How would I show that $\mathbb R(u)/\mathbb R$ is a Galois extension?

Question

Where $\mathbb R(u)$ is the rational function field

Answer 1

The usual definition of a Galois extension requires the extension to be algebraic, so according to this definition $\mathbb{R}(u)$ is not Galois. However, I'm guessing you are asked to prove that the field fixed by $\textrm{Gal}(\mathbb{R}(u)/\mathbb{R})$ is exactly $\mathbb{R}$.

To do that, just notice that for $x \in \mathbb{R}$, the map $\lambda(x) : F(u) \mapsto F(u+x)$ is an element of $\textrm{Gal}(\mathbb{R}(u)/\mathbb{R})$. Now assume $F \in \mathbb{R}(u)$ is fixed by all the $\lambda(x)$, then the real function $t \mapsto F(t)$ is constant, so $F \in \mathbb{R}$ (converserly constants are obviously fixed by the $\lambda(x)$).

Answer 2

A Galois extension is a field extension that is both separable and normal. Separability is always clear in charactaristic zero. Normality means that there is a family of polynomials such that $\mathbb R(u)$ is the splitting field of these polynomials, i.e. the smallest field where all these polynomials split into linear factors. The only nontrivial algebraic extension of $\mathbb R$ is $\mathbb C$, hence we would need $\mathbb C\subseteq \mathbb R(u)$ (or more precisely: $\mathbb C\hookrightarrow\mathbb R(u)$), especially, $\mathbb R(u)$ should contain an element $\frac{f(u)}{g(u)}$ (where $f,g$ are polynomials and $g\ne0$) with $\left(\frac{f(u)}{g(u)}\right)^2=-1$. But this is not the case: It would impliy $f(u)^2=-g(u)^2$ and plugging in any $\alpha\in \mathbb R$ for which $g(\alpha)\ne 0$ (such $\alpha$ exists becaue $g\ne0$) the right hand side becomes negative while the left hand side does not.