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Ring isomorphism

  1. Nov 25, 2005 #1
    what is the easiest way to show that
    Q[x]/<x^2-2> is ring isomorphic to
    Q[sqrt2]={a+b(sqrt2)|a,b in Q}

    just give me a hint how to start
  2. jcsd
  3. Nov 25, 2005 #2

    do I have to show that x^2 - 2 is in the kernal?
  4. Nov 25, 2005 #3

    matt grime

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    why don't you just write down the (obvious) isomorphism? (obvious in the sense of one side only has x as a special quantity, the other sqrt(2), and anything in Q[x]/(x^2-2) is of the form a+bx, isn't it....)
    Last edited: Nov 25, 2005
  5. Nov 25, 2005 #4
    yes! thank you. the example in the book goes into too much detail and I was trying follow that, but yes the function f(a+bx)=a+b(sqrt2) is a ring isomorphism.
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