Computing $phi(frac32)$ where $phi$ is an automorphism of $mathbb Q[sqrt2]$ such that $phi(1)=1$ and...












6












$begingroup$


This question is a followup 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 $phileft(frac{3}{2}right)$. I can do the following:



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



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






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







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$endgroup$

















    6












    $begingroup$


    This question is a followup 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 $phileft(frac{3}{2}right)$. I can do the following:



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



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











    $endgroup$















      6












      6








      6





      $begingroup$


      This question is a followup 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 $phileft(frac{3}{2}right)$. I can do the following:



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



      I am unsure how to proceed from here. I would assume that it is true that $$phileft(frac{1}{1 + 1}right) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ 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 followup 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 $phileft(frac{3}{2}right)$. I can do the following:



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



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













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      edited Apr 17 at 9:47









      Asaf Karagila♦

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      asked Apr 17 at 1:08









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          2 Answers
          2






          active

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          6












          $begingroup$

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






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            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
            Apr 17 at 5:28






          • 1




            $begingroup$
            @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
            $endgroup$
            – ancientmathematician
            Apr 17 at 7:11





















          5












          $begingroup$

          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









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            2 Answers
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            2 Answers
            2






            active

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            6












            $begingroup$

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






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              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
              Apr 17 at 5:28






            • 1




              $begingroup$
              @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
              $endgroup$
              – ancientmathematician
              Apr 17 at 7:11


















            6












            $begingroup$

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






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              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
              Apr 17 at 5:28






            • 1




              $begingroup$
              @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
              $endgroup$
              – ancientmathematician
              Apr 17 at 7:11
















            6












            6








            6





            $begingroup$

            $$
            2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
            implies
            phi(frac{3}{2}) =frac{3}{2}
            $$
            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 Apr 17 at 1:12









            lhflhf

            168k11173405




            168k11173405








            • 1




              $begingroup$
              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
              Apr 17 at 5:28






            • 1




              $begingroup$
              @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
              $endgroup$
              – ancientmathematician
              Apr 17 at 7:11
















            • 1




              $begingroup$
              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
              Apr 17 at 5:28






            • 1




              $begingroup$
              @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
              $endgroup$
              – ancientmathematician
              Apr 17 at 7:11










            1




            1




            $begingroup$
            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
            Apr 17 at 5:28




            $begingroup$
            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
            Apr 17 at 5:28




            1




            1




            $begingroup$
            @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
            $endgroup$
            – ancientmathematician
            Apr 17 at 7:11






            $begingroup$
            @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $phi(1)=1$. .
            $endgroup$
            – ancientmathematician
            Apr 17 at 7:11













            5












            $begingroup$

            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









            $endgroup$


















              5












              $begingroup$

              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









              $endgroup$
















                5












                5








                5





                $begingroup$

                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 Apr 17 at 1:30









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