Find the quotient field of $mathbb{Z}[sqrt {2}]$












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My idea: how to make a connection between this with isomorphism.






abstract-algebra quotient-spaces







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    $begingroup$
    Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
    $endgroup$
    – Dane
    May 17 '14 at 1:46












  • $begingroup$
    can we replace root of 2 with any number outside Q. @Dane
    $endgroup$
    – Danial Jeo
    May 17 '14 at 1:47


















1












$begingroup$


enter image description here



My idea: how to make a connection between this with isomorphism.






abstract-algebra quotient-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
    $endgroup$
    – Dane
    May 17 '14 at 1:46












  • $begingroup$
    can we replace root of 2 with any number outside Q. @Dane
    $endgroup$
    – Danial Jeo
    May 17 '14 at 1:47
















1












1








1





$begingroup$


enter image description here



My idea: how to make a connection between this with isomorphism.






abstract-algebra quotient-spaces







share|cite|improve this question











$endgroup$




enter image description here



My idea: how to make a connection between this with isomorphism.






abstract-algebra quotient-spaces




abstract-algebra quotient-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 17 '14 at 2:12









Benjamin Dickman

10.3k22968




10.3k22968










asked May 17 '14 at 1:44









Danial JeoDanial Jeo

508




508








  • 1




    $begingroup$
    Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
    $endgroup$
    – Dane
    May 17 '14 at 1:46












  • $begingroup$
    can we replace root of 2 with any number outside Q. @Dane
    $endgroup$
    – Danial Jeo
    May 17 '14 at 1:47
















  • 1




    $begingroup$
    Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
    $endgroup$
    – Dane
    May 17 '14 at 1:46












  • $begingroup$
    can we replace root of 2 with any number outside Q. @Dane
    $endgroup$
    – Danial Jeo
    May 17 '14 at 1:47










1




1




$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46






$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46














$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47






$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47












1 Answer
1






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$begingroup$

$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.



$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.



$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.



Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.



More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.






share|cite|improve this answer











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    1 Answer
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    $begingroup$

    $mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.



    $mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.



    $mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.



    Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.



    More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      $mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.



      $mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.



      $mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.



      Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.



      More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        $mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.



        $mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.



        $mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.



        Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.



        More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.






        share|cite|improve this answer











        $endgroup$



        $mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.



        $mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.



        $mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.



        Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.



        More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 17 '14 at 2:21

























        answered May 17 '14 at 2:05









        lhflhf

        163k10169393




        163k10169393






























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