MHB Basic Theory of Field Extensions - Exercise from D&F ....

[Math Amateur]

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 of Section 13.1 ... ...

Exercise 1 reads as follows:
View attachment 6597
My attempt at a solution is as follows:$$p(x) = x^3 + 9x + 6$$ is irreducible by Eisenstein ... ...Now consider $$(x^3 + 9x + 10) = (x + 1) ( x^2 - x + 10) $$
and note that $$(x^3 + 9x + 10) = (x^2 + 9x + 6) + 4$$ ... ...

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?

... 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 ... ...
Can someone please help to clarify this issue ... ...Peter

 

[Euge]

Hi Peter,

Your answer is correct. The irreducibility of $p$ ensures that $\Bbb Q(\theta) \neq \Bbb Q$. The inverse you found lies in $\Bbb Q(\theta)$, not $\Bbb Q$.

 

[Math Amateur]

Hi Peter,

Your answer is correct. The irreducibility of $p$ ensures that $\Bbb Q(\theta) \neq \Bbb Q$. The inverse you found lies in $\Bbb Q(\theta)$, not $\Bbb Q$.

Thanks Euge ... appreciate the help ...

Peter