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Ring Automorphisms that fix 1.



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?










2












$begingroup$


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









$endgroup$
















    2












    $begingroup$


    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









    $endgroup$














      2












      2








      2





      $begingroup$


      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






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      asked 3 hours ago









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          3












          $begingroup$

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

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






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            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









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              2 Answers
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              2 Answers
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              3












              $begingroup$

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

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






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

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

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






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

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

                  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 3 hours ago









                  lhflhf

                  168k11172405




                  168k11172405





















                      1












                      $begingroup$

                      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









                      $endgroup$

















                        1












                        $begingroup$

                        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









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          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 2 hours ago









                          60056005

                          37.1k752127




                          37.1k752127



























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