1. Non-algebraic Galois Extension

Say $\displaystyle E/F$ is Galois*.
Must it be the case that $\displaystyle E/F$ is algebraic?
(This was not addressed in my book).

*)I just realized how bad my question is. A Galois extension $\displaystyle E/F$ is an algebraic extension such that $\displaystyle F = E^{\text{Gal}(E/F)}$. Note we use "algebraic" within the definition itself. My question would be more appropriately asked if $\displaystyle F = E^{\text{Gal}(E/F)}$ then must it follow that $\displaystyle E/F$ is algebraic?

2. Originally Posted by ThePerfectHacker
My question would be more appropriately asked if $\displaystyle F = E^{\text{Gal}(E/F)}$ then must it follow that $\displaystyle E/F$ is algebraic?
this is a good question! the answer is No! because, for example, if $\displaystyle x$ is transcendental over $\displaystyle F$ and $\displaystyle E=F(x)$, then

it can be proved (not very easily though!) that $\displaystyle E^{\text{Gal}(E/F)}=F.$ the main part of the proof is to show that $\displaystyle \text{Gal}(E/F)$

consists of all maps $\displaystyle \sigma$ defined by $\displaystyle \sigma(f(x))=f(\frac{ax+b}{cx+d}), \ \forall f \in E,$ where $\displaystyle a,b,c,d \in F$ and $\displaystyle ad - bc \neq 0.$

3. Originally Posted by NonCommAlg
this is a good question! the answer is No! because, for example, if $\displaystyle x$ is transcendental over $\displaystyle F$ and $\displaystyle E=F(x)$, then

it can be proved (not very easily though!) that $\displaystyle E^{\text{Gal}(E/F)}=F.$ the main part of the proof is to show that $\displaystyle \text{Gal}(E/F)$

consists of all maps $\displaystyle \sigma$ defined by $\displaystyle \sigma(f(x))=f(\frac{ax+b}{cx+d}), \ \forall f \in E,$ where $\displaystyle a,b,c,d \in F$ and $\displaystyle ad - bc \neq 0.$
I actually was thinking about $\displaystyle \mathbb{Q}(\pi)/\mathbb{Q}$, but I never did any of the details to prove it is Galois.

Now this leads me to my next question: is there anything interesting about a non-algebraic Galois extension?
but it would be interesting to me to know ahead of time).

4. Originally Posted by ThePerfectHacker
Now this leads me to my next question: is there anything interesting about a non-algebraic Galois extension?
i'm not sure, probably it's more complicated than interesting! for example we know that in algebraic Galois extensions,

if E/F is Galois and K is a subfield of E which contains F, then E/K would be Galois too. now in order to have this nice

property in non-algebraic extensions we may define that E/F to be Galois if $\displaystyle E^{\text{Gal}(E/K)}=K$, for any subfield K of E which

contains F. but then we'd have a problem: with this definition there is no non-algebraic Galois extension of any field of

characteristic p > 0 (this is easy to prove!)