Mixing
Dual of module isomorphic to module itself?
$begingroup$
If $A$ is a finitely generated $B$-module, is it true that dual of $A$ is isomorphic to $A$,i.e.,$Hom(A,B)cong A$?
I guess given $f in Hom(A,B)$, if $f(a)=1$, then I can identify $f$ to $a$. Am I right?
modules duality-theorems
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
$endgroup$
add a comment |
$begingroup$
If $A$ is a finitely generated $B$-module, is it true that dual of $A$ is isomorphic to $A$,i.e.,$Hom(A,B)cong A$?
I guess given $f in Hom(A,B)$, if $f(a)=1$, then I can identify $f$ to $a$. Am I right?
modules duality-theorems
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
$endgroup$
3
$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
1
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36
add a comment |
$begingroup$
If $A$ is a finitely generated $B$-module, is it true that dual of $A$ is isomorphic to $A$,i.e.,$Hom(A,B)cong A$?
I guess given $f in Hom(A,B)$, if $f(a)=1$, then I can identify $f$ to $a$. Am I right?
modules duality-theorems
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
$endgroup$
If $A$ is a finitely generated $B$-module, is it true that dual of $A$ is isomorphic to $A$,i.e.,$Hom(A,B)cong A$?
I guess given $f in Hom(A,B)$, if $f(a)=1$, then I can identify $f$ to $a$. Am I right?
modules duality-theorems
modules duality-theorems
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
edited Feb 1 at 11:20
Yuyi Zhang
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
asked Feb 1 at 9:54
Yuyi ZhangYuyi Zhang
17117
17117
3
$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
1
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36
add a comment |
3
$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
1
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36
3
3
$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
1
1
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is false in general. Indeed, take the $mathbb Z$ module $mathbb Z/nmathbb Z$ for $n geq 2 $. While $|mathbb Z/n mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $mathbb Z$-linear map) $mathbb Z/n mathbb Z longrightarrow mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.
Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a in mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M longrightarrow M'$ which vanishes on a submodule $N subseteq M$ factors uniquely through $M/N$.
The result is, however, true for free modules of finite rank by the same proof as the vector space case.
answered Feb 1 at 11:34
user571438user571438
1,0418
$endgroup$
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
add a comment |
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$begingroup$
This is false in general. Indeed, take the $mathbb Z$ module $mathbb Z/nmathbb Z$ for $n geq 2 $. While $|mathbb Z/n mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $mathbb Z$-linear map) $mathbb Z/n mathbb Z longrightarrow mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.
Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a in mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M longrightarrow M'$ which vanishes on a submodule $N subseteq M$ factors uniquely through $M/N$.
The result is, however, true for free modules of finite rank by the same proof as the vector space case.
answered Feb 1 at 11:34
user571438user571438
1,0418
$endgroup$
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
add a comment |
$begingroup$
This is false in general. Indeed, take the $mathbb Z$ module $mathbb Z/nmathbb Z$ for $n geq 2 $. While $|mathbb Z/n mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $mathbb Z$-linear map) $mathbb Z/n mathbb Z longrightarrow mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.
Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a in mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M longrightarrow M'$ which vanishes on a submodule $N subseteq M$ factors uniquely through $M/N$.
The result is, however, true for free modules of finite rank by the same proof as the vector space case.
answered Feb 1 at 11:34
user571438user571438
1,0418
$endgroup$
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
add a comment |
$begingroup$
This is false in general. Indeed, take the $mathbb Z$ module $mathbb Z/nmathbb Z$ for $n geq 2 $. While $|mathbb Z/n mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $mathbb Z$-linear map) $mathbb Z/n mathbb Z longrightarrow mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.
Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a in mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M longrightarrow M'$ which vanishes on a submodule $N subseteq M$ factors uniquely through $M/N$.
The result is, however, true for free modules of finite rank by the same proof as the vector space case.
answered Feb 1 at 11:34
user571438user571438
1,0418
$endgroup$
This is false in general. Indeed, take the $mathbb Z$ module $mathbb Z/nmathbb Z$ for $n geq 2 $. While $|mathbb Z/n mathbb Z| = n$, one can show that there is exactly one group homomorphism (i.e. $mathbb Z$-linear map) $mathbb Z/n mathbb Z longrightarrow mathbb Z$, which is the zero morphism. Thus, these modules are certainly not isomorphic.
Indeed, such a map is determined entirely by its image on 1 and exists if and only if 1 is sent to an element $a in mathbb Z$ such that $na = 0$, and the only such $a$ is $0$. This is because of the universal property of quotients of $R$ modules, which states that an $R$ linear map $M longrightarrow M'$ which vanishes on a submodule $N subseteq M$ factors uniquely through $M/N$.
The result is, however, true for free modules of finite rank by the same proof as the vector space case.
answered Feb 1 at 11:34
user571438user571438
1,0418
answered Feb 1 at 11:34
user571438user571438
1,0418
answered Feb 1 at 11:34
user571438user571438
1,0418
answered Feb 1 at 11:34
user571438user571438
1,0418
1,0418
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
add a comment |
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
Thank you for your explanation. But how could the the proof of vector spaces be applied to free modules of finite rank? The ring of coefficient may not be division ring, thus free modules may not be identical to vector spaces.
$endgroup$
– Yuyi Zhang
Feb 2 at 7:55
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
The same proof works because the essential element of the proof was the existence of bases, not the divisibility of the coefficients. That was just a mechanism that allowed bases to exist, but bases were the key. Note, however, that not all properties of vector spaces extend to free modules. For example, the quotient of vector spaces is a vector space (hence free), but the quotient of free modules is not (take $mathbb Z /2 mathbb Z $).
$endgroup$
– user571438
Feb 2 at 7:59
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
$begingroup$
I got your point, thanks!
$endgroup$
– Yuyi Zhang
Feb 2 at 8:15
add a comment |
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$begingroup$
It will very rarely happen that such an $f$ exists; e.g. if $B=mathbb{Z}$, and $A$ is a finite abelian group, $hom(A,B) = 0$
$endgroup$
– Max
Feb 1 at 9:56
$begingroup$
Thank you for your example. But now if $A$ is a finitely generated $B$-module , is that statement right?
$endgroup$
– Yuyi Zhang
Feb 1 at 11:22
1
$begingroup$
A finite abelian group is a finitely generated $mathbb Z$ module, as a finitely generated $mathbb Z$ module is precisely a finitely generated abelian group, and finite groups are certainly finitely generated.
$endgroup$
– user571438
Feb 1 at 11:36