Adjoints

Context?

I am adjoining the rational field to those numbers

Aha, I should have looked at it more closely before asking. The title is misleading - this has nothing to with adjoints. I was thinking that this was going to be some kind of adjoint function and from that point of view it wouldn't make any sense.

Also, you are adjoining those numbers to the rationals - not the other way round.

Anyway, to answer the question, it is technically correct to say that:

$\mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+j \omega^2+e\omega\alpha+f\omega \alpha^2+g\omega^{2}\alpha+h \omega^{2}{\alpha^2}\mid a,b,c,d,e,f,g,h,j \in \mathbb{Q}\}$

but it is also misleading and even disingenuous to put it like that since the extension is only of degree six, not nine.

I only put it like that because that is the way I think. I like to think concretely, and by putting it like that, I can see what that actually is. But ok, so it is technically correct, and thats all I need I guess :P

Well, presumably this example arises in the context of working out the Galois group or something and in that case, or indeed any case, it is much better to work with a $\mathbb{Q}$-basis, it really will simplify and clarify anything you want to do with it.

I mean, it is technically correct to say that

$\mathbb{R}^2= \{a(1,0)+b(3,5)+c(12,333)+d(\pi,13.67)+e(\frac{\mathrm{sin}(34)}{27.8888},3)+f(0,1)\mid a,b,c,d,e,f \in \mathbb{R}\}$

but you aren't really specifying $\mathbb{R}^2$ any more clearly than the obvious alternative.

Or for an example closer to the spirit of the example, imagine wiriting:

$\mathbb{C}=\{a + bi + ci^2 + di^3 \mid a,b,c,d \in \mathbb{R}\}$

Hmm. ok, i get that in your exmaples, you have redundant terms. So in my expression, which ones are the redundant terms? And how would you write it?

Like in your C example, ci^2 = -c, which I guess can be 'absorbed' by the a, making it redundant. Which terms of mine can be absorbed?

$\omega$ is a primitive cube root of unity so in particular:

$1+\omega +\omega^2 = 0$

i.e. $\omega^2 = -1-\omega$

So it would be more appropiate to write this:

$[br]\mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+e \omega\alpha+\omega\alpha^2+f\mid a,b,c,d,e,f \in \mathbb{Q}\}[br]$