Compute in a field extension

Show that p(x) = x^3 - 2x - 2 is irreducible over \mathbb{Q}, and let \theta be a root of p(x). Compute (1+\theta)(1+\theta+\theta^2) and \dfrac{1+\theta}{1+\theta+\theta^2}.

p(x) is Eisenstein at 2, and so is irreducible over \mathbb{Q}.

In \mathbb{Q}(\theta), we have \theta^3 = 2\theta+2. We can easily see that (1+\theta)(1+\theta+\theta^2) = 3+4\theta + 2\theta^2.

Next we compute (1+\theta+\theta^2)^{-1}; to this end, we use the extended Euclidean Algorithm to find polynomials a(x) and b(x) in \mathbb{Q}[x] such that a(x)(x^2+x+1) + b(x)(x^3-2x-2) = 1; evidently, a(x) = \frac{1}{3}(-2x^2+x+5) and b(x) = \frac{1}{3}(2x+1) work. (WolframAlpha agrees.) Thus, mod p(x), we have (1+\theta+\theta^2)^{-1} = \frac{1}{3}(-2\theta^2 + \theta + 5). Now (1+\theta)(1+\theta+\theta^2)^{-1} = \frac{1}{3}(-\theta^2 + 2\theta + 1).

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