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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Dummit and Foote Chapter 13, Exercise 2, page 519 reads as follows:

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

Compute [TEX] (1 + \theta ) ( 1 + \theta + {\theta}^2) [/TEX] and [TEX] \frac{(1 + \theta )}{ ( 1 + \theta + {\theta}^2)} [/TEX] in [TEX] \mathbb{Q} (\theta)[/TEX]

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My attempt at this problem so far is as follows:

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

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

Proceeding this way one finds the following:

[TEX] (1 + \theta ) ( 1 + \theta + {\theta}^2) = 1 + 2{\theta} + 2{\theta}^2 + {\theta}^3 [/TEX]

[TEX] = ({\theta}^3 - 2{\theta} - 2) + (2{\theta}^2 + 4{\theta} + 3) [/TEX]

[TEX] 2{\theta}^2 + 4{\theta} + 3 [/TEX]

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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.