# Thread: t^3-3t+1 splits or is irreducible

1. ## t^3-3t+1 splits or is irreducible

Prove that for any field K, $\displaystyle t^3-3t+1$ splits or is irreducible in K.

Hint: show that any zero is a rational expression in any other zero.

Using the formula for the sum of roots: $\displaystyle \sum \alpha_i = \frac{-a_{n-1}}{a_n}$ where $\displaystyle \alpha_i$ are the roots, and $\displaystyle a_i$ the coefficients of the i-th power term in the polynomial, we get $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$

Also using the formula for the product of the roots: $\displaystyle \prod \alpha_i=(-1)^n (\frac{a_0}{a_n}) \$ we get $\displaystyle \alpha_1\alpha_2\alpha_3=-1$

I only seem to be able to combine these equations, and the equations resulting from the fact that each $\displaystyle \alpha_i$ is a root of our polynomial to get equations of two variables where a power higher than 1 is present in each variable, and quadratic formulas don't seem to resolve those nicely. I think if I'm reading the problem right, we don't want any roots in our relating equation so I guess we need another rational expression (other than $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$) involving the three roots to combine with the previous to get a rational expression involving two roots.

Ideas?

2. Assuming this is the right path to go down, you could get a third equation from the coefficient of $\displaystyle t$, namely $\displaystyle \alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3 =-3$.

Now you have three equations and three unknowns. See if that does the trick. See here for more info.

3. Ok, using that information relating coefficients of our polynomial to symmetric polynomials in the roots I have the equations:

$\displaystyle 0=\alpha_1+\alpha_2+\alpha_3$

$\displaystyle -3=\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha _3$

$\displaystyle -1=\alpha_1\alpha_2\alpha_3$

or solving for $\displaystyle \alpha_1$ so I can directly relate two variables:
$\displaystyle \alpha_1=-\alpha_2-\alpha_3=-\frac{1}{\alpha_2 \alpha3}=\frac{-3-\alpha_2 \alpha_3}{\alpha_2+\alpha_3}$

When I manipulate those equations, I just seem to get things like $\displaystyle \alpha_2^n*\alpha_3^m*p(\alpha_2)=0$ where p is the original polynomial and this just confirms that my variables are roots ( I already know they're nonzero).

It's not enough to simply solve for one root in terms of another. I have to relate them using only add/subtr, mult/div (no radicals).

Am I perhaps starting off with the wrong approach?

4. The problem became much easier once I actually solved for the roots and got them in a good looking form, namely $\displaystyle \alpha_1=2cos(2\pi/9),\alpha_2=2sin(\pi/18),\alpha_3=-2cos(\pi/9)$
I ended up only using one of the prior formulas $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$ and the double angle formulas.