How to prove $mathbb Q(i)=mathbb Q(frac{2i+1}{i-1})$?












2












$begingroup$


I know the elements in these two groups, but how to prove that $mathbb Q(i)=mathbb Q(frac{2i+1}{i-1})$ ?



Can the second thing be replaced by other complex numbers?






abstract-algebra







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$endgroup$








  • 1




    $begingroup$
    I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
    $endgroup$
    – Arthur
    Jan 3 at 11:37






  • 1




    $begingroup$
    What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
    $endgroup$
    – lulu
    Jan 3 at 11:37
















2












$begingroup$


I know the elements in these two groups, but how to prove that $mathbb Q(i)=mathbb Q(frac{2i+1}{i-1})$ ?



Can the second thing be replaced by other complex numbers?






abstract-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
    $endgroup$
    – Arthur
    Jan 3 at 11:37






  • 1




    $begingroup$
    What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
    $endgroup$
    – lulu
    Jan 3 at 11:37














2












2








2





$begingroup$


I know the elements in these two groups, but how to prove that $mathbb Q(i)=mathbb Q(frac{2i+1}{i-1})$ ?



Can the second thing be replaced by other complex numbers?






abstract-algebra







share|cite|improve this question











$endgroup$




I know the elements in these two groups, but how to prove that $mathbb Q(i)=mathbb Q(frac{2i+1}{i-1})$ ?



Can the second thing be replaced by other complex numbers?






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 11:36









Jennifer

8,41721737




8,41721737










asked Jan 3 at 11:34









yLcccyLccc

233




233








  • 1




    $begingroup$
    I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
    $endgroup$
    – Arthur
    Jan 3 at 11:37






  • 1




    $begingroup$
    What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
    $endgroup$
    – lulu
    Jan 3 at 11:37














  • 1




    $begingroup$
    I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
    $endgroup$
    – Arthur
    Jan 3 at 11:37






  • 1




    $begingroup$
    What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
    $endgroup$
    – lulu
    Jan 3 at 11:37








1




1




$begingroup$
I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
$endgroup$
– Arthur
Jan 3 at 11:37




$begingroup$
I am unfamiliar with the notation $Bbb Q(i)$ when speaking about groups. What does it mean?
$endgroup$
– Arthur
Jan 3 at 11:37




1




1




$begingroup$
What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
$endgroup$
– lulu
Jan 3 at 11:37




$begingroup$
What does your notation mean? Are you looking at field extensions of the rationals? If so, then just note that $frac {2i+1}{i-1}=frac {1-3i}2$ so $i$ is contained in the field on the right.
$endgroup$
– lulu
Jan 3 at 11:37










3 Answers
3






active

oldest

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3












$begingroup$

Notice that $frac{2i+1}{i-1} = frac12 - frac32 i$.



Hence $$frac{2i+1}{i-1} = underbrace{frac12}_{inmathbb{Q}} - underbrace{frac32 i}_{inmathbb{Q}(i)} in mathbb{Q}(i)$$
so $mathbb{Q}left(frac{2i+1}{i-1}right) subseteq mathbb{Q}(i)$.



Conversely, we have $$i = underbrace{frac13}_{inmathbb{Q}} - underbrace{frac23frac{2i+1}{i-1}}_{inmathbb{Q}left(frac{2i+1}{i-1}right)} in mathbb{Q}left(frac{2i+1}{i-1}right)$$
so $mathbb{Q}(i) subseteq mathbb{Q}left(frac{2i+1}{i-1}right)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that $mathbb{Q}(frac{2i+1}{i-1})=mathbb{Q}(2+frac{3}{i-1})=mathbb{Q}(frac{i+2}{-2})=mathbb{Q}(-frac{i}{2})=mathbb{Q}(frac i 2)$



    Can you continue from here?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y neq 0$ (if $x=0, yneq 0$ it's trivial and if $y =0$ it's impossible) and suppose $Bbb Q(i) = Bbb Q(x+iy)$. So $(a+bx) + icdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y in Bbb Q$ is our choice and with $Bbb Q(i) supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $Bbb Q(x+iy) supseteq Q(i)$ so they're equal.



      So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) in Bbb Q times Bbb Q_*$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
        $endgroup$
        – Lucas Henrique
        Jan 3 at 12:35











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Notice that $frac{2i+1}{i-1} = frac12 - frac32 i$.



      Hence $$frac{2i+1}{i-1} = underbrace{frac12}_{inmathbb{Q}} - underbrace{frac32 i}_{inmathbb{Q}(i)} in mathbb{Q}(i)$$
      so $mathbb{Q}left(frac{2i+1}{i-1}right) subseteq mathbb{Q}(i)$.



      Conversely, we have $$i = underbrace{frac13}_{inmathbb{Q}} - underbrace{frac23frac{2i+1}{i-1}}_{inmathbb{Q}left(frac{2i+1}{i-1}right)} in mathbb{Q}left(frac{2i+1}{i-1}right)$$
      so $mathbb{Q}(i) subseteq mathbb{Q}left(frac{2i+1}{i-1}right)$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Notice that $frac{2i+1}{i-1} = frac12 - frac32 i$.



        Hence $$frac{2i+1}{i-1} = underbrace{frac12}_{inmathbb{Q}} - underbrace{frac32 i}_{inmathbb{Q}(i)} in mathbb{Q}(i)$$
        so $mathbb{Q}left(frac{2i+1}{i-1}right) subseteq mathbb{Q}(i)$.



        Conversely, we have $$i = underbrace{frac13}_{inmathbb{Q}} - underbrace{frac23frac{2i+1}{i-1}}_{inmathbb{Q}left(frac{2i+1}{i-1}right)} in mathbb{Q}left(frac{2i+1}{i-1}right)$$
        so $mathbb{Q}(i) subseteq mathbb{Q}left(frac{2i+1}{i-1}right)$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Notice that $frac{2i+1}{i-1} = frac12 - frac32 i$.



          Hence $$frac{2i+1}{i-1} = underbrace{frac12}_{inmathbb{Q}} - underbrace{frac32 i}_{inmathbb{Q}(i)} in mathbb{Q}(i)$$
          so $mathbb{Q}left(frac{2i+1}{i-1}right) subseteq mathbb{Q}(i)$.



          Conversely, we have $$i = underbrace{frac13}_{inmathbb{Q}} - underbrace{frac23frac{2i+1}{i-1}}_{inmathbb{Q}left(frac{2i+1}{i-1}right)} in mathbb{Q}left(frac{2i+1}{i-1}right)$$
          so $mathbb{Q}(i) subseteq mathbb{Q}left(frac{2i+1}{i-1}right)$.






          share|cite|improve this answer









          $endgroup$



          Notice that $frac{2i+1}{i-1} = frac12 - frac32 i$.



          Hence $$frac{2i+1}{i-1} = underbrace{frac12}_{inmathbb{Q}} - underbrace{frac32 i}_{inmathbb{Q}(i)} in mathbb{Q}(i)$$
          so $mathbb{Q}left(frac{2i+1}{i-1}right) subseteq mathbb{Q}(i)$.



          Conversely, we have $$i = underbrace{frac13}_{inmathbb{Q}} - underbrace{frac23frac{2i+1}{i-1}}_{inmathbb{Q}left(frac{2i+1}{i-1}right)} in mathbb{Q}left(frac{2i+1}{i-1}right)$$
          so $mathbb{Q}(i) subseteq mathbb{Q}left(frac{2i+1}{i-1}right)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 11:42









          mechanodroidmechanodroid

          27.1k62446




          27.1k62446























              1












              $begingroup$

              Note that $mathbb{Q}(frac{2i+1}{i-1})=mathbb{Q}(2+frac{3}{i-1})=mathbb{Q}(frac{i+2}{-2})=mathbb{Q}(-frac{i}{2})=mathbb{Q}(frac i 2)$



              Can you continue from here?






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that $mathbb{Q}(frac{2i+1}{i-1})=mathbb{Q}(2+frac{3}{i-1})=mathbb{Q}(frac{i+2}{-2})=mathbb{Q}(-frac{i}{2})=mathbb{Q}(frac i 2)$



                Can you continue from here?






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that $mathbb{Q}(frac{2i+1}{i-1})=mathbb{Q}(2+frac{3}{i-1})=mathbb{Q}(frac{i+2}{-2})=mathbb{Q}(-frac{i}{2})=mathbb{Q}(frac i 2)$



                  Can you continue from here?






                  share|cite|improve this answer









                  $endgroup$



                  Note that $mathbb{Q}(frac{2i+1}{i-1})=mathbb{Q}(2+frac{3}{i-1})=mathbb{Q}(frac{i+2}{-2})=mathbb{Q}(-frac{i}{2})=mathbb{Q}(frac i 2)$



                  Can you continue from here?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 11:43









                  Test123Test123

                  2,762828




                  2,762828























                      0












                      $begingroup$

                      I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y neq 0$ (if $x=0, yneq 0$ it's trivial and if $y =0$ it's impossible) and suppose $Bbb Q(i) = Bbb Q(x+iy)$. So $(a+bx) + icdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y in Bbb Q$ is our choice and with $Bbb Q(i) supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $Bbb Q(x+iy) supseteq Q(i)$ so they're equal.



                      So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) in Bbb Q times Bbb Q_*$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                        $endgroup$
                        – Lucas Henrique
                        Jan 3 at 12:35
















                      0












                      $begingroup$

                      I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y neq 0$ (if $x=0, yneq 0$ it's trivial and if $y =0$ it's impossible) and suppose $Bbb Q(i) = Bbb Q(x+iy)$. So $(a+bx) + icdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y in Bbb Q$ is our choice and with $Bbb Q(i) supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $Bbb Q(x+iy) supseteq Q(i)$ so they're equal.



                      So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) in Bbb Q times Bbb Q_*$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                        $endgroup$
                        – Lucas Henrique
                        Jan 3 at 12:35














                      0












                      0








                      0





                      $begingroup$

                      I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y neq 0$ (if $x=0, yneq 0$ it's trivial and if $y =0$ it's impossible) and suppose $Bbb Q(i) = Bbb Q(x+iy)$. So $(a+bx) + icdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y in Bbb Q$ is our choice and with $Bbb Q(i) supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $Bbb Q(x+iy) supseteq Q(i)$ so they're equal.



                      So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) in Bbb Q times Bbb Q_*$.






                      share|cite|improve this answer









                      $endgroup$



                      I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y neq 0$ (if $x=0, yneq 0$ it's trivial and if $y =0$ it's impossible) and suppose $Bbb Q(i) = Bbb Q(x+iy)$. So $(a+bx) + icdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y in Bbb Q$ is our choice and with $Bbb Q(i) supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $Bbb Q(x+iy) supseteq Q(i)$ so they're equal.



                      So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) in Bbb Q times Bbb Q_*$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 3 at 12:32









                      Lucas HenriqueLucas Henrique

                      999414




                      999414












                      • $begingroup$
                        And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                        $endgroup$
                        – Lucas Henrique
                        Jan 3 at 12:35


















                      • $begingroup$
                        And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                        $endgroup$
                        – Lucas Henrique
                        Jan 3 at 12:35
















                      $begingroup$
                      And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                      $endgroup$
                      – Lucas Henrique
                      Jan 3 at 12:35




                      $begingroup$
                      And now note that $frac{2i+1}{i-1} = frac{-2 - 2i + i - 1}{2} = -3/2 + 1/2cdot i$. So $(-3/2, 1/2)$ is our choice and it works perfectly.
                      $endgroup$
                      – Lucas Henrique
                      Jan 3 at 12:35


















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