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Quick question regarding isomorphic groups?

  1. Sep 2, 2013 #1
    Let F be a field. R is an element of Mat(2,2)
    [a -b
    b a]
    for a, b in F with matrix operations.
    a. Show that R is a commutative ring with 1 and the set of diagonal matrices are
    naturally isomorphic to F .
    b. For which of the fields Q , R , C , F5
    , F7
    is R a field?
    c. Characterize which elements of R have a multiplicative inverse.
    d. Characterize the fields F for which R will be a fi eld.
    e. For which Fp
    ( p prime) is R a field?
     
  2. jcsd
  3. Sep 2, 2013 #2

    CompuChip

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    Homework Helper

    That looks like an entire homework problem.
    So where do you get stuck? What have you tried already?
     
  4. Sep 2, 2013 #3
    Just wanted to confirm my answer.only F2 would be a prime subfield such that R is a field right?

    There's just one problem I can't figure out. Give examples of fields F such that
    f is isomorphic to f(t) where f(t) is the set of formal fractions.

    And just one more thing. How would you show Q[sqrt2 sqrt3] is a field? I'm having trouble showing multiplicative inverse.

    Thanks
     
  5. Sep 2, 2013 #4

    HallsofIvy

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    If you want to "confirm an answer", then tell us what your answer is!
     
  6. Sep 2, 2013 #5
    inverse element for field a + bsqrt(2) + c(sqrt(3)

    Yes there is an inverse element, you just have to work at it a little. Let's say your element is a + b√2 + c√3 and you want to show that 1/( a + b√2 + c√3) is in the field. As a first step multiply top and bottom of that fraction by a + b√2 - c√3 giving you

    ( a + b√2 + c√3)/((a + b√2)[itex]^2[/itex] +3c[itex]^{2}[/itex])

    the denominator becomes a[itex]^{2}[/itex] + 2b[itex]^{2}[/itex] +3c[itex]^{2}[/itex] + ab√2.

    Now multiply numerator and denominator by
    a[itex]^{2}[/itex] + 2b[itex]^{2}[/itex] +3c[itex]^{2}[/itex] - ab√2

    Your denominator will be all real and the numerator will have various terms that can be simplified down to something in your field.
     
    Last edited: Sep 2, 2013
  7. Sep 5, 2013 #6
    For some reason the answer up above got placed in this topic. But it was the answer to another question. I wonder if I managed to actually do this, or is there a bug somewhere?
     
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