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Toughie about multiple roots

  1. May 6, 2004 #1
    hello everyone, this is my first post.. I've got myself a problem.

    say I have a field F, and that f(x) is a polynomial with coefficients in F.
    (all the normal differentiation rules apply like always.. product rule, quotient rule, etc.) Now, I know that a polynomial f(x) has a "multiple root" if there exists some field E (containing F) and some element, alpha, of E such that (x-alpha)^2 divides f(x) in E[x].

    my questoin is:
    how can I prove that a non-constant polynomial f(x) has a multiple root if and only if f(x) is not relatively prime to its derivative?

    Thank you to anyone who even takes a look at this. :smile:
     
    Last edited: May 7, 2004
  2. jcsd
  3. May 6, 2004 #2

    NateTG

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    Well, let's say you have n degree polynomial:
    [tex]k(x-a_1)(x-a_2)...(x-a_n)=k \prod_{i=1}{n} (x-a_i)[/tex]

    Then if you apply the product rule you get
    [tex] k\sum_{i=1}^{n}\prod_{1\leq j \leq n, j \neq i}(x-a_j)[/tex]

    Now, consider which terms of the lower sum are divisible by [tex](x-a_k)[/tex], and what happens if any of the [tex]a_k[/tex] are repeated.
     
  4. May 6, 2004 #3

    matt grime

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    let (x-a) be a repeated factor of f(x), ie, by the division alogrithms, f(x)= (x-a)^2g(x).

    now differentiate.
     
  5. May 7, 2004 #4
    I'm sorry if this is a stupid question, but I thought the general equation of a polynomial was something more like:

    [tex](a_n)(x^n)+...+(a_3)(x^3)+(a_2)(x^2)+(a_1)(x)+a_0[/tex]

    what gives?? and where'd the "k" come from? :confused:
     
  6. May 7, 2004 #5

    matt grime

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    the other form is after you've factorized it (presuming the roots are in the field)
     
  7. May 7, 2004 #6

    NateTG

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    Well, what I gave is not exactly a general form for polynomials, but let's say we have a polynomial:
    [tex]p(x)=a_nx^n+a_{n-1}x^{n-1}...a_1x+a_0[/tex]

    With roots [tex]r_1.r_2,...r_j[/tex]
    then p is divisible by [tex](x-r_i)[/tex] so I can rewrite it as:
    [tex](x-r_1)(x-r_2)....(x-r_j) q(x)[/tex]
    where [tex]q(x)[/tex] doesn't have any roots in the field.
    If I assume that all of the roots are in the field, then [tex]q(x)=k[/tex]
     
  8. May 7, 2004 #7
    I wrote the polynomial as a product of linear factors and used the product rule. I think that worked ok.
     
  9. May 8, 2004 #8
    I've got it! Thanks everyone!!
     
  10. Sep 19, 2004 #9

    mathwonk

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    I think these solutions are not correct. I.e. the problem is to show that there is a multiple root in an extension field, if and only if the polynomial and its derivative have a common factor already in the given field.

    Thus the point is that if they are relatively prime over the given field, then there is an equation of form Af + Bf' = 1 holding in the given field. Then it follows that it holds also in any extension field and hence they cannot have a common root in any extension field, hence f cannot have a multiple root in any extension field.

    Conversely if f and f' have common prime factor h of degree at least one in the given field, then they are also not relatively prime in a splitting field, and hence f has a multiple root there as argued above.

    what do you think?
     
    Last edited: Sep 20, 2004
  11. Sep 21, 2004 #10

    mathwonk

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    i.e. the point which is a little subtle is that if two functions f,g over a field have a common linear factor in an extension field, then they must already have a common factor, possibly non linear, over the base field. This is not intuitively obvious to me, but the magic equation Af+Bg = 1 in the base field proves it.

    This is the clever fact, known to the greeks, that (in a euclidean domain) the greatest length that measures two given lengths (gcd), is also the least length they can measure together (smallest linear combination).

    this is intuitively true because, from their point of view, in which "divides" in euclid translates as "measures", the point is that you can measure anything with the least common linear combination of two things that you can measure with both of them.

    I.e. supose L can be measured with both A and B, but not with their smallest linear combination g. then the remainder of L after measuring with g, is smaller than g and can be measured with both A and B, a contradiction.

    this result which Always seemed clever to me in algebra, is a little less so in geometry.

    the point the greeks missed at first it seems, was that some pairs of lengths A,B, admit no smallest linear combination.

    i apologize for carrying on this moribund string, but it brings out a point which was always puzzling to me until i read the original discussion in euclid and noticed the word "measures" instead of "divides". this notion is broader than the one over the rationals, since it applies also to integral submodules of the reals generated by real numbers, something lost in modern elementary bowdlerizations.
     
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