how to determine End$(mathbb{Q},+)$












4












$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







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$endgroup$








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38
















4












$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38














4












4








4


1



$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







share|cite|improve this question











$endgroup$




I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













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edited Jan 18 at 3:57









user549397

1,5101418




1,5101418










asked Dec 27 '14 at 15:26









litter boylitter boy

261




261








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38














  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38








2




2




$begingroup$
Both are correct, so you just need to prove the claims.
$endgroup$
– Tobias Kildetoft
Dec 27 '14 at 15:38




$begingroup$
Both are correct, so you just need to prove the claims.
$endgroup$
– Tobias Kildetoft
Dec 27 '14 at 15:38










2 Answers
2






active

oldest

votes


















2












$begingroup$

Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
$$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
$$c=underbrace{r+ldots+r}_{ntext{ times}}$$
suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



    On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
      $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
      where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
      $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
      suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
        $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
        where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
        $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
        suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
          $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
          where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
          $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
          suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






          share|cite|improve this answer









          $endgroup$



          Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
          $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
          where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
          $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
          suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '14 at 16:23









          Milo BrandtMilo Brandt

          39.8k476140




          39.8k476140























              1












              $begingroup$

              Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



              On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                  On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






                  share|cite|improve this answer









                  $endgroup$



                  Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                  On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 27 '14 at 16:25









                  KrishKrish

                  6,32411020




                  6,32411020






























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