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On Algebraic Integers

  1. Mar 29, 2009 #1
    The problem statement, all variables and given/known data
    If a is an algebraic integer with a^3 + a + 1 = 0 and b is an algebraic integer with b^2 + b - 3 = 0, prove that both a + b and ab are algebraic integers.


    2. Relevant equations
    An algebraic number is said to be an algebraic integer if it satisfies an equation of the form x^m + c_{m-1}x^{m-1} + ... + c_0 = 0, where the c's are integers.


    3. The attempt at a solution
    Since the algebraic numbers form a field, ab and a + b satisfy some polynomial of the form c_mx^m + ... + c_0, where the c's are integers and m <= 6. The problem here is that c_m may not equal 1. I don't know how to get around this. Any tips?
     
  2. jcsd
  3. Mar 30, 2009 #2

    Hurkyl

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    You could compute it.
     
  4. Mar 30, 2009 #3
    What do you mean compute it? Are you implicitly suggest that c_m will always be 1?
     
  5. Mar 30, 2009 #4

    matt grime

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    No, he's suggesting that you explicitly compute a polynomial that is monic that has a+b as a root (and one for ab). It is obviously not true that c_m will always be 1, but it is true that c_m can be 1 if you do it properly.
     
  6. Mar 30, 2009 #5
    OK. I guess the next question is how do I compute such a thing. Expanding (a+b)^6 + c_5(a+b)^5 + ... + c_0 and then guessing at the possible c's doesn't seem like the smart way to do this.
     
  7. Mar 30, 2009 #6

    Hurkyl

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    Then use a non-guessing method to solve the equation.
     
  8. Mar 30, 2009 #7
    Ah, of course! ... But I don't know of any non-guessing method.
     
  9. Mar 30, 2009 #8

    matt grime

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    Find a monic poly with sqrt(2)+sqrt(3) as a root.
     
  10. Mar 30, 2009 #9
    Why sqrt(2) + sqrt(3), may I ask?
     
  11. Mar 30, 2009 #10

    matt grime

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    Why not? If you don't understand a more general case try an example. This is the simplest example I could think of.
     
  12. Mar 30, 2009 #11
    Good idea. You should have mentioned that in the first place: Let u = (sqrt{2} + sqrt{3})^2 and then solve u^2 + bu + c = 0 to get that x^4 - 10x^2 + 1.

    So with my problem, I let u = (alpha + beta)^3 and solve u^2 + bu + c = 0 ... I don't know any integers b and c that will make this work.
     
  13. Mar 31, 2009 #12

    matt grime

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    Probably because there aren't any. But why would u have to satisfy a quadratic equation? Try thinking about (a+b)^6 and smaller powers of a+b (i.e. not u).


    Incidentally, why should *I* have to tell you that if you're struggling with a problem then you should try an easier example? Bye (as in that's the end of my contribution).
     
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