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












24












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















24












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















24












24








24


8



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 9 '18 at 23:12









Al Jebr

4,18343176




4,18343176










asked Apr 8 '12 at 15:57









EmilyEmily

426513




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
















  • 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












3 Answers
3






active

oldest

votes


















27












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    12












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 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



















    1












    $begingroup$

    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 $mathbb{Z}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    27












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      27












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        27












        27








        27





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 8 '12 at 17:11









        Arturo MagidinArturo Magidin

        262k34586906




        262k34586906























            12












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 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
















            12












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 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














            12












            12








            12





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 2 '13 at 5:09

























            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














            • 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











            1












            $begingroup$

            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 $mathbb{Z}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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
















            1












            $begingroup$

            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 $mathbb{Z}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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














            1












            1








            1





            $begingroup$

            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 $mathbb{Z}$.






            share|cite|improve this answer











            $endgroup$



            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 $mathbb{Z}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 27 '18 at 11:25

























            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


















            • $begingroup$
              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












            $begingroup$
            @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




            $begingroup$
            @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




            $begingroup$
            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




            $begingroup$
            @ 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


















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