Field Theory _ Dummit and Foote - Example 4 - page 516 - Simple Computation

[Math Amateur]

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Example 4 [pages 515 - 516]

I need some help with what D&F call a simple computation.

Example 4 on pages 515-516 reads as follows:View attachment 2731
View attachment 2732

Now in the above example, D&F write the following:

" ... ... In this case, a simple computation shows that we can take \(\displaystyle a(x) = \frac{1}{3} ( x^2 - x + 1) \text{ (and } b(x) = - \frac{1}{3} )\) so that

\(\displaystyle {(1 + \theta)}^{-1} = \frac{ \theta^2 - \theta +1}{3} \) ... ... "Although I can verify that the expressions for a(x) and b(x) are correct I cannot see how D&F arrived at (or calculated) a(x) and b(x).

Can someone please help me by explaining the process by which one arrives at the explicit expressions for a(x) and b(x)?

Peter

***NOTE*** Some time ago Deveno helped me with the theory of this example, but we did not discuss the details of the computation mentioned above ... now that I am revising this example I am happy with the theory but cannot see how D&F did their computation ...

 

[Opalg]

The "simple computation" is Euclid's algorithm. What that says here is that to find $a(x)$ and $b(x)$ we need to compute the greatest common divisor of $x^3-2$ and $x+1$. To do that, divide $x^3-2$ by $x+1$, getting a quotient and a remainder: $$x^3 - 2 = (x+1)(x^2 - x + 1) - 3.$$ Then rearrange that as $-\frac13(x^3-2) + \frac13(x^2 - x + 1)(x+1) = 1.$

 

[Math Amateur]

The "simple computation" is Euclid's algorithm. What that says here is that to find $a(x)$ and $b(x)$ we need to compute the greatest common divisor of $x^3-2$ and $x+1$. To do that, divide $x^3-2$ by $x+1$, getting a quotient and a remainder: $$x^3 - 2 = (x+1)(x^2 - x + 1) - 3.$$ Then rearrange that as $-\frac13(x^3-2) + \frac13(x^2 - x + 1)(x+1) = 1.$

Thanks Opalg ... Appreciate the help

Peter

 

[Deveno]

There is another way to find $\dfrac{1}{1 + \theta}$ using the fact that:

$\theta^3 = 2$. (1)

We know such an inverse must be of the form $a + b\theta + c\theta^2$, since any higher powers of $\theta$ can be reduced using (1).

We thus have:

$1 = (1 + \theta)(a + b\theta + c\theta^2) = a + (a + b)\theta + (b + c)\theta^2 + c\theta^3$

$= (a + 2c) + (a + b)\theta + (b + c)\theta^2$, so that:

$a + 2c = 1$
$a + b= 0$
$b + c = 0$.

So $a = c$, and thus $a = \frac{1}{3}$, so that:

$\dfrac{1}{1 + \theta} = \dfrac{1 - \theta + \theta^2}{3}$

This (I hope) illustrates the utility of replacing the coset $x + (x^3 - 2)$ with a single symbol, $\theta$.

 

[Math Amateur]

There is another way to find $\dfrac{1}{1 + \theta}$ using the fact that:

$\theta^3 = 2$. (1)

We know such an inverse must be of the form $a + b\theta + c\theta^2$, since any higher powers of $\theta$ can be reduced using (1).

We thus have:

$1 = (1 + \theta)(a + b\theta + c\theta^2) = a + (a + b)\theta + (b + c)\theta^2 + c\theta^3$

$= (a + 2c) + (a + b)\theta + (b + c)\theta^2$, so that:

$a + 2c = 1$
$a + b= 0$
$b + c = 0$.

So $a = c$, and thus $a = \frac{1}{3}$, so that:

$\dfrac{1}{1 + \theta} = \dfrac{1 - \theta + \theta^2}{3}$

This (I hope) illustrates the utility of replacing the coset $x + (x^3 - 2)$ with a single symbol, $\theta$.

Thanks Deveno ... Appreciate the help and the insights ...

Peter