Dummit and Foote Problem 13.4.1 (in Section 13.4 which simply proves the existence and very basic properties of splitting fields and algebraic closures) asks me to
Determine the splitting field and its degree over $ \Bbb{Q} $ for $ f(x) = x^4 – 2 $.
Additional Context and my actual question: I'm well aware that the splitting field is $ \Bbb{Q}(\sqrt[4]{2},i) $ with degree $ 8 $ over $ \Bbb{Q} $ but would like to prove it "properly" and from the basic principles of field extension theory discussed so far in the text. I am loath [edit, not loathe] to use any reference to embedding in $ \Bbb{R} $ or $ \Bbb{C} $ since the fundamental theorem of algebra is unproven by this point in the text and at any rate seems too powerful and not in the spirit of this question or section. I propose a solution below. Please read it and tell me if I missed something far simpler.
What I've done so far: $ f(x) $ is Eisenstein at $ 2 $ so is irreducible. Let $ \alpha $ be a root. In the spirit of doing things "the right way" I'd like to insist on not calling it $ \sqrt[4]{2} $ so as to avoid smuggling in any notion of embedding into $\Bbb{R}$ or $\Bbb{C}$. Now $ \Bbb{Q}(\alpha) $ is a degree 4 extension over $ \Bbb{Q} $ containing two roots $ \alpha $ and $ -\alpha $ of $ f(x) $. I need to ascertain that this extension does not contain any more roots. I can factor $ f(x) = (x-\alpha)(x+\alpha)(x^2 + \alpha^2) $, so my task would be to check that $ x^2 + \alpha^2 $ is irreducible in $ \Bbb{Q}(\alpha) $. In fact if I can show $ x^2 + 1 $ is irreducible then this result would follow since, if $ \beta $ is a root of $ x^2 + 1 $ then $ \alpha \beta $ would be a root of $ x^2 + \alpha^2 $. But it's not immediately clear why $ x^2 + 1 $ should be irreducible in $ \Bbb{Q}(\alpha) $ (again, I emphasize, using purely algebraic notions and avoiding imbedding into $ \Bbb{C} $).
After several attempts, here's the simplest argument I could come up with: consider the degree 2 extension $ E = \Bbb{Q}(\alpha^2) $. In $ E $, $ f(x) $ factors
$$
f(x) = (x^2 – \alpha^2)(x^2 + \alpha^2)
$$
and we know these factors are irreducible (or else it would contradict that $ \alpha $ is degree $4$). Letting $ K $ be the splitting field of $ f(x) $ over $E$, it suffices to show that $ K $ is a biquadratic extension of $ E $. I've solved this problem
Problem 13.2.8: Let $ F $ be a field of characteristic $ \neq 2 $. Let $ D_1, D_2 \in F $ where none of $ D_1 $, $ D_2 $, and $ D_1 D_2 $ are perfect squares. Then $ F(\sqrt{D_1}, \sqrt{D_2}) $ is a degree 4 extension, called a biquadratic extension.
I can apply this with $ F = E = \Bbb{Q}(\alpha^2) $, $ D_1 = \alpha^2 $, $ D_2 = -\alpha^2 $. Indeed $ D_1 D_2 = -2 $ is readily checked (by direct computation) to not be a perfect square in $ E = \{ a + b \alpha^2 \mid a,b \in \Bbb{Q} \} $. I conclude that
$$
K = E(\alpha, i\alpha) = \Bbb{Q}(\alpha^2)(\sqrt{D_1}, \sqrt{D_2})
$$
is a biquadratic extension of $ E = \Bbb{Q}(\alpha^2) $ and conclude that the splitting field of $ x^4 – 2 $ is a degree $ 8 $ extension of $ \Bbb{Q} $ generated by $ \alpha $ (a root of $ x^4 – 2 $) and $ \beta $ (a root of $ x^2 + 1 $).
A meta-question, if you'll indulge me: Proving Fundamental Theorem of Algebra is not hard to do with a bit of topology. I could easily write down a proof using winding number. For a student like me, what should be my philosophy toward using embedding into $ \Bbb{C} $ to think about and solve problems in field extension theory? On the one hand, using overly-powerful machinery may cause me to miss more general techniques and subtleties of Galois theory (such as over finite fields). On the other hand, $ \Bbb{C} $ supplies such rich intuition. A great example of this is that the embedding into $ \Bbb{C} $ makes it "obvious" that there exists a primitive $ n^{\text{th}} $ root of unity $ \zeta_n $ which we can identify with $ \exp(i\tau /n) $, whereas proving this purely algebraically is not-at-all straightforward. It also would have made this problem so much easier. Was this even a useful exercise, working so hard to avoid $ \Bbb{C} $, or was my intution (that I should even look for such a solution) misguided?
Best Answer
Here's my philosophy: to solve this problem, one shouldn't use facts that haven't been proven yet in the text. You especially shouldn't use any facts from other fields of math --- you're just hindering your algebraic development. However, a student should be able to use any intuition they've so far acquired in their learning path. So, here's a solution I believe the average student at this level has the understanding to come up with.
Starting from here, I believe it's reasonably intuitive to attempt to develop the notion of a conjugate in $\mathbb{Q}(\alpha)$. So, define $$\overline{a + b\alpha + c\alpha^2 + d\alpha^3} := a - b\alpha + c\alpha^2 - d\alpha^3.$$ I've simply sent $\alpha$ to $-\alpha$, so it's completely analogous to the familiar concept of complex conjugation. Then it can be checked that $z^2 + 1 =: f(z) = 0$ iff $f(\bar{z}) = 0$ by purely computational means. Then, assuming $f$ is reducible in $\mathbb{Q}(\alpha)$, $f(x) = (x - a - b\alpha - c\alpha^2 - d\alpha^3)(x - a + b\alpha - c\alpha^2 + d\alpha^3)$. Again, by pure computation
$$x^2 + 1 = f(x) = x^2 + (-2a - 2c\alpha^2)x + a^2 + 2c^2 - 4bd + (2ac - b^2 - 2d^2)\alpha^2$$
and so $c\alpha^2 = a$, from which $a = c = 0$ since $\alpha$ is of degree $4$. Then $b^2 + 2d^2 = 0$ from which $b = d = 0$ since $b, d$ are rational. But that's clearly not a root.
The ideas here are related to Galois theory, but I think that's the point of the exercise, to get you to develop some introductory approaches by yourself. Properties of $\mathbb{C}$ aren't actually utilized, they just serve as motivation. We could've also applied the same technique to the polynomial $x^2 + \alpha^2$, so your simplification wasn't strictly necessary, although it's perhaps easier to spot the method of finding the conjugate when the polynomial has purely rational coefficients.