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Roots of Cubic Polynomials over R

  1. Dec 28, 2011 #1
    I'm trying to prove the following, which is left unproven in something I'm reading on ruler-and-compass constructions:

    If [itex]ax^3+bx^2+cx+d[/itex] is a polynomial over a subfield F of ℝ, and [itex]p+q\sqrt{r}[/itex] is a root (with [itex]\sqrt{r}\notin F[/itex]) then [itex]p-q\sqrt{r}[/itex] is also a root.

    The theorem immediately before this made use of expanding [itex]a(x-r_1)(x-r_2)(x-r_3)[/itex] and equating coefficients to get a system of three equations in the roots and coefficients. The book suggested that this proof was similar, but I was't able to derive anything useful from that.

    I also tried defining [itex]g(t)=f(p-t\sqrt{r})-f(p+t\sqrt{r})[/itex], and noting that [itex]g(q)=f(p-q\sqrt{r})[/itex] and [itex]-g(-t)=g(t)[/itex].

    Then we see that

    [itex]f(p-q\sqrt{r})=g(q)=-g(-q)=-f(p+q\sqrt{r})=0[/itex].

    However, I know this must be flawed, because I don't believe I ever used the assumption that [itex]\sqrt{r}\notin F[/itex]. Any thoughts?
     
  2. jcsd
  3. Dec 28, 2011 #2

    micromass

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    This step is wrong.

    Basically, you know that you can write your polynomial in the form

    [tex]a(x-p-q\sqrt{r})(x-r_2)(x-r_3)[/tex]

    What do you get when you work that out??

    For example, you must get that

    [tex]-(p+q\sqrt{r})r_2r_3\in F[/tex]

    Do you see why?? Does that imply something for r2 and r3??
     
  4. Dec 28, 2011 #3
    I think I got it but I just want to check.

    From the first and third coefficient equations, we have [itex]\prod r_i\in F[/itex] and [itex]\sum r_i\in F[/itex]. Therefore, [itex]r_1r_2=f(p-q\sqrt r)[/itex] and [itex]r_1 + r_2 = t-q\sqrt r[/itex].

    Then we use the middle equation to get

    [itex]r_{2}r_{3}+r_{1}r_{3}+r_{1}r_{2} =
    \left(r_{2}+r_{1}\right)r_{3}+r_{1}r_{2} =
    \left(t-q\sqrt{r}\right)\left(p+q\sqrt{r}\right)+x\left(p-q\sqrt{r}\right) =
    pt-q^{2}r+xp+\left(t-x-p\right)q\sqrt{r}[/itex]

    Therefore, [itex]t-x-p = f\sqrt r[/itex]. Since all three terms are in F, it must be that f = 0. This gives us t = x+p, so [itex]r_1 + r_2 = x+p-q\sqrt r[/itex].

    Of course, the unique pair of numbers with that sum and product are x and [itex]p-q\sqrt r[/itex]
     
  5. Dec 28, 2011 #4

    micromass

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    I think it's ok.
     
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