Why must an automorphism of an extension of $mathbb{Q}$ send 1 to a rational number?












2












$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







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$endgroup$








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58
















2












$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58














2












2








2


2



$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







share|cite|improve this question









$endgroup$




This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field




field-theory extension-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 23:46









anonanon444anonanon444

1756




1756








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58














  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58








3




3




$begingroup$
Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
$endgroup$
– J. W. Tanner
Jan 10 at 23:54




$begingroup$
Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
$endgroup$
– J. W. Tanner
Jan 10 at 23:54












$begingroup$
$phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
$endgroup$
– reuns
Jan 10 at 23:58




$begingroup$
$phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
$endgroup$
– reuns
Jan 10 at 23:58










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let



$phi:F to F tag 1$



be any field automorphism of the field



$F supset Bbb Q; tag 2$



then



$(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



which implies



$phi(1) in {0, 1 }; tag 4$



now



$phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



which we rule out since $phi$ is a field automorphism; thus we see just why



$phi(1) = 1; tag 6$



it follows that



$n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



and also for $n in N$,



$n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



$z in Bbb Z Longrightarrow phi(z) = z; tag 9$



thus, for



$dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



$s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



whence



$phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



and we have shown that



$forall q in Bbb Q, ; phi(q) = q, tag{13}$



as per request. $OEDelta$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let



      $phi:F to F tag 1$



      be any field automorphism of the field



      $F supset Bbb Q; tag 2$



      then



      $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



      which implies



      $phi(1) in {0, 1 }; tag 4$



      now



      $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



      which we rule out since $phi$ is a field automorphism; thus we see just why



      $phi(1) = 1; tag 6$



      it follows that



      $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



      and also for $n in N$,



      $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



      since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



      $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



      thus, for



      $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



      $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



      whence



      $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



      and we have shown that



      $forall q in Bbb Q, ; phi(q) = q, tag{13}$



      as per request. $OEDelta$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Let



        $phi:F to F tag 1$



        be any field automorphism of the field



        $F supset Bbb Q; tag 2$



        then



        $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



        which implies



        $phi(1) in {0, 1 }; tag 4$



        now



        $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



        which we rule out since $phi$ is a field automorphism; thus we see just why



        $phi(1) = 1; tag 6$



        it follows that



        $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



        and also for $n in N$,



        $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



        since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



        $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



        thus, for



        $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



        $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



        whence



        $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



        and we have shown that



        $forall q in Bbb Q, ; phi(q) = q, tag{13}$



        as per request. $OEDelta$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Let



          $phi:F to F tag 1$



          be any field automorphism of the field



          $F supset Bbb Q; tag 2$



          then



          $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



          which implies



          $phi(1) in {0, 1 }; tag 4$



          now



          $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



          which we rule out since $phi$ is a field automorphism; thus we see just why



          $phi(1) = 1; tag 6$



          it follows that



          $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



          and also for $n in N$,



          $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



          since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



          $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



          thus, for



          $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



          $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



          whence



          $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



          and we have shown that



          $forall q in Bbb Q, ; phi(q) = q, tag{13}$



          as per request. $OEDelta$.






          share|cite|improve this answer











          $endgroup$



          Let



          $phi:F to F tag 1$



          be any field automorphism of the field



          $F supset Bbb Q; tag 2$



          then



          $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



          which implies



          $phi(1) in {0, 1 }; tag 4$



          now



          $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



          which we rule out since $phi$ is a field automorphism; thus we see just why



          $phi(1) = 1; tag 6$



          it follows that



          $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



          and also for $n in N$,



          $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



          since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



          $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



          thus, for



          $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



          $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



          whence



          $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



          and we have shown that



          $forall q in Bbb Q, ; phi(q) = q, tag{13}$



          as per request. $OEDelta$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 3:46

























          answered Jan 11 at 0:35









          Robert LewisRobert Lewis

          47.7k23067




          47.7k23067























              0












              $begingroup$

              I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






                  share|cite|improve this answer









                  $endgroup$



                  I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 23:49









                  ArthurArthur

                  117k7116200




                  117k7116200






























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