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

What exactly is a "Meth" in Altered Carbon?

Naming the result of a source block

How do pianists reach extremely loud dynamics?

Fundamental Solution of the Pell Equation

How to find all the available tools in mac terminal?

Why is "Consequences inflicted." not a sentence?

Why did the IBM 650 use bi-quinary?

What's the purpose of writing one's academic biography in the third person?

Why did the Falcon Heavy center core fall off the ASDS OCISLY barge?

When were vectors invented?

Using audio cues to encourage good posture

How to call a function with default parameter through a pointer to function that is the return of another function?

How does the particle を relate to the verb 行く in the structure「A を + B に行く」?

At the end of Thor: Ragnarok why don't the Asgardians turn and head for the Bifrost as per their original plan?

When do you get frequent flier miles - when you buy, or when you fly?

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

What is the logic behind the Maharil's explanation of why we don't say שעשה ניסים on Pesach?

Is the Standard Deduction better than Itemized when both are the same amount?

What does this icon in iOS Stardew Valley mean?

What causes the vertical darker bands in my photo?

If a contract sometimes uses the wrong name, is it still valid?

How to align text above triangle figure

Can an alien society believe that their star system is the universe?

What is the role of the transistor and diode in a soft start circuit?



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




      share|cite|improve this question










      asked 3 hours ago









      Solarflare0Solarflare0

      9813




      9813






















          2 Answers
          2






          active

          oldest

          votes


















          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









          $endgroup$





















            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









            $endgroup$














              Your Answer








              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190546%2fring-automorphisms-that-fix-1%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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












                $begingroup$

                $$
                2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
                implies
                phi(frac{3}{2}) =frac{3}{2}
                $$
                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






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190546%2fring-automorphisms-that-fix-1%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Can't compile dgruyter and caption packagesLaTeX templates/packages for writing a patent specificationLatex...

                              Image
                              Why does Arabsat 6A need a Falcon Heavy to launch Etiquette around loan refinance - decision is going to cost first broker a lot of money What is the PIE reconstruction for word-initial alpha with rough breathing? Is "remove commented out code" correct English? Why do I get two different answers for this counting problem? Blender 2.8 I can't see vertices, edges or faces in edit mode Where does SFDX store details about scratch orgs? 1960's book about a plague that kills all white people How can saying a song's name be a copyright violation? UK: Is there precedent for the governments e-petition site changing the ...
                              Read more

                              Schneeberg (Smreczany) Bibliografia | Menu...

                              Image
                              Szczyty w Niemczech paśmieSmreczanyn.p.m.Ochrza Schneeberg Państwo   Niemcy Położenie Bawaria Pasmo Fichtelgebirge Wysokość 1 051 m n.p.m. Położenie na mapie Bawarii Schneeberg Położenie na mapie Niemiec Schneeberg 50°03′N   11°51′E / 50,050000   11,850000 Multimedia w Wikimedia Commons Schneeberg – szczyt w paśmie Smreczany o wysokości 1051 m n.p.m., na zboczach którego znajdują się źródła rzeki Ochrza. Bibliografia | Rainer H. Schmeisser: Der Schneeberg, Beiträge zur Geschichts- und Landeskunde des Fichtelgebirges Nr. 1, Regensburg 1979 This page is only for reference, If you need detailed information, please check here
                              Read more

                              Hans Bellmer Spis treści Życiorys | Upamiętnienie | Przypisy | Bibliografia | Linki zewnętrzne |...

                              Image
                              Ludzie związani z III Liceum Ogólnokształcącym im. Adama Mickiewicza w KatowicachNiemieccy rzeźbiarzeNiemieccy malarze współcześniNiemieccy graficyLudzie urodzeni w KatowicachUrodzeni w 1902Zmarli w 1975 13 marca1902Katowicach23 lutego1975Paryżumalarzrzeźbiarzgrafikul. A. Mickiewiczaul. T. KościuszkiWyższej Szkole TechnicznejinstalacjiPaula EluardaBerlinie„tableaux vivants”NiemczechnazistówzdegenerowanąBerlinsurrealistówII wojnie światowejCamp des MillesAix-en-ProvenceParyżuUnicę ZürnPère-Lachaiseul. T. KościuszkiDział Grafiki im. Pawła Stellera Muzeum Historii Katowic2008Muzeum Historii KatowicNatalia KruszynaBerlinie Tablica pamiątkowa z 2008 roku na domu, w którym mieszkał artysta przy ul. T. Kościuszki 47 w Katowicach (2017) Grób Bellmera na cmentarzu Père-Lachaise Tablica pamiątkowa przy Ehrenfelsstraße 8a w Berlinie Budynek, w którym mieszkał artysta przy ul. T. Kościuszki 47 w Katowicach (2017) Hans Bellmer (ur. 13 marca 1902 w Katowicach [1] , zm. 23 lut...
                              Read more