Ring Automorphisms that fix 1. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Automorphisms of $mathbb Q(sqrt 2)$Automorphisms of $mathbbR^n$group of automorphisms of the ring $mathbbZtimesmathbbZ$Trying to understand a proof for the automorphisms of a polynomial ringAll automorphisms of splitting fieldsDetermining automorphisms of this extensionRing automorphisms of $mathbbQ[sqrt[3]5]$Automorphism of ring and isomorphism of quotient ringsThe automorphisms of the extension $mathbbQ(sqrt[4]2)/mathbbQ$.Extension theorem for field automorphismsAre all verbal automorphisms inner power automorphisms?

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Ring Automorphisms that fix 1.



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Automorphisms of $mathbb Q(sqrt 2)$Automorphisms of $mathbbR^n$group of automorphisms of the ring $mathbbZtimesmathbbZ$Trying to understand a proof for the automorphisms of a polynomial ringAll automorphisms of splitting fieldsDetermining automorphisms of this extensionRing automorphisms of $mathbbQ[sqrt[3]5]$Automorphism of ring and isomorphism of quotient ringsThe automorphisms of the extension $mathbbQ(sqrt[4]2)/mathbbQ$.Extension theorem for field automorphismsAre all verbal automorphisms inner power automorphisms?










4












$begingroup$


This question is a follow - up to this question about Field Automorphisms of $mathbbQ[sqrt2]$.



Since $mathbbQ[sqrt2]$ is a vector space over $mathbbQ$ with basis $1, sqrt2$, I naively understand why it is the case that automorphisms $phi$ of $mathbbQ[sqrt2]$ are determined wholly by the image of $1$ and $sqrt2$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt2) = sqrt2$, and I want to compute the value of $phi(frac32)$. I can do the following:



$$ phi(frac32) = phi(3) phi(frac12) = [phi(1) + phi(1) + phi(1)] phi(frac12) = 3phi(frac12).$$



I am unsure how to proceed from here. I would assume that it is true that $$phi(frac11 + 1) = fracphi(1)phi(1) + phi(1) = frac12,$$ but I don't know what property of ring isomorphisms would allow me to do this.










share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    This question is a follow - up to this question about Field Automorphisms of $mathbbQ[sqrt2]$.



    Since $mathbbQ[sqrt2]$ is a vector space over $mathbbQ$ with basis $1, sqrt2$, I naively understand why it is the case that automorphisms $phi$ of $mathbbQ[sqrt2]$ are determined wholly by the image of $1$ and $sqrt2$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt2) = sqrt2$, and I want to compute the value of $phi(frac32)$. I can do the following:



    $$ phi(frac32) = phi(3) phi(frac12) = [phi(1) + phi(1) + phi(1)] phi(frac12) = 3phi(frac12).$$



    I am unsure how to proceed from here. I would assume that it is true that $$phi(frac11 + 1) = fracphi(1)phi(1) + phi(1) = frac12,$$ but I don't know what property of ring isomorphisms would allow me to do this.










    share|cite|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      This question is a follow - up to this question about Field Automorphisms of $mathbbQ[sqrt2]$.



      Since $mathbbQ[sqrt2]$ is a vector space over $mathbbQ$ with basis $1, sqrt2$, I naively understand why it is the case that automorphisms $phi$ of $mathbbQ[sqrt2]$ are determined wholly by the image of $1$ and $sqrt2$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt2) = sqrt2$, and I want to compute the value of $phi(frac32)$. I can do the following:



      $$ phi(frac32) = phi(3) phi(frac12) = [phi(1) + phi(1) + phi(1)] phi(frac12) = 3phi(frac12).$$



      I am unsure how to proceed from here. I would assume that it is true that $$phi(frac11 + 1) = fracphi(1)phi(1) + phi(1) = frac12,$$ but I don't know what property of ring isomorphisms would allow me to do this.










      share|cite|improve this question









      $endgroup$




      This question is a follow - up to this question about Field Automorphisms of $mathbbQ[sqrt2]$.



      Since $mathbbQ[sqrt2]$ is a vector space over $mathbbQ$ with basis $1, sqrt2$, I naively understand why it is the case that automorphisms $phi$ of $mathbbQ[sqrt2]$ are determined wholly by the image of $1$ and $sqrt2$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt2) = sqrt2$, and I want to compute the value of $phi(frac32)$. I can do the following:



      $$ phi(frac32) = phi(3) phi(frac12) = [phi(1) + phi(1) + phi(1)] phi(frac12) = 3phi(frac12).$$



      I am unsure how to proceed from here. I would assume that it is true that $$phi(frac11 + 1) = fracphi(1)phi(1) + phi(1) = frac12,$$ but I don't know what property of ring isomorphisms would allow me to do this.







      abstract-algebra ring-theory field-theory galois-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      Solarflare0Solarflare0

      10813




      10813




















          2 Answers
          2






          active

          oldest

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          5












          $begingroup$

          $$
          2phi(frac32) = phi(3) = 3phi(1) = 3
          implies
          phi(frac32) =frac32
          $$

          Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
            $endgroup$
            – Eevee Trainer
            1 hour ago


















          2












          $begingroup$

          Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.



          For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:



          • $phi$ fixes $0$ and $1$, by definition.


          • $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.


          • $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.


          • $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.



          More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            $$
            2phi(frac32) = phi(3) = 3phi(1) = 3
            implies
            phi(frac32) =frac32
            $$

            Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
              $endgroup$
              – Eevee Trainer
              1 hour ago















            5












            $begingroup$

            $$
            2phi(frac32) = phi(3) = 3phi(1) = 3
            implies
            phi(frac32) =frac32
            $$

            Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
              $endgroup$
              – Eevee Trainer
              1 hour ago













            5












            5








            5





            $begingroup$

            $$
            2phi(frac32) = phi(3) = 3phi(1) = 3
            implies
            phi(frac32) =frac32
            $$

            Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






            share|cite|improve this answer









            $endgroup$



            $$
            2phi(frac32) = phi(3) = 3phi(1) = 3
            implies
            phi(frac32) =frac32
            $$

            Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 5 hours ago









            lhflhf

            168k11172405




            168k11172405











            • $begingroup$
              In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
              $endgroup$
              – Eevee Trainer
              1 hour ago
















            • $begingroup$
              In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
              $endgroup$
              – Eevee Trainer
              1 hour ago















            $begingroup$
            In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
            $endgroup$
            – Eevee Trainer
            1 hour ago




            $begingroup$
            In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $phi(xy)=phi(x)phi(y)$, under the consideration $3 = 3cdot 1$.
            $endgroup$
            – Eevee Trainer
            1 hour ago











            2












            $begingroup$

            Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.



            For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:



            • $phi$ fixes $0$ and $1$, by definition.


            • $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.


            • $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.


            • $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.



            More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.



              For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:



              • $phi$ fixes $0$ and $1$, by definition.


              • $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.


              • $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.


              • $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.



              More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.



                For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:



                • $phi$ fixes $0$ and $1$, by definition.


                • $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.


                • $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.


                • $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.



                More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.






                share|cite|improve this answer









                $endgroup$



                Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.



                For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:



                • $phi$ fixes $0$ and $1$, by definition.


                • $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.


                • $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.


                • $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.



                More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 5 hours ago









                60056005

                37.1k752127




                37.1k752127



























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