Results 1 to 6 of 6

Math Help - Isomorphic

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    21

    Isomorphic

    Show that Z[root 7] is not isomorphic to Z[root 13].

    How can i show this by contradiction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Hi

    Assume there is a ring isomorphism f:\mathbb{Z}[\sqrt{7}]\rightarrow\mathbb{Z}[\sqrt{13}].

    Using the morphism properties, try to find a contradiction with f(\sqrt{7}). (Remember that, by definition, f(1)=1 hence f(7)=? )
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    21
    What are the morphism properties?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    What appears in its definition, for instance, for any x,y,\ f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    Posts
    21
    Could you start me off please?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    We started a proof by contradiction, assuming there was an isomorphism f:\mathbb{Z}[\sqrt{7}]\rightarrow\mathbb{Z}[\sqrt{13}]
    Well, consider f(\sqrt{7}), you can see it is an element of \mathbb{Z}[\sqrt{13}] whose square is 7 (why?)
    What we have to do now is to prove there is no squared root of 7 in \mathbb{Z}[\sqrt{13}], assume we already know that, then it's over because we obtained a contradiction.

    To show the missing part, we can also use a proof by contradiction: suppose there is a x\in\mathbb{Z}[\sqrt{13}] such that x^2=7.
    x=a+b\sqrt{13} for some a,b\in\mathbb{Z} and we get 7=a^2+13b^2+2ab\sqrt{13}\ ; since 7-a^2+13b^2 belongs to \mathbb{Z} while 2ab\sqrt{13} is irrational or 0, that means it is 0, i.e. ab=0.
    x being irrational, we can say a=0 and we finally have x=b\sqrt{13}\ ; impossible: consider their squares and conclude.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 10
    Last Post: November 8th 2010, 08:19 PM
  2. is U14 isomorphic to U18?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 21st 2010, 10:52 AM
  3. Isomorphic
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 13th 2008, 09:55 PM
  4. Isomorphic or not?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 13th 2008, 05:41 PM
  5. Isomorphic?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 15th 2007, 05:42 PM

Search Tags


/mathhelpforum @mathhelpforum