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:\(\displaystyle p(x) = x^3 + 9x + 6\) is irreducible by Eisenstein ... ...Now consider \(\displaystyle (x^3 + 9x + 10) = (x + 1) ( x^2 - x + 10) \)
and note that \(\displaystyle (x^3 + 9x + 10) = (x^2 + 9x + 6) + 4\) ... ...

Now \(\displaystyle \theta\) is a root of \(\displaystyle (x^3 + 9x + 6)\) so that ...\(\displaystyle ( \theta + 1) ( \theta^2 - \theta +10) = ( \theta^3 + 9 \theta + 6) + 4 = 0 +4 = 4 \)Thus \(\displaystyle ( \theta + 1)^{-1} = \frac{ ( \theta^2 - \theta +10) }{4}\) ... ...Is that correct?

... BUT ... if it is correct I am most unsure of exactly where in the calculation we depend on \(\displaystyle p(x)\) being irreducible ...Can someone please explain where exactly in the above calculation we depend on \(\displaystyle p(x)\) being irreducible?Note that I am vaguely aware that we are calculating in \(\displaystyle \mathbb{Q} ( \theta )\) ... which is isomorphic to \(\displaystyle \mathbb{Q} [x] / ( p(x) )\) ... if \(\displaystyle p(x)\) is irreducible ...
... BUT ...I cannot specify the exact point(s) in the above calculation above where the calculation would break down if \(\displaystyle p(x)\) 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