I want to see if my reasoning is correct. I am given a list o field extensions and asked to determine which ones are normal extensions:
(a) $\mathbb{Q}(\alpha) : \mathbb{Q}$ where $\alpha$ is the real seventh root of 5
(b) $\mathbb{Q}(\sqrt{-5}) : \mathbb{Q}$
(c) $\mathbb{Q}(\sqrt{5}, \alpha) : \mathbb{Q}$ where $\alpha$ is as described in (a)
(d) $\mathbb{R}(\sqrt{-7}) : \mathbb{R}$
(e) $\mathbb{Q}(t) : \mathbb{Q}$
For extensions (b) and (d), the argument seems pretty straightforward: the splitting polynomial for (b) is $x^2 + 5$, and the extension $\mathbb{Q}(\sqrt{-5}) : \mathbb{Q}$ contains all the roots of the polynomial, so the extension is normal. For (d), the splitting polynomial is (I think) $x – \sqrt{-7}$, and the extension $\mathbb{R}(\sqrt{-7}) : \mathbb{R}$ contains all the roots of the polynomial, so the extension is normal (there is only one root, since square roots are in the field of real numbers).
For (a), I noted that since the splitting polynomial $(x^7 – 5)$ have multiple roots in the complex field, the extension cannot be normal. I'm not really sure where to start with (c) or (e), particularly (e) because it seems like I would need more information about $t$ to know anything about splitting polynomials, roots, etc.
Best Answer
For c), the same argument that for $a)$ works: $\mathbb{Q}(\sqrt{5},\alpha)\subset \mathbb{R}$, so you won't get the non-real conjugates of $\alpha$ (which are $\alpha$ times some $7^{\text{th}}$ root of unity).
For e), the extension is not even algebraic, so automatically not normal (it's normal that you cannot find the minimal polynomial of $t$ because there is none, $t$ is transcendental).