How can we show that $mathbb Q$ is not a free $mathbb Z$-module?

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.

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

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 |

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.

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

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 |

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.

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

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

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

edited Aug 9 '18 at 23:12

Al Jebr

4,18343176

asked Apr 8 '12 at 15:57

Emily

426513

asked Apr 8 '12 at 15:57

Emily

426513

asked Apr 8 '12 at 15:57

Emily

426513

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

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).

– David Wheeler
Apr 8 '12 at 16:09

add a comment |

3 Answers
3

active

oldest

votes

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 Magidin

262k34586906

add a comment |

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.

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

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 |

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 $mathbb{Z}$.

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

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 |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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 $mathbb{Q}$ cannot be free.

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

add a comment |

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 $mathbb{Q}$ cannot be free.

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

add a comment |

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 $mathbb{Q}$ cannot be free.

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

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 $mathbb{Q}$ cannot be free.

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

answered Apr 8 '12 at 17:11

Arturo Magidin

262k34586906

add a comment |

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

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 |

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

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 |

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

answered Apr 8 '12 at 22:13

Michael Hardy

answered Apr 8 '12 at 22:13

Michael Hardy

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

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.

– Michael Hardy
Apr 8 '12 at 22:15

add a comment |

It follows from the definition of free modules.

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

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 |

It follows from the definition of free modules.

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

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 |

It follows from the definition of free modules.

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

375

It follows from the definition of free modules.

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

375

edited Jul 27 '18 at 11:25

answered Jul 26 '18 at 9:14

eyp

375

answered Jul 26 '18 at 9:14

eyp

375

answered Jul 26 '18 at 9:14

eyp

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

If the only homomorphism $mathbb{Q}to mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence.

– Arnaud D.
Jul 26 '18 at 9:31

@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.

– eyp
Jul 27 '18 at 5:08

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.

– Arnaud D.
Jul 27 '18 at 9:46

@ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.

– eyp
Jul 27 '18 at 11:27

add a comment |

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