_ZZ Hom_ZZ(M, QQ/ZZ), QQ/ZZ)
$$
where $F_ZZ Hom_ZZ(M,QQ/ZZ)$ is the free abelian group over the abelian group $Hom_ZZ(M,QQ/ZZ)$.
The group $I(M)$ is an injective (divisible) group because of the following equation
$$
Hom_ZZ(N^bullet, Hom_ZZ (F_ZZHom_ZZ(M,QQ/ZZ), QQ/ZZ)) =
Hom_ZZ(N^bullet otimes_ZZ F_ZZ(Hom_ZZ(M, QQ/ZZ)), QQ/ZZ)
$$
where $N^bullet$ is the exact sequence $0 to N' to N to N'' to 0$.
Now tensoring with a free group is exact and $QQ/ZZ$ is an injective group.
So $Hom_ZZ(N^bullet, I(M))$ preserves exactness, especially on $0 to N' to N$ and so $I(M)$ is an injective group.
Now the map $M to I(M)$ is given as
$$
(*) quad m mapsto ((sum_i n_i phi_i) mapsto sum_i n_i phi_i(m))
$$
where $sum_i n_i phi_i$ is an element of $F_ZZ Hom_ZZ(M, QQ/ZZ)$.
Now for every $m in M$ with $m neq 0$ one can construct a map $phi_m:M to QQ/ZZ$ with $phi_m(m) neq 0$ (because $QQ/ZZ$ is an injective group).
So the map $(*)$ is injective by choosing for given $m in M$ the map $phi_m$ as $sum_i n_i phi_i$ in $(*)$.
Two questions
1) Is this proof valid? It seems so simple that it is somewhat embarassing to ask such a question, but:
going through all my commutative algebra books I could not find this proof, although it seems obvious and simple and has the advantage of being functorial in $M$. So
2) Is there a place in the literature where a proof of enough injective objects in the abelian groups (or for modules over a commutative ring) is done with this idea (especially forming a free group with the intent of moving it to the left and tensor with it)?
reference-request commutative-algebra abelian-groups injective-module
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
$endgroup$
add a comment |
$begingroup$
$newcommand{ZZ}{mathbb{Z}}$
$newcommand{QQ}{mathbb{Q}}$
$newcommand{Hom}{mathrm{Hom}}$
Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which should be an injective object (that is a divisible abelian group) and provide an injection $0 to M hookrightarrow I(M)$.
I formed
$$
I(M) = Hom_{ZZ}(F_ZZ Hom_ZZ(M, QQ/ZZ), QQ/ZZ)
$$
where $F_ZZ Hom_ZZ(M,QQ/ZZ)$ is the free abelian group over the abelian group $Hom_ZZ(M,QQ/ZZ)$.
The group $I(M)$ is an injective (divisible) group because of the following equation
$$
Hom_ZZ(N^bullet, Hom_ZZ (F_ZZHom_ZZ(M,QQ/ZZ), QQ/ZZ)) =
Hom_ZZ(N^bullet otimes_ZZ F_ZZ(Hom_ZZ(M, QQ/ZZ)), QQ/ZZ)
$$
where $N^bullet$ is the exact sequence $0 to N' to N to N'' to 0$.
Now tensoring with a free group is exact and $QQ/ZZ$ is an injective group.
So $Hom_ZZ(N^bullet, I(M))$ preserves exactness, especially on $0 to N' to N$ and so $I(M)$ is an injective group.
Now the map $M to I(M)$ is given as
$$
(*) quad m mapsto ((sum_i n_i phi_i) mapsto sum_i n_i phi_i(m))
$$
where $sum_i n_i phi_i$ is an element of $F_ZZ Hom_ZZ(M, QQ/ZZ)$.
Now for every $m in M$ with $m neq 0$ one can construct a map $phi_m:M to QQ/ZZ$ with $phi_m(m) neq 0$ (because $QQ/ZZ$ is an injective group).
So the map $(*)$ is injective by choosing for given $m in M$ the map $phi_m$ as $sum_i n_i phi_i$ in $(*)$.
Two questions
1) Is this proof valid? It seems so simple that it is somewhat embarassing to ask such a question, but:
going through all my commutative algebra books I could not find this proof, although it seems obvious and simple and has the advantage of being functorial in $M$. So
2) Is there a place in the literature where a proof of enough injective objects in the abelian groups (or for modules over a commutative ring) is done with this idea (especially forming a free group with the intent of moving it to the left and tensor with it)?
reference-request commutative-algebra abelian-groups injective-module
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
$endgroup$
add a comment |
$begingroup$
$newcommand{ZZ}{mathbb{Z}}$
$newcommand{QQ}{mathbb{Q}}$
$newcommand{Hom}{mathrm{Hom}}$
Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which should be an injective object (that is a divisible abelian group) and provide an injection $0 to M hookrightarrow I(M)$.
I formed
$$
I(M) = Hom_{ZZ}(F_ZZ Hom_ZZ(M, QQ/ZZ), QQ/ZZ)
$$
where $F_ZZ Hom_ZZ(M,QQ/ZZ)$ is the free abelian group over the abelian group $Hom_ZZ(M,QQ/ZZ)$.
The group $I(M)$ is an injective (divisible) group because of the following equation
$$
Hom_ZZ(N^bullet, Hom_ZZ (F_ZZHom_ZZ(M,QQ/ZZ), QQ/ZZ)) =
Hom_ZZ(N^bullet otimes_ZZ F_ZZ(Hom_ZZ(M, QQ/ZZ)), QQ/ZZ)
$$
where $N^bullet$ is the exact sequence $0 to N' to N to N'' to 0$.
Now tensoring with a free group is exact and $QQ/ZZ$ is an injective group.
So $Hom_ZZ(N^bullet, I(M))$ preserves exactness, especially on $0 to N' to N$ and so $I(M)$ is an injective group.
Now the map $M to I(M)$ is given as
$$
(*) quad m mapsto ((sum_i n_i phi_i) mapsto sum_i n_i phi_i(m))
$$
where $sum_i n_i phi_i$ is an element of $F_ZZ Hom_ZZ(M, QQ/ZZ)$.
Now for every $m in M$ with $m neq 0$ one can construct a map $phi_m:M to QQ/ZZ$ with $phi_m(m) neq 0$ (because $QQ/ZZ$ is an injective group).
So the map $(*)$ is injective by choosing for given $m in M$ the map $phi_m$ as $sum_i n_i phi_i$ in $(*)$.
Two questions
1) Is this proof valid? It seems so simple that it is somewhat embarassing to ask such a question, but:
going through all my commutative algebra books I could not find this proof, although it seems obvious and simple and has the advantage of being functorial in $M$. So
2) Is there a place in the literature where a proof of enough injective objects in the abelian groups (or for modules over a commutative ring) is done with this idea (especially forming a free group with the intent of moving it to the left and tensor with it)?
reference-request commutative-algebra abelian-groups injective-module
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
$endgroup$
$newcommand{ZZ}{mathbb{Z}}$
$newcommand{QQ}{mathbb{Q}}$
$newcommand{Hom}{mathrm{Hom}}$
Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which should be an injective object (that is a divisible abelian group) and provide an injection $0 to M hookrightarrow I(M)$.
I formed
$$
I(M) = Hom_{ZZ}(F_ZZ Hom_ZZ(M, QQ/ZZ), QQ/ZZ)
$$
where $F_ZZ Hom_ZZ(M,QQ/ZZ)$ is the free abelian group over the abelian group $Hom_ZZ(M,QQ/ZZ)$.
The group $I(M)$ is an injective (divisible) group because of the following equation
$$
Hom_ZZ(N^bullet, Hom_ZZ (F_ZZHom_ZZ(M,QQ/ZZ), QQ/ZZ)) =
Hom_ZZ(N^bullet otimes_ZZ F_ZZ(Hom_ZZ(M, QQ/ZZ)), QQ/ZZ)
$$
where $N^bullet$ is the exact sequence $0 to N' to N to N'' to 0$.
Now tensoring with a free group is exact and $QQ/ZZ$ is an injective group.
So $Hom_ZZ(N^bullet, I(M))$ preserves exactness, especially on $0 to N' to N$ and so $I(M)$ is an injective group.
Now the map $M to I(M)$ is given as
$$
(*) quad m mapsto ((sum_i n_i phi_i) mapsto sum_i n_i phi_i(m))
$$
where $sum_i n_i phi_i$ is an element of $F_ZZ Hom_ZZ(M, QQ/ZZ)$.
Now for every $m in M$ with $m neq 0$ one can construct a map $phi_m:M to QQ/ZZ$ with $phi_m(m) neq 0$ (because $QQ/ZZ$ is an injective group).
So the map $(*)$ is injective by choosing for given $m in M$ the map $phi_m$ as $sum_i n_i phi_i$ in $(*)$.
Two questions
1) Is this proof valid? It seems so simple that it is somewhat embarassing to ask such a question, but:
going through all my commutative algebra books I could not find this proof, although it seems obvious and simple and has the advantage of being functorial in $M$. So
2) Is there a place in the literature where a proof of enough injective objects in the abelian groups (or for modules over a commutative ring) is done with this idea (especially forming a free group with the intent of moving it to the left and tensor with it)?
reference-request commutative-algebra abelian-groups injective-module
reference-request commutative-algebra abelian-groups injective-module
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
asked Dec 11 '18 at 11:40
Jürgen BöhmJürgen Böhm
2,303512
2,303512
add a comment |
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