Inverse of $1+\theta$?

Question

I am trying to understand an example in Dummit and Foote on page 515. Let $F=\mathbb{Q}$ and $p(x)=x^3-2$, which is irreducible. Denote a root of $p(x)$ by $\theta$. We wish to find the inverse of the element $1+\theta$. By the Euclidean Algorithm in $\mathbb{Q}[x]$ there are polynomials $a(x),b(x)$ with $a(x)(1+x)+b(x)(x^3-2)=1$ (since $p(x)$ is irreducible, it is relatively prime to $a(x)$ which has a smaller degree). Thus, in the quotient field, $a(x)$ is the inverse of $1+\theta$.

The book says that a "simple calculation" shows that we can take $a(x)=1/3(x^2-x+1)$ and $b=1/3$, so that $(1+\theta)^{-1}=\frac{\theta^2-\theta+1}{3}$. My question is how to arrive at this calculation.

Answer 1

A good trick would be this : $$\frac{1}{1+\theta} = \frac{1}{1-(-\theta)}=\sum_{i=0}^\infty (-\theta)^i = 1+(-\theta)+(-\theta)^2 + (-\theta)^3\cdot \frac{1}{1+\theta}=1-\theta+\theta^2-2\cdot \frac{1}{1+\theta}$$

so $(1+\theta)^{-1}=\frac{1}{1+\theta} = \frac{1-\theta+\theta^2}{3}$ so you don't have to use the Euclidean algorithm.

Answer 2

In the general case, use the extended Euclidean algorithm.

In this case, just use polynomial division: $$ x^3 - 2 = (x^2 - x + 1)(x + 1) -3 $$ Now plug $x=\theta$ to get $$ 0 = (\theta^2 - \theta + 1)(\theta + 1) -3 $$