Ring Automorphisms that fix 1. Announcing the arrival of Valued Associate #679: Cesar Manara ...

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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 $mathbb{R}^n$group of automorphisms of the ring $mathbb{Z}timesmathbb{Z}$Trying to understand a proof for the automorphisms of a polynomial ringAll automorphisms of splitting fieldsDetermining automorphisms of this extensionRing automorphisms of $mathbb{Q}[sqrt[3]{5}]$Automorphism of ring and isomorphism of quotient ringsThe automorphisms of the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}$.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 $mathbb{Q}[sqrt{2}]$.



Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:



$$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$



I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ 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









$endgroup$

















    2












    $begingroup$


    This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.



    Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:



    $$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$



    I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ 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









    $endgroup$















      2












      2








      2





      $begingroup$


      This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.



      Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:



      $$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$



      I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ 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









      $endgroup$




      This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.



      Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:



      $$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$



      I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ but I don't know what property of ring isomorphisms would allow me to do this.






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




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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 3 hours ago









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          $begingroup$

          $$
          2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
          implies
          phi(frac{3}{2}) =frac{3}{2}
          $$
          Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

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



            For the proof, we assume WLOG that $mathbb{Q} subseteq K$. 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 $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.




            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 $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              $begingroup$

              $$
              2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
              implies
              phi(frac{3}{2}) =frac{3}{2}
              $$
              Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                $$
                2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
                implies
                phi(frac{3}{2}) =frac{3}{2}
                $$
                Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $$
                  2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
                  implies
                  phi(frac{3}{2}) =frac{3}{2}
                  $$
                  Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.






                  share|cite|improve this answer









                  $endgroup$



                  $$
                  2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
                  implies
                  phi(frac{3}{2}) =frac{3}{2}
                  $$
                  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 $mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbb{Q}$.



                      For the proof, we assume WLOG that $mathbb{Q} subseteq K$. 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 $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.




                      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 $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

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



                        For the proof, we assume WLOG that $mathbb{Q} subseteq K$. 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 $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.




                        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 $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

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



                          For the proof, we assume WLOG that $mathbb{Q} subseteq K$. 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 $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.




                          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 $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.






                          share|cite|improve this answer









                          $endgroup$



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



                          For the proof, we assume WLOG that $mathbb{Q} subseteq K$. 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 $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.




                          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 $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, 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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