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

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Here $R=\mathbb R$? And $u$ is an indeterminate? Then this is a transcendental extension. –  Hagen von Eitzen Nov 7 '12 at 18:56
Yes, R is the reals and u is an indeterminate. Is it a Galois extension too? –  Mathematician_in_Training Nov 7 '12 at 19:04
You need to stop posted your PMATH 442 assignment question on the stack exchange :P Anyways, what Hagen von Eitzen says is incorrect. By our definition in class, a FINITE extension is Galois extension if and only if it is normal and separable. However, $\mathbb{R}(u)/\mathbb{R}$ is not a finite extension. Therefore, you can't use that theorem. Edit: I'm stuck too but I think you're suppose to show that Fix(Gal($\mathbb{R}(u)/\mathbb{R}))=\mathbb{R}$. What Joel Cohen wrote is correct. So write that down. –  user44322 Nov 8 '12 at 17:14
What have you tried? If this is homework, you need to tag your post as such; people will still answer, don't worry. –  Najib Idrissi Nov 8 '12 at 17:23

2 Answers 2

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 $\lambda(x)$).

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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.

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doesn't this show that it's not a Galois extension? –  Mathematician_in_Training Nov 8 '12 at 2:50
This is not what normality means. A finite (or algebraic, I don't remember) extension is normal iff it's the splitting field of a polynomial. $\mathbb{R} \hookrightarrow \mathbb{R}(u)$ is not finite (or even algebraic). You need to use the actual definition of normality in this case. –  Najib Idrissi Nov 8 '12 at 17:20
@nik: That's why I said family of polynomials instead of polynomial. That definition agrees with both Wikipedia and Serge Lang. –  Hagen von Eitzen Nov 8 '12 at 17:52
@Mathematician_in_Training: Yes, that's what it shows. –  Hagen von Eitzen Nov 8 '12 at 17:52
I'm reading Lang's algebra right now, he specifically deals with algebraic extensions. And if you read Wikipedia, there's a disagreement between the editors; it seems to me that the author of the current version copied Lang's book but forgot the "algebraic" adjective. –  Najib Idrissi Nov 8 '12 at 20:24

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