Computation of $\frac{1+t}{1+t+t^2}$ for $x^3-2x-2$ in $\mathbb{Q}(t)$.

Question

Show that $x^3-2x-2$ is irreducible over $\mathbb{Q}$. Let $t$ be a root of this polynomial, then compute $$\frac{1+t}{1+t+t^2}$$ in $\mathbb{Q}(t)$.

My attempt: It is easy to show that given polynomial is irreducible over $\mathbb{Q}$ by Eisenstein's Criterion with $p=2$. Now for the second part....by applying Euclidean Algorithm we get that $x^3-2x-2=(1+x+x^2)(x-1)-2x-1$. By putting $x=t$ we get $1+t+t^2=(2t+1)/(t-1)$. Since $t$ is a root then $t^3-2t-2=0$ implies $(1+t)=t^3/2$. Consequently, our value of $(1+t)/(1+t+t^2)$ is $t^3(2t+1)/2(t-1)$ which is not correct according to the answer. The answer is given as a quadratic polynomial in $t$. Please help me to solve this in good way.

Answer 1

You need to use the extended Euclidean Algorithm to get $$ 1 = \frac13(2 x + 1)(x^3-2x-2) + \frac13(-2 x^2 + x + 5)(1+x+x^2) $$ which implies $$ \frac{1+t}{1+t+t^2} = \frac13(1+t)(-2 t^2 + t + 5) $$ Now expand the RHS and use $t^3=2t+2$ to reduce it to a quadratic.

Answer 2

If you apply the Euclidean Algorithm you get

$$(2x+1)(x^3-2x-2)=(2x^2-x-5)(x^2+x+1)+3$$

(in which I probably made a mistake) and this gives $$0=(2t^2-t-5)(t^2+t+1)+3$$ and so

$$\frac{1}{1+t+t^2}=\frac{1}{3}(5+t-2t^2)$$

Answer 3

You dont need Euclidean algorithm here. Because $t$ is a root of the given polynomial you know already that $t^3-2t-2=0$ which gives you $t^3=2t+2$. And so this gives you a way to reduce all polynomials of degree more than two to lower degree. Now if you write $$ \frac{1+t}{1+t+t^2}=a+bt+ct^2 $$ And multiply this out and compare coefficients in both sides of the equations you get a linear system of three equations that will give you a correct answer i.e. $a=1/3$, $b=2/3$ and $c=-2/3$.