Roots of Cubic Polynomials over R

[alexfloo]

I'm trying to prove the following, which is left unproven in something I'm reading on ruler-and-compass constructions:

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

The theorem immediately before this made use of expanding a(x-r_1)(x-r_2)(x-r_3) 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 g(t)=f(p-t\sqrt{r})-f(p+t\sqrt{r}), and noting that g(q)=f(p-q\sqrt{r}) and -g(-t)=g(t).

Then we see that

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

However, I know this must be flawed, because I don't believe I ever used the assumption that \sqrt{r}\notin F. Any thoughts?

 

[micromass]

-g(-q)=-f(p+q\sqrt{r})

This step is wrong.

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

a(x-p-q\sqrt{r})(x-r_2)(x-r_3)

What do you get when you work that out??

For example, you must get that

-(p+q\sqrt{r})r_2r_3\in F

Do you see why?? Does that imply something for r₂ and r₃??

 

[alexfloo]

I think I got it but I just want to check.

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

Then we use the middle equation to get

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

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

Of course, the unique pair of numbers with that sum and product are x and p-q\sqrt r

 

[micromass]

I think it's ok.