How can we show that $mathbb Q$ is not a free $mathbb Z$-module?
$begingroup$
I am really confused from the definition.
How do we know that $mathbb Q$ is not a free $mathbb Z$-module?
In class people use it as a trivial fact, but I don't seem to understand.
abstract-algebra modules
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
$endgroup$
add a comment |
$begingroup$
I am really confused from the definition.
How do we know that $mathbb Q$ is not a free $mathbb Z$-module?
In class people use it as a trivial fact, but I don't seem to understand.
abstract-algebra modules
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
$endgroup$
13
$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
$endgroup$
– David Wheeler
Apr 8 '12 at 16:09
add a comment |
$begingroup$
I am really confused from the definition.
How do we know that $mathbb Q$ is not a free $mathbb Z$-module?
In class people use it as a trivial fact, but I don't seem to understand.
abstract-algebra modules
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
$endgroup$
I am really confused from the definition.
How do we know that $mathbb Q$ is not a free $mathbb Z$-module?
In class people use it as a trivial fact, but I don't seem to understand.
abstract-algebra modules
abstract-algebra modules
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
edited Aug 9 '18 at 23:12
Al Jebr
4,25343277
4,25343277
asked Apr 8 '12 at 15:57
EmilyEmily
431513
asked Apr 8 '12 at 15:57
EmilyEmily
431513
asked Apr 8 '12 at 15:57
EmilyEmily
431513
431513
13
$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
$endgroup$
– David Wheeler
Apr 8 '12 at 16:09
add a comment |
13
$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
$endgroup$
– David Wheeler
Apr 8 '12 at 16:09
13
13
$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
$endgroup$
– David Wheeler
Apr 8 '12 at 16:09
$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
$endgroup$
– David Wheeler
Apr 8 '12 at 16:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Any two nonzero rationals are linearly dependent: if $a,binmathbb{Q}$, $aneq 0 neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.
So if $mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $mathbb{Q}$ is not a cyclic $mathbb{Z}$ module (it is divisible, so it is not isomorphic to $mathbb{Z}$, the only infinite cyclic $mathbb{Z}$-module.
So $mathbb{Q}$ cannot be free.
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
add a comment |
$begingroup$
Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=operatorname{lcm}(b,d)$ and write both fractions as $(text{something}/e$). Then
$$
frac a b = frac 1 e + cdots + frac 1 etext{ and }frac c d = frac 1 e + cdots + frac 1 e,
$$
where in general the numbers of terms in the two sums will be different.
Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $mathbb{Q}$ must be generated by just one generator, so $mathbb{Q} = { 0, pm f, pm 2f, pm 3f, ldots }$. But that fails to include the average of $f$ and $2f$, which is rational.
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
$endgroup$
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
add a comment |
$begingroup$
It follows from the definition of free modules.
Let us suppose to the contradictory that $mathbb{Q}$ is a free $mathbb{Z}$ module, so by definition of free modules, for a given injective map $alpha: X rightarrow mathbb{Q}$ and for any map $f : X rightarrow mathbb{Z}$, there exist a unique $mathbb{Z}$-homomorphism $g: mathbb{Q} rightarrow mathbb{Z}$ such that $f=galpha$. Every $mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $mathbb{Q}$ to $mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $mathbb{Z}$ and we don't have any homomorphism from $mathbb{Q}$ to $mathbb{Z}$ corresponding to non-zero maps $f:X rightarrow mathbb{Z}$, thus $mathbb{Q}$ is not a free module over $mathbb{Z}$.
answered Jul 26 '18 at 9:14
eypeyp
375
$endgroup$
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
$endgroup$
– eyp
Jul 27 '18 at 5:08
$begingroup$
That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
$endgroup$
– Arnaud D.
Jul 27 '18 at 9:46
$begingroup$
@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
$endgroup$
– eyp
Jul 27 '18 at 11:27
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any two nonzero rationals are linearly dependent: if $a,binmathbb{Q}$, $aneq 0 neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.
So if $mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $mathbb{Q}$ is not a cyclic $mathbb{Z}$ module (it is divisible, so it is not isomorphic to $mathbb{Z}$, the only infinite cyclic $mathbb{Z}$-module.
So $mathbb{Q}$ cannot be free.
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
add a comment |
$begingroup$
Any two nonzero rationals are linearly dependent: if $a,binmathbb{Q}$, $aneq 0 neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.
So if $mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $mathbb{Q}$ is not a cyclic $mathbb{Z}$ module (it is divisible, so it is not isomorphic to $mathbb{Z}$, the only infinite cyclic $mathbb{Z}$-module.
So $mathbb{Q}$ cannot be free.
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
add a comment |
$begingroup$
Any two nonzero rationals are linearly dependent: if $a,binmathbb{Q}$, $aneq 0 neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.
So if $mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $mathbb{Q}$ is not a cyclic $mathbb{Z}$ module (it is divisible, so it is not isomorphic to $mathbb{Z}$, the only infinite cyclic $mathbb{Z}$-module.
So $mathbb{Q}$ cannot be free.
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
Any two nonzero rationals are linearly dependent: if $a,binmathbb{Q}$, $aneq 0 neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.
So if $mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $mathbb{Q}$ is not a cyclic $mathbb{Z}$ module (it is divisible, so it is not isomorphic to $mathbb{Z}$, the only infinite cyclic $mathbb{Z}$-module.
So $mathbb{Q}$ cannot be free.
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
answered Apr 8 '12 at 17:11
Arturo MagidinArturo Magidin
263k34587911
263k34587911
add a comment |
add a comment |
$begingroup$
Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=operatorname{lcm}(b,d)$ and write both fractions as $(text{something}/e$). Then
$$
frac a b = frac 1 e + cdots + frac 1 etext{ and }frac c d = frac 1 e + cdots + frac 1 e,
$$
where in general the numbers of terms in the two sums will be different.
Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $mathbb{Q}$ must be generated by just one generator, so $mathbb{Q} = { 0, pm f, pm 2f, pm 3f, ldots }$. But that fails to include the average of $f$ and $2f$, which is rational.
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
$endgroup$
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
add a comment |
$begingroup$
Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=operatorname{lcm}(b,d)$ and write both fractions as $(text{something}/e$). Then
$$
frac a b = frac 1 e + cdots + frac 1 etext{ and }frac c d = frac 1 e + cdots + frac 1 e,
$$
where in general the numbers of terms in the two sums will be different.
Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $mathbb{Q}$ must be generated by just one generator, so $mathbb{Q} = { 0, pm f, pm 2f, pm 3f, ldots }$. But that fails to include the average of $f$ and $2f$, which is rational.
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
$endgroup$
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
add a comment |
$begingroup$
Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=operatorname{lcm}(b,d)$ and write both fractions as $(text{something}/e$). Then
$$
frac a b = frac 1 e + cdots + frac 1 etext{ and }frac c d = frac 1 e + cdots + frac 1 e,
$$
where in general the numbers of terms in the two sums will be different.
Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $mathbb{Q}$ must be generated by just one generator, so $mathbb{Q} = { 0, pm f, pm 2f, pm 3f, ldots }$. But that fails to include the average of $f$ and $2f$, which is rational.
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
$endgroup$
Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=operatorname{lcm}(b,d)$ and write both fractions as $(text{something}/e$). Then
$$
frac a b = frac 1 e + cdots + frac 1 etext{ and }frac c d = frac 1 e + cdots + frac 1 e,
$$
where in general the numbers of terms in the two sums will be different.
Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $mathbb{Q}$ must be generated by just one generator, so $mathbb{Q} = { 0, pm f, pm 2f, pm 3f, ldots }$. But that fails to include the average of $f$ and $2f$, which is rational.
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
edited May 2 '13 at 5:09
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
answered Apr 8 '12 at 22:13
Michael HardyMichael Hardy
1
1
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
add a comment |
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
2
2
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
$begingroup$
Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style.
$endgroup$
– Michael Hardy
Apr 8 '12 at 22:15
add a comment |
$begingroup$
It follows from the definition of free modules.
Let us suppose to the contradictory that $mathbb{Q}$ is a free $mathbb{Z}$ module, so by definition of free modules, for a given injective map $alpha: X rightarrow mathbb{Q}$ and for any map $f : X rightarrow mathbb{Z}$, there exist a unique $mathbb{Z}$-homomorphism $g: mathbb{Q} rightarrow mathbb{Z}$ such that $f=galpha$. Every $mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $mathbb{Q}$ to $mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $mathbb{Z}$ and we don't have any homomorphism from $mathbb{Q}$ to $mathbb{Z}$ corresponding to non-zero maps $f:X rightarrow mathbb{Z}$, thus $mathbb{Q}$ is not a free module over $mathbb{Z}$.
answered Jul 26 '18 at 9:14
eypeyp
375
$endgroup$
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
$endgroup$
– eyp
Jul 27 '18 at 5:08
$begingroup$
That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
$endgroup$
– Arnaud D.
Jul 27 '18 at 9:46
$begingroup$
@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
$endgroup$
– eyp
Jul 27 '18 at 11:27
add a comment |
$begingroup$
It follows from the definition of free modules.
Let us suppose to the contradictory that $mathbb{Q}$ is a free $mathbb{Z}$ module, so by definition of free modules, for a given injective map $alpha: X rightarrow mathbb{Q}$ and for any map $f : X rightarrow mathbb{Z}$, there exist a unique $mathbb{Z}$-homomorphism $g: mathbb{Q} rightarrow mathbb{Z}$ such that $f=galpha$. Every $mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $mathbb{Q}$ to $mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $mathbb{Z}$ and we don't have any homomorphism from $mathbb{Q}$ to $mathbb{Z}$ corresponding to non-zero maps $f:X rightarrow mathbb{Z}$, thus $mathbb{Q}$ is not a free module over $mathbb{Z}$.
answered Jul 26 '18 at 9:14
eypeyp
375
$endgroup$
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
$endgroup$
– eyp
Jul 27 '18 at 5:08
$begingroup$
That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
$endgroup$
– Arnaud D.
Jul 27 '18 at 9:46
$begingroup$
@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
$endgroup$
– eyp
Jul 27 '18 at 11:27
add a comment |
$begingroup$
It follows from the definition of free modules.
Let us suppose to the contradictory that $mathbb{Q}$ is a free $mathbb{Z}$ module, so by definition of free modules, for a given injective map $alpha: X rightarrow mathbb{Q}$ and for any map $f : X rightarrow mathbb{Z}$, there exist a unique $mathbb{Z}$-homomorphism $g: mathbb{Q} rightarrow mathbb{Z}$ such that $f=galpha$. Every $mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $mathbb{Q}$ to $mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $mathbb{Z}$ and we don't have any homomorphism from $mathbb{Q}$ to $mathbb{Z}$ corresponding to non-zero maps $f:X rightarrow mathbb{Z}$, thus $mathbb{Q}$ is not a free module over $mathbb{Z}$.
answered Jul 26 '18 at 9:14
eypeyp
375
$endgroup$
It follows from the definition of free modules.
Let us suppose to the contradictory that $mathbb{Q}$ is a free $mathbb{Z}$ module, so by definition of free modules, for a given injective map $alpha: X rightarrow mathbb{Q}$ and for any map $f : X rightarrow mathbb{Z}$, there exist a unique $mathbb{Z}$-homomorphism $g: mathbb{Q} rightarrow mathbb{Z}$ such that $f=galpha$. Every $mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $mathbb{Q}$ to $mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $mathbb{Z}$ and we don't have any homomorphism from $mathbb{Q}$ to $mathbb{Z}$ corresponding to non-zero maps $f:X rightarrow mathbb{Z}$, thus $mathbb{Q}$ is not a free module over $mathbb{Z}$.
answered Jul 26 '18 at 9:14
eypeyp
375
edited Jul 27 '18 at 11:25
answered Jul 26 '18 at 9:14
eypeyp
375
answered Jul 26 '18 at 9:14
eypeyp
375
answered Jul 26 '18 at 9:14
eypeyp
375
375
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
$endgroup$
– eyp
Jul 27 '18 at 5:08
$begingroup$
That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
$endgroup$
– Arnaud D.
Jul 27 '18 at 9:46
$begingroup$
@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
$endgroup$
– eyp
Jul 27 '18 at 11:27
add a comment |
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
$endgroup$
– eyp
Jul 27 '18 at 5:08
$begingroup$
That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
$endgroup$
– Arnaud D.
Jul 27 '18 at 9:46
$begingroup$
@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
$endgroup$
– eyp
Jul 27 '18 at 11:27
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
$begingroup$
If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.
$endgroup$
– Arnaud D.
Jul 26 '18 at 9:31
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@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
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– eyp
Jul 27 '18 at 5:08
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@Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X rightarrow mathbb{Z}$, so if $mathbb{Q}$ is free $mathbb{Z}$ module, then there must be more than one homomorphism from $mathbb{Q}$ to $mathbb{Z}$, which is not the case.
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– eyp
Jul 27 '18 at 5:08
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That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
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– Arnaud D.
Jul 27 '18 at 9:46
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That's what I was saying : the problem is that there are too few homomorphisms $mathbb{Q}to mathbb{Z}$ in comparison with maps $Xto mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $mathbb{Q}to mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique.
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– Arnaud D.
Jul 27 '18 at 9:46
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@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
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– eyp
Jul 27 '18 at 11:27
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@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
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– eyp
Jul 27 '18 at 11:27
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$begingroup$
If $Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $Bbb{Z}$ (so not of rank > 1).
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– David Wheeler
Apr 8 '12 at 16:09