# Homework Help: Basic Theory of Field Extensions - Exercise from D&F ...

1. May 11, 2017

### Math Amateur

1. The problem statement, all variables and given/known data

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

I am currently studying Section 13.1 : Basic Theory of Field Extensions

I need some help with an aspect of Exercise 1 ... ...

2. Relevant equations

Relevant theorems for this exercise seem to be Theorems 4 and 6 of Section 13.1.

3. The attempt at a solution

$p(x) = x^2 + 9x + 6$ is irreducible by Eisenstein ... ...

Now consider $(x^2 + 9x + 10) = (x + 1) ( x^2 - x + 10)$

and note that $(x^2 + 9x + 10) = (x^2 + 9x + 6) + 4$ ... ...

Now $\theta$ is a root of $(x^2 + 9x + 6)$ so that ...

$( \theta + 1) ( \theta^2 - \theta +10) = ( \theta^2 + 9 \theta + 6) + 4 = 0 +4 = 4$

Thus $( \theta + 1)^{-1} = \frac{ ( \theta^2 - \theta +10) }{4}$ ... ...

Is that correct?

... BUT ... if it is correct I am most unsure of exactly where in the calculation we depend on $p(x)$ being irreducible ...

Can someone please explain where exactly in the above calculation we depend on $p(x)$ being irreducible?

Note that I am vaguely aware that we are calculating in $\mathbb{Q} ( \theta )$ ... which is isomorphic to $\mathbb{Q} [x] / ( p(x) )$ ... if $p(x)$ is irreducible ...

... BUT ...

I cannot specify the exact point(s) in the above calculation above where the calculation would break down if p(x) was not irreducible ... .. ... .. in fact, I cannot specify any specific points where the calculation would break down ... so I am not understanding the connection of the theory to this exercise ... ...

Peter

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• ###### D&F - Theorem 6, Ch 13 ... ....png
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2. May 11, 2017

### Staff: Mentor

If you already have Eisenstein, then yes. Which prime did you use?
You can easily check the result by multiplying $\theta \cdot (1+\theta)$.

On the first view, I wonder where the cube has gone? The minimal polynomial is $x^3 + \ldots$.
This means $\theta^3 +9 \theta + 6 = 0$ or $\theta^3 = -9\theta -6$. Therefore you can use $x^3=-9x-6$ in your method. Myself I did it more stupidly: $\theta \cdot (a\cdot 1 + b \cdot \theta + c \cdot \theta^2) = 1$ and then I used the above equation to get rid of $\theta^3$ and did a comparison of the coefficients of the basis vectors $\{1,\theta,\theta^2\}$ and solved the linear equation system for $a,b,c$.
Maybe less elegant than your method, but it works. Yours should work, too, with the correct power of $x$.

3. May 11, 2017

### Math Amateur

My apologies to readers of the above post ... especially apologies to fresh_42 ... there are some silly typos which I will now correct ...

(I wrote out the solution and then typed it in to the post making some errors ...)

The (hopefully) correct solution to the exercise is as follows ... but first I will provide the Exercise again ... it reads as follows:

Now ... my attempt at a solution follows ... with the typos corrected ...

=============================================================================================================

$p(x) = x^3 + 9x + 6$ is irreducible by Eisenstein ... ... use prime = 3

Now consider $(x^3 + 9x + 10) = (x + 1) ( x^2 - x + 10)$

and note that $(x^3 + 9x + 10) = (x^3 + 9x + 6) + 4$ ... ...

Now $\theta$ is a root of $(x^3 + 9x + 6)$ so that ...

$( \theta + 1) ( \theta^2 - \theta +10) = ( \theta^3 + 9 \theta + 6) + 4 = 0 +4 = 4$

Thus $( \theta + 1)^{-1} = \frac{ ( \theta^2 - \theta +10) }{4}$ ... ...

Is that correct? (I certainly hope that now it is correct!)

=================================================================================================================

... ... and now the questions I am REALLY interested in ...

... BUT ... if it is correct I am most unsure of exactly where in the calculation we depend on $p(x)$ being irreducible ...

Can someone please explain where exactly in the above calculation we depend on $p(x)$ being irreducible?

Note that I am vaguely aware that we are calculating in $\mathbb{Q} ( \theta )$ ... which is isomorphic to $\mathbb{Q} [x] / ( p(x) )$ ... if $p(x)$ is irreducible ...

... BUT ...

I cannot specify the exact point(s) in the above calculation above where the calculation would break down if p(x) was not irreducible ... .. ... .. in fact, I cannot specify any specific points where the calculation would break down ... so I am not understanding the connection of the theory to this exercise ... ...

Peter[/QUOTE]

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• ###### D&F - Exercise 1 - Section 13.1 ... ....png
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Last edited: May 11, 2017
4. May 11, 2017

### Math Amateur

Thanks for the help fresh_42 ... BUT ... my apologies for some silly typos ...

I wrote out my attempt before typing it in to the forum and made some typos when copying it ...

I have now corrected these in a new post which I hope you will find is correct ... many apologies ...

I have followed that by the questions that I am really interested in ... and puzzled by ...

Hope you can help ...

Peter

5. May 11, 2017

### Staff: Mentor

No need for apologies. Yes, it is correct now.
Nowhere directly. It is implicitely used by the fact that $\{1,\theta,\theta^2\}$ is a basis in which we can express $(1+\theta)^{-1}$. Otherwise we would have $p(x)=(x-a)q(x)$ with a quadratic polynomial $q(x)$ and an element $a \in \mathbb{Q}$. This way we would get $\mathbb{Q}(\theta) = \mathbb{Q}(\varphi)$ for a $\varphi$ with $q(\varphi)=0$ and $\{1,\varphi\}$ would be a basis of $\mathbb{Q}(\theta)$ and no quadratic terms would be needed.

Try the same exercise with $r(x)=x^3+2x^2-2x-4$ and a root $r(\theta)=0$ and see what happens. What is $(1+\theta)^{-1}$ in this case?

6. May 12, 2017

### Math Amateur

Thanks fresh_42 ...

$r(x)=x^3+2x^2-2x-4$ is reducible in $\mathbb{Q}$ since ...

... $r(x)=x^3+2x^2-2x-4 = (x + 2) ( x^2 - 2)$

BUT ... the approach I used before still seems to work!!! ... see below ... ( but I have the feeling that it should not work ...)

... now consider $( x^3 + 2 x^2 - 2x - 3 ) = (x + 1 ) ( x^2 + x - 3)$

Also note that $( x^3+2x^2-2x-4 ) = ( x^3+2x^2-2x-3 ) - 1 = (x + 1 ) ( x^2 + x - 3) - 1$

But we know $r( \theta ) = ( \theta^3 + 2 \theta^2 - 2 \theta - 4 ) = 0$

so that ... ...

$( \theta^3 + 2 \theta^2 - 2 \theta - 4 ) = 0 = ( \theta + 1) ( \theta^2 + \theta - 3 ) - 1$

Thus $0 + 1 = ( \theta + 1) ( \theta^2 + \theta - 3 )$

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

So, it appears that $r(x)$ being reducible makes no difference to the mechanics of finding the inverse of $( \theta + 1)$ ... doesn't it matter to this process that $r(x)$ is reducible ...???

Can you help explain how the theory of field extensions squares with the above ...?

Peter

7. May 12, 2017

### Staff: Mentor

Yes, it makes no difference and all three roots $\{-2,\pm \sqrt{2}\}$ satisfy this equation. It is simply not the shortest way to express them, as $\{1,\theta,\theta^2\}$ isn't a basis anymore: $\theta^2=2$ or $\theta =-2$ are in $\mathbb{Q}$ and $\theta^2+\theta -3 = \theta -1$ or $\theta^2+\theta -3 = \theta +1$. One can always add new vectors to a basis and get different linear combinations to the expense of losing linear independency and the uniqueness of expression.

The irreducibilty of $p(x)$ guaranteed, that $\{1,\theta,\ldots,\theta^{\deg p -1}\}$ is a basis. If it wasn't, then we would have a linear equation among them and $p(x)$ would be reducible.

We could also consider $\mathbb{Q}(x)/\langle x \rangle$ or $\mathbb{Q}(x)/\langle x - 1\rangle$, call $\theta$ a root of $r(x)=x$ or $r(x)=x-1$ and express rational numbers in terms of $\{1,\theta\}$. Would be a bit inconvenient, but why not?