# Thread: Field Extensions - Dummit and Foote Chapter 13

1. ## Field Extensions - Dummit and Foote Chapter 13

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

2. ## Re: Field Extensions - Dummit and Foote Chapter 13

Don't just try to simplify immediately. First, find more relationships. Note that you can reduce any term of $\theta^n$ for $n>2$ to an expression where the highest power of $\theta$ is 2.

$\theta^3 = 2\theta+2$
$\theta^4 = 2\theta^2 + 2\theta$
\begin{align*}\theta^5 & = 2\theta^3 + 2\theta^2 \\ & = 2(2\theta+2) + 2\theta^2 \\ & = 2\theta^2 + 4\theta+4\end{align*}

Also note that if you do this for $\theta^6$, the expression you get is exactly the one you would get from $(\theta^3)^2$. So, these expressions become exhaustive fairly quickly.

Next, remember that you are using the rationals, so you need to figure out inverse powers of theta, too.

$2 = \theta^3-2\theta$ so
$\theta^{-1} = \dfrac{\theta^2-2}{2}$
\begin{align*}\theta^{-2} & = \dfrac{\theta}{2} - \theta^{-1} \\ & = \dfrac{2 + \theta -\theta^2}{2}\end{align*}

Again, these expressions become exhaustive fairly quickly.

Does that help at all?

3. ## Re: Field Extensions - Dummit and Foote Chapter 13

To offer a bit more assistance, $1+\theta = \dfrac{\theta^3}{2}$ and $\theta^2 = \dfrac{\theta^5-2\theta^3}{2}$. So, adding those together, we get: $1+\theta+\theta^2 = \dfrac{\theta^3}{2} + \dfrac{\theta^5-2\theta^3}{2} = \dfrac{\theta^5-\theta^3}{2} = \dfrac{\theta^3}{2}(\theta^2-1) = (1+\theta)(\theta^2-1)$. This will help with the second part of your problem.

4. ## Re: Field Extensions - Dummit and Foote Chapter 13

For the second part of the problem, first find the multiplicative inverse of

$\theta ^2+\theta +1$ in $Q(\theta )$

This can be done by using the division algorithm to find f(x) and g(x) such that

$\left(x^3-2x-2\right)f(x)+\left(x^2+x+1\right)g(x)=1$

5. ## Re: Field Extensions - Dummit and Foote Chapter 13

Thanks SlipEternal, most helpful however I need a bit more help:

Can you explain how you proceed in the following:

$2 = {\theta}^3 - 2 \theta \ \Longrightarrow \ {\theta}^{-1} = \frac{{\theta}^2 - 2}{2}$

How does the above follow?

Peter

6. ## Re: Field Extensions - Dummit and Foote Chapter 13

Originally Posted by Bernhard
Thanks SlipEternal, most helpful however I need a bit more help:

Can you explain how you proceed in the following:

$2 = {\theta}^3 - 2 \theta \ \Longrightarrow \ {\theta}^{-1} = \frac{{\theta}^2 - 2}{2}$

How does the above follow?

Peter
Divide both sides by $\theta$ then divide both sides by 2.