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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


    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
    This step is wrong.

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


    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) =

    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
    I think it's ok.
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