Prove that $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(i sqrt{5})$ are not isomorphic.












1












$begingroup$


The question is :




Prove that $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(i sqrt{5})$ are not isomorphic (I'm talking about ring isomorphism).




What I have done : suppose there is an isomorphism $f:mathbb{Q}(sqrt{5})to mathbb{Q}(isqrt{5})$.



We have $f(1)=1$ and
$$f(n+1)=f(n)+1, qquad f(n-1)=f(n)-1.$$
Which leads to $f(x)=x$ for all $xinmathbb{Z}$, but $f(sqrt{5})^2 =f(5)=5.$ then $f(sqrt{5}) in {sqrt{5},-sqrt{5}}$, which is not possible since $pm sqrt{5} notin mathbb{Q}(isqrt{5})$. Then, there is no such morphism $f$.



Am I right?






abstract-algebra ring-theory proof-verification







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  • $begingroup$
    What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
    $endgroup$
    – ÍgjøgnumMeg
    May 28 '16 at 13:16












  • $begingroup$
    What is $x$ in "Which leads to $f(x)=x$,..."?
    $endgroup$
    – russoo
    May 28 '16 at 13:18










  • $begingroup$
    @Ed_4434 I'm talking about ring isomorphism.
    $endgroup$
    – aziiri
    May 28 '16 at 13:18






  • 1




    $begingroup$
    Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
    $endgroup$
    – awllower
    May 28 '16 at 13:22






  • 1




    $begingroup$
    @awllower : no I can't.
    $endgroup$
    – aziiri
    May 28 '16 at 13:23
















1












$begingroup$


The question is :




Prove that $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(i sqrt{5})$ are not isomorphic (I'm talking about ring isomorphism).




What I have done : suppose there is an isomorphism $f:mathbb{Q}(sqrt{5})to mathbb{Q}(isqrt{5})$.



We have $f(1)=1$ and
$$f(n+1)=f(n)+1, qquad f(n-1)=f(n)-1.$$
Which leads to $f(x)=x$ for all $xinmathbb{Z}$, but $f(sqrt{5})^2 =f(5)=5.$ then $f(sqrt{5}) in {sqrt{5},-sqrt{5}}$, which is not possible since $pm sqrt{5} notin mathbb{Q}(isqrt{5})$. Then, there is no such morphism $f$.



Am I right?






abstract-algebra ring-theory proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
    $endgroup$
    – ÍgjøgnumMeg
    May 28 '16 at 13:16












  • $begingroup$
    What is $x$ in "Which leads to $f(x)=x$,..."?
    $endgroup$
    – russoo
    May 28 '16 at 13:18










  • $begingroup$
    @Ed_4434 I'm talking about ring isomorphism.
    $endgroup$
    – aziiri
    May 28 '16 at 13:18






  • 1




    $begingroup$
    Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
    $endgroup$
    – awllower
    May 28 '16 at 13:22






  • 1




    $begingroup$
    @awllower : no I can't.
    $endgroup$
    – aziiri
    May 28 '16 at 13:23














1












1








1


1



$begingroup$


The question is :




Prove that $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(i sqrt{5})$ are not isomorphic (I'm talking about ring isomorphism).




What I have done : suppose there is an isomorphism $f:mathbb{Q}(sqrt{5})to mathbb{Q}(isqrt{5})$.



We have $f(1)=1$ and
$$f(n+1)=f(n)+1, qquad f(n-1)=f(n)-1.$$
Which leads to $f(x)=x$ for all $xinmathbb{Z}$, but $f(sqrt{5})^2 =f(5)=5.$ then $f(sqrt{5}) in {sqrt{5},-sqrt{5}}$, which is not possible since $pm sqrt{5} notin mathbb{Q}(isqrt{5})$. Then, there is no such morphism $f$.



Am I right?






abstract-algebra ring-theory proof-verification







share|cite|improve this question











$endgroup$




The question is :




Prove that $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(i sqrt{5})$ are not isomorphic (I'm talking about ring isomorphism).




What I have done : suppose there is an isomorphism $f:mathbb{Q}(sqrt{5})to mathbb{Q}(isqrt{5})$.



We have $f(1)=1$ and
$$f(n+1)=f(n)+1, qquad f(n-1)=f(n)-1.$$
Which leads to $f(x)=x$ for all $xinmathbb{Z}$, but $f(sqrt{5})^2 =f(5)=5.$ then $f(sqrt{5}) in {sqrt{5},-sqrt{5}}$, which is not possible since $pm sqrt{5} notin mathbb{Q}(isqrt{5})$. Then, there is no such morphism $f$.



Am I right?






abstract-algebra ring-theory proof-verification




abstract-algebra ring-theory proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 15:28









F.A.

9612719




9612719










asked May 28 '16 at 13:15









aziiriaziiri

2,48311430




2,48311430












  • $begingroup$
    What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
    $endgroup$
    – ÍgjøgnumMeg
    May 28 '16 at 13:16












  • $begingroup$
    What is $x$ in "Which leads to $f(x)=x$,..."?
    $endgroup$
    – russoo
    May 28 '16 at 13:18










  • $begingroup$
    @Ed_4434 I'm talking about ring isomorphism.
    $endgroup$
    – aziiri
    May 28 '16 at 13:18






  • 1




    $begingroup$
    Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
    $endgroup$
    – awllower
    May 28 '16 at 13:22






  • 1




    $begingroup$
    @awllower : no I can't.
    $endgroup$
    – aziiri
    May 28 '16 at 13:23


















  • $begingroup$
    What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
    $endgroup$
    – ÍgjøgnumMeg
    May 28 '16 at 13:16












  • $begingroup$
    What is $x$ in "Which leads to $f(x)=x$,..."?
    $endgroup$
    – russoo
    May 28 '16 at 13:18










  • $begingroup$
    @Ed_4434 I'm talking about ring isomorphism.
    $endgroup$
    – aziiri
    May 28 '16 at 13:18






  • 1




    $begingroup$
    Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
    $endgroup$
    – awllower
    May 28 '16 at 13:22






  • 1




    $begingroup$
    @awllower : no I can't.
    $endgroup$
    – aziiri
    May 28 '16 at 13:23
















$begingroup$
What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
$endgroup$
– ÍgjøgnumMeg
May 28 '16 at 13:16






$begingroup$
What kind of isomorphism/structures are you talking about? Vector Spaces or Field Extensions? It makes a big difference.
$endgroup$
– ÍgjøgnumMeg
May 28 '16 at 13:16














$begingroup$
What is $x$ in "Which leads to $f(x)=x$,..."?
$endgroup$
– russoo
May 28 '16 at 13:18




$begingroup$
What is $x$ in "Which leads to $f(x)=x$,..."?
$endgroup$
– russoo
May 28 '16 at 13:18












$begingroup$
@Ed_4434 I'm talking about ring isomorphism.
$endgroup$
– aziiri
May 28 '16 at 13:18




$begingroup$
@Ed_4434 I'm talking about ring isomorphism.
$endgroup$
– aziiri
May 28 '16 at 13:18




1




1




$begingroup$
Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
$endgroup$
– awllower
May 28 '16 at 13:22




$begingroup$
Can you find an element $x$ in $mathbb Q[sqrt5]$ with $x^2=-5?$
$endgroup$
– awllower
May 28 '16 at 13:22




1




1




$begingroup$
@awllower : no I can't.
$endgroup$
– aziiri
May 28 '16 at 13:23




$begingroup$
@awllower : no I can't.
$endgroup$
– aziiri
May 28 '16 at 13:23










2 Answers
2






active

oldest

votes


















8












$begingroup$

They are not isomorphic as fields: While $mathbb{Q}(i sqrt{5})$ contains a square root of $-5$ (namely $i sqrt{5}$), the field $mathbb{Q}(sqrt{5})$ does not (as it is a subfield of $mathbb{R}$).



They are however isomorphic as $mathbb{Q}$-vector spaces, as they are both two-dimensional over $mathbb{Q}$ (because $mathbb{Q}(sqrt{5})$ has ${1, sqrt{5}}$ as a $mathbb{Q}$-basis, and $mathbb{Q}(isqrt{5})$ has ${1, isqrt{5}}$ as a $mathbb{Q}$-basis).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Short, simple, accurate and elegant. +1
    $endgroup$
    – DonAntonio
    May 28 '16 at 13:36



















0












$begingroup$

Consider the fields $K = Bbb{Q}(sqrt{5})$ and $L = Bbb{Q}(isqrt{5})$ and suppose that $varphi$ is a field homomorphism.



If we look at $varphi(sqrt{5}^2)$ then we have



$$varphi(sqrt{5}^2) = varphi(sqrt{5})^2 = varphi(5) = 5,$$



so for $varphi$ to exist, there must exist an element $varphi(sqrt{5})$ in $L$ such that $varphi(sqrt{5})^2 - 5 = 0$, a contradiction. So $K$ and $L$ cannot be isomorphic.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    They are not isomorphic as fields: While $mathbb{Q}(i sqrt{5})$ contains a square root of $-5$ (namely $i sqrt{5}$), the field $mathbb{Q}(sqrt{5})$ does not (as it is a subfield of $mathbb{R}$).



    They are however isomorphic as $mathbb{Q}$-vector spaces, as they are both two-dimensional over $mathbb{Q}$ (because $mathbb{Q}(sqrt{5})$ has ${1, sqrt{5}}$ as a $mathbb{Q}$-basis, and $mathbb{Q}(isqrt{5})$ has ${1, isqrt{5}}$ as a $mathbb{Q}$-basis).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Short, simple, accurate and elegant. +1
      $endgroup$
      – DonAntonio
      May 28 '16 at 13:36
















    8












    $begingroup$

    They are not isomorphic as fields: While $mathbb{Q}(i sqrt{5})$ contains a square root of $-5$ (namely $i sqrt{5}$), the field $mathbb{Q}(sqrt{5})$ does not (as it is a subfield of $mathbb{R}$).



    They are however isomorphic as $mathbb{Q}$-vector spaces, as they are both two-dimensional over $mathbb{Q}$ (because $mathbb{Q}(sqrt{5})$ has ${1, sqrt{5}}$ as a $mathbb{Q}$-basis, and $mathbb{Q}(isqrt{5})$ has ${1, isqrt{5}}$ as a $mathbb{Q}$-basis).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Short, simple, accurate and elegant. +1
      $endgroup$
      – DonAntonio
      May 28 '16 at 13:36














    8












    8








    8





    $begingroup$

    They are not isomorphic as fields: While $mathbb{Q}(i sqrt{5})$ contains a square root of $-5$ (namely $i sqrt{5}$), the field $mathbb{Q}(sqrt{5})$ does not (as it is a subfield of $mathbb{R}$).



    They are however isomorphic as $mathbb{Q}$-vector spaces, as they are both two-dimensional over $mathbb{Q}$ (because $mathbb{Q}(sqrt{5})$ has ${1, sqrt{5}}$ as a $mathbb{Q}$-basis, and $mathbb{Q}(isqrt{5})$ has ${1, isqrt{5}}$ as a $mathbb{Q}$-basis).






    share|cite|improve this answer









    $endgroup$



    They are not isomorphic as fields: While $mathbb{Q}(i sqrt{5})$ contains a square root of $-5$ (namely $i sqrt{5}$), the field $mathbb{Q}(sqrt{5})$ does not (as it is a subfield of $mathbb{R}$).



    They are however isomorphic as $mathbb{Q}$-vector spaces, as they are both two-dimensional over $mathbb{Q}$ (because $mathbb{Q}(sqrt{5})$ has ${1, sqrt{5}}$ as a $mathbb{Q}$-basis, and $mathbb{Q}(isqrt{5})$ has ${1, isqrt{5}}$ as a $mathbb{Q}$-basis).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 28 '16 at 13:22









    Jendrik StelznerJendrik Stelzner

    7,90621440




    7,90621440












    • $begingroup$
      Short, simple, accurate and elegant. +1
      $endgroup$
      – DonAntonio
      May 28 '16 at 13:36


















    • $begingroup$
      Short, simple, accurate and elegant. +1
      $endgroup$
      – DonAntonio
      May 28 '16 at 13:36
















    $begingroup$
    Short, simple, accurate and elegant. +1
    $endgroup$
    – DonAntonio
    May 28 '16 at 13:36




    $begingroup$
    Short, simple, accurate and elegant. +1
    $endgroup$
    – DonAntonio
    May 28 '16 at 13:36











    0












    $begingroup$

    Consider the fields $K = Bbb{Q}(sqrt{5})$ and $L = Bbb{Q}(isqrt{5})$ and suppose that $varphi$ is a field homomorphism.



    If we look at $varphi(sqrt{5}^2)$ then we have



    $$varphi(sqrt{5}^2) = varphi(sqrt{5})^2 = varphi(5) = 5,$$



    so for $varphi$ to exist, there must exist an element $varphi(sqrt{5})$ in $L$ such that $varphi(sqrt{5})^2 - 5 = 0$, a contradiction. So $K$ and $L$ cannot be isomorphic.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Consider the fields $K = Bbb{Q}(sqrt{5})$ and $L = Bbb{Q}(isqrt{5})$ and suppose that $varphi$ is a field homomorphism.



      If we look at $varphi(sqrt{5}^2)$ then we have



      $$varphi(sqrt{5}^2) = varphi(sqrt{5})^2 = varphi(5) = 5,$$



      so for $varphi$ to exist, there must exist an element $varphi(sqrt{5})$ in $L$ such that $varphi(sqrt{5})^2 - 5 = 0$, a contradiction. So $K$ and $L$ cannot be isomorphic.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Consider the fields $K = Bbb{Q}(sqrt{5})$ and $L = Bbb{Q}(isqrt{5})$ and suppose that $varphi$ is a field homomorphism.



        If we look at $varphi(sqrt{5}^2)$ then we have



        $$varphi(sqrt{5}^2) = varphi(sqrt{5})^2 = varphi(5) = 5,$$



        so for $varphi$ to exist, there must exist an element $varphi(sqrt{5})$ in $L$ such that $varphi(sqrt{5})^2 - 5 = 0$, a contradiction. So $K$ and $L$ cannot be isomorphic.






        share|cite|improve this answer









        $endgroup$



        Consider the fields $K = Bbb{Q}(sqrt{5})$ and $L = Bbb{Q}(isqrt{5})$ and suppose that $varphi$ is a field homomorphism.



        If we look at $varphi(sqrt{5}^2)$ then we have



        $$varphi(sqrt{5}^2) = varphi(sqrt{5})^2 = varphi(5) = 5,$$



        so for $varphi$ to exist, there must exist an element $varphi(sqrt{5})$ in $L$ such that $varphi(sqrt{5})^2 - 5 = 0$, a contradiction. So $K$ and $L$ cannot be isomorphic.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 28 '16 at 13:50









        ÍgjøgnumMegÍgjøgnumMeg

        2,86011129




        2,86011129






























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