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Algebraic Integers

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    if a is an algebraic number satisfying a^3+a+1 = 0 and b is an algebraic number satisfying b^2+b-3 = 0 prove that both a+b and ab are algebraic

    2. Relevant equations



    3. The attempt at a solution

    a is root of equation x^3+x+1 = 0 and similarly b, so there exists a x = ab,and also x = a+b. so, ( x- ab)(x - a+b) is the required one.

    Its from I N Herstein's Topics in Algebra 5.1.13
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Dick

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    ( x- ab)(x - a+b) is the required one what? It may be a polynomial that has x=ab as a root. But to show ab is algebraic you have to find a polynomial with integer coefficients that has ab as a root.
     
  4. Oct 18, 2009 #3
    Dick, thanks for the reply.

    Sorry I am confused between polynomial and algebraic.
     
  5. Oct 18, 2009 #4

    Dick

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    I just told you the difference. An algebraic number is the root of a polynomial with integer coefficients. sqrt(2) is a root of x^2-2=0. That shows it algebraic. The coefficients of the polynomial are integers. x=pi is a root of x^2-pi^2=0. That doesn't show it's algebraic. pi^2 isn't an integer.
     
  6. Oct 18, 2009 #5
    a, b are integers, a+b and ab also. so.

    (x- a-b)*(x-ab) = x^2 - x*(ab+a+b) - ab*(a+b) = f(x)

    f(ab) = f (a+b) = 0, so, ab, a+b satisfies polynomial f(x) and hence, ab and a+b are algebraic.
     
  7. Oct 18, 2009 #6

    Dick

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    If you were given that a and b are integers, that would be fine. You aren't given that a and b are integers. So it's not fine. I think you'd better go back and review the definition of 'algebraic' and any examples about them in your book.
     
  8. Oct 18, 2009 #7
    hey, its given as algebraic integers, I typed that as number.

    Thanks for your replies.
     
  9. Oct 18, 2009 #8
    but I couldn't find any integer satisfying b^2+b-3 = 0, but Herstein had mentioned that b is an Algebraic integer.
     
  10. Oct 18, 2009 #9

    HallsofIvy

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    Yes, your problem says that a and b are "algebraic integers". It does NOT follow that a and b are integers. Do you know the definition of "algebraic integer"?
     
  11. Oct 18, 2009 #10
    yea, all ai s should be integers, but not necessarily roots right.
     
  12. Oct 19, 2009 #11
    I think this should do.

    f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0.

    h(a+b) = ci*xi. This should be zero since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., a^i*b^i = a^j*(a^3+a+1)*b^k*( b^2+b-3) = 0. Similarly k(ab) = 0
     
  13. Oct 19, 2009 #12

    Office_Shredder

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    You need to work on expressing yourself clearly. What are the xis? What is h(x), and why is h(a+b)=ci*xi (the c's here aren't defined either). And then you have k(ab) without saying what k is!
     
  14. Oct 19, 2009 #13
    Say f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0. [Given]

    There exists , h(x) a polynomial with in the form of Ci*Xi where Cis are integer coefficients and h(a+b) = 0. since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., for each a^i*b^i = a^j*(a^3+a+1)*b^k*( b^2+b-3) = 0. Similarly there exists a polynomial k(x) with integer coefficients such that k(ab) = 0.
     
  15. Oct 19, 2009 #14
    Office thanks for your reply.

    More over, h(x) will be of degree 6 since, a and b are of degree 3,2 respectively with gcd 1. so, a+b will be of degree 6.
     
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