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Proving an axiom

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data

    I was asked to prove the axioms of a field.
    so, if we look at the first one:
    commutativity: a+b=b+a and a*b=b*a where a and b belong in the set of the field

    2. Relevant equations



    3. The attempt at a solution

    it's tempting to just substitute values but i know this is the wrong approach since this wouldnt prove the axiom for all values in the set. so im stuck.
     
  2. jcsd
  3. Oct 2, 2009 #2
    It would help a lot if we knew what the set was.
     
  4. Oct 2, 2009 #3
    the question just says: prove the axioms of a field.
    but the chapter we are studying is complex numbers so maybe i have to prove them for complex numbers. if so, would this proof be good enough:
    for: a+b=b+a: where a=u+vi and b=x+yi
    u+vi+x+yi=x+yi+u+vi
    ??
     
  5. Oct 2, 2009 #4
    mmmmm. I suppose you can't just say that C is isomorphic to R[x]/(x2 + 1) and that because (x2 + 1) is maximal, it must be a field?

    How you write the proof depends on how you've defined complex numbers and how you've defined addition and multiplication. Is the definition something like a + bi = (a, b)?
     
  6. Oct 2, 2009 #5
    what do you mean '(x^2+1) i maximal' here?
    yes that's the definition we use for complex numbers.
     
  7. Oct 2, 2009 #6
    Ok, then what you want to do for commutativity of addition is (a, b) + (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) + (a, b).

    Do you notice how we "got a free ride" once we got a and c in the same tuple? That just reduces to commutativity of the reals, which I assume you can take for granted.

    As far as the maximal thing, it just means that the ideal generated by x1 + 1 in the polynomial ring over R isn't contained in any others. You'll probably learn about it if you're taking an algebra class. It was mainly just a joke.
     
  8. Oct 2, 2009 #7
    :) ok
    I understand the proof now.
    for a*b=b*a, would this be convincing:
    (a,b)*(c,d)=(a+bi)(c+di)=(ac-bd, bc+ad)=(ca-db, cb+da)=(c,d)*(a,b)
     
  9. Oct 2, 2009 #8
    You don't need to stick the term that I struck out in there. Remember, i is just (0, 1). Other than that, it looks great though.
     
  10. Oct 2, 2009 #9
    thank you. now i can prove the other axioms.
     
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