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Homework Help: Finding the units (number theory)

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that the ring R of polynomials with real coefficients (i.e. f(x) = a0 + a1x + ... + anxn, ai real, are elements of R) has only the constant term a0 as the group of units, providing the constant term isn't zero.


    2. Relevant equations

    u is a unit if there exists a v such that uv=1


    3. The attempt at a solution

    Clearly all the constant terms are units of the ring as we are dealing with the real numbers.

    Claim: There are more than just these, and that a polynomial f(x) = a0 + a1x + ... + anxn is invertible.

    => there must exist a g(x) = b0 + b1x + ... + bnxn

    such that f(x).g(x) = 1

    f(x).g(x) = b0(a0 + a1x + ...) + b1x(a0 + a1x + ...) + b2x2(a0 + a1x + ...) + ... = 1

    Not too sure where to go from here.. Sorry if my notation is a little off but hopefully you can understand

    Thanks
     
  2. jcsd
  3. Mar 5, 2010 #2
    R[x] is the ring of polynomials over a field: R. In this case, what is the degree of the product of two polynomials? What must be the degree of a unit?
     
  4. Mar 5, 2010 #3

    Dick

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    If f(x) and g(x) aren't constants, then each one has a highest degree term with a coefficient that's nonzero, right? What's the coefficient of the highest degree term of f(x)*g(x)?
     
  5. Mar 5, 2010 #4
    If f has degree n and g has degree m then the degree of the product is n+m which must be greater than or equal to 2, but the the degree of 1 is 0.

    Surely some polynomial can equal 1 though can't it?

    coefficient anbm, not too sure how this helps though
     
  6. Mar 5, 2010 #5

    Dick

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    If an*bm is not zero, then f(x)*g(x) contains a term of the form an*bm*x^(n+m) and that's the only term of that degree. So the product is probably not 1.
     
  7. Mar 5, 2010 #6
    But if these are real coefficients then maybe they're small enough so we can choose an x such that the product is equal to 1?
     
  8. Mar 5, 2010 #7

    Dick

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    "1" has coefficient 0 for all powers of x greater than 0. Not "close to zero". Zero. I'm having a hard time picturing what you are thinking about here. The reals don't have any zero divisors.
     
  9. Mar 5, 2010 #8

    Dick

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    "x" isn't a number. It's a symbol. This question isn't about roots of a polynomial. Is that what you are thinking?
     
  10. Mar 5, 2010 #9
    ye i was thinking that, guess that's why I'm having a hard time understanding

    I can just go by the degrees of 1 and the product of the polynomials are different then as above?

    Thanks
     
  11. Mar 5, 2010 #10

    Dick

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    Yes, you can. Two polynomials in the "ring of polynomials" are equal only if all of their coefficients are equal. Look back at the definition of "ring of polynomials". Nothing to do with whether they are equal for some particular value of x.
     
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