Field Extensions - Dummit and Foote Chapter 13 - Exercise 2, page 519

[Math Amateur]

Dummit and Foote Chapter 13, Exercise 2, page 519 reads as follows:

"Show that $$x^3 - 2x - 2$$ is irreducible over $$\mathbb{Q}$$ and let $$\theta$$ be a root.

Compute $$(1 + \theta ) ( 1 + \theta + {\theta}^2)$$ and $$\frac{(1 + \theta )}{ ( 1 + \theta + {\theta}^2)}$$ in $$\mathbb{Q} (\theta)$$

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My attempt at this problem so far is as follows:

$$p(x) = x^3 - 2x - 2$$ is irreducible over $$\mathbb{Q}$$ by Eisenstein's Criterion.

To compute $$(1 + \theta ) ( 1 + \theta + {\theta}^2)$$ I adopted the simple (but moderately ineffective) strategy of multiplying out and trying to use the fact that $$\theta$$ is a root of p(x) - that is to use the fact that $${\theta}^3 - 2{\theta} - 2 = 0$$.

Proceeding this way one finds the following:

$$(1 + \theta ) ( 1 + \theta + {\theta}^2) = 1 + 2{\theta} + 2{\theta}^2 + {\theta}^3$$

$$= ({\theta}^3 - 2{\theta} - 2) + (2{\theta}^2 + 4{\theta} + 3)$$

$$2{\theta}^2 + 4{\theta} + 3$$

Well, that does not seem to be going anywhere really! I must be missing something!

Can someone please help with the above and also help with the second part of the question ...

Peter

[Note: The above has also been posted on MHF]

 

[Deveno]

Your answer to the first question is correct. The fact that:

$\theta^3 = 2\theta + 2$ serves to "knock down" any powers of $\theta$ higher than 2 in $\Bbb Q(\theta)$.

To compute the quotient, what we need to do is compute the multiplicative inverse of $1 + \theta + \theta^2$ in $\Bbb Q(\theta)$. The easiest way to do this is to compute the gcd of $x^3 - 2x - 2$ and $x^2 + x + 1$ using the division algorithm:

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

$x^2 + x + 1 = (-2x - 1)\left(-\dfrac{x}{2} - \dfrac{1}{4}\right) + \dfrac{3}{4}$

Therefore:

$1 = \left(\dfrac{4}{3}\right)\left(\dfrac{3}{4}\right) = \left(\dfrac{4}{3}\right)\left(x^2 + x + 1 + (-2x - 1)\left(\dfrac{x}{2} + \dfrac{1}{4}\right)\right)$

$= \left(\dfrac{4}{3}\right)\left(x^2 + x + 1 + [x^3 - 2x - 2 - (x - 1)(x^2 + x + 1)]\left(\dfrac{x}{2} + \dfrac{1}{4}\right)\right)$

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

Taking this last equation mod $x^3 - 2x - 2$, we see that in $\Bbb Q(\theta)$:

$\dfrac{1}{1+\theta+\theta^2} = \frac{1}{3}(5 + \theta - 2\theta^2)$.

Now, multiply the two.