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I'm suppossed to use an example to show the following statement. If F over K is galois but not algebraic and L is an intermediate field between K and F, then F over L is not galois. Any help at all would be greatly appreciated.

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The most common definition of "Galois extension" requires that the extension be algebraic. What definition are you using? –  Brad Mar 28 '11 at 0:22
    
@ Brad I think that F must have char 0 –  user8771 Mar 28 '11 at 0:32
    
@user8771, a Galois extension may well be infinite but it still needs to be algebraic under the common definition. Also, if F is an extension of K and L is an intermediate field, then "K over L" doesn't make sense. –  Alon Amit Mar 28 '11 at 0:34
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Ok, that still leaves the question of what "Galois but not algebraic" means. Whatever the definition, the question seems odd - F/L is not Galois for all L except L=K? –  Alon Amit Mar 28 '11 at 0:40
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@All Hungerford's Algebra book defines a Galois extension as an extension $F/K$ such that the fixed field of $\text{Aut}_K F$ is $K$, without requiring it to be algebraic, although he explicitly notes that it is commonly required to be algebraic. –  Adrián Barquero Mar 28 '11 at 0:43

2 Answers 2

If you use the definition of Galois extension that I quoted in the comments then you can look at the field extension $K(x) / K$ where $K$ is an infinite field. In this case this field extension is Galois but not algebraic over $K$. So my suggestion is for you to look at exercises $6$ and $9$ in section $2$, chapter $5$ of Thomas Hungerford's algebra book.

In exercise $6$ you are guided in a series of steps to prove some important facts about this field extension and about the automorphism group. And exercise $9$ guides you to prove that this extension is Galois.

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@ Adrian..thanks..pg. 256..looks helpful –  user8771 Mar 28 '11 at 1:01
    
@user8771 Yes it does =) –  Adrián Barquero Mar 28 '11 at 1:11
    
@ Adrian,those exercises are helpful..but I will need to still produce a concrete example for the problem. I'll try to post one when I get through those exercises. –  user8771 Mar 28 '11 at 1:12
    
@ pete thanks. I kept thinking it was a transcendental extension. –  user8771 Mar 28 '11 at 23:22

First let me say that I am interpreting your question according to the definition of Galois for transcendental extensions given by Adrian Barquero (this is not my preferred definition: see below). Second, interpreting it that way it seems that you have a quantifier error: you ask for a Galois transcendental extension $F/K$ such that for every subextension $L$ of $F/K$, $F/L$ is not Galois. But this can't be: take for instance $L = F$ or more generally let $G$ be any finite subgroup of $\operatorname{Aut}(F/K)$ and let $L$ be the fixed field $F^G$. Then $F/F^G$ is a finite Galois extension. If $F/K$ is transcendental and Galois then $\operatorname{Aut}(F/K)$ must be infinite and thus there are infinitely many examples like this.

I will assume you meant to ask: find some Galois transcendental extension $F/K$ and *some subextension $L$ such that $F/L$ is not Galois.

For this: $K = \mathbb{Q}$ and $F = \mathbb{Q}(t)$. Then the only elements of $F$ fixed by every element of $\operatorname{Aut}(K/\mathbb{Q}) = \operatorname{Aut}(K) = PGL_2(\mathbb{Q})$ acting by linear fractional transformations are the elements of $\mathbb{Q}$. So $F/K$ is Galois according to the definition Adrian provides above.

However, consider the intermediate field $L = \mathbb{Q}(t^3)$. Then $F/L$ is not Galois because $\mathbb{Q}$ does not contain a primitive $3$rd root of unity.

You can create similar examples with ground field $K = \mathbb{C}$ by choosing a finite subgroup $G$ of $PGL_2(\mathbb{C}) = \operatorname{Aut}(\mathbb{C}(t)/\mathbb{C})$ which contains a non-normal subgroup $H$. Then take $F = \mathbb{C}(t)^H$ and $L = \mathbb{C}(t)$. For instance, you can take $G$ to the be icosahedral group $A_5$, which is nonabelian simple, so has plenty of non-normal subgroups.

Note that in this rough note I attempt to study transcendental Galois extensions. The property of an extension $F/K$ that $F^{\operatorname{Aut}(F/K)} = K$ I call weakly Galois. My definition of a Galois extension is precisely that the examples you seek do not occur: namely for every intermediate extension $L$, also $F^{\operatorname{Aut}(F/L)} = L$. So in my terminology, for any infinite field $K$, the rational function field $K(t)$ is weakly Galois but not Galois.

In fact I have made a conjecture that a transcendental extension $F/K$ is Galois iff $F$ is algebraically closed of characteristic $0$. One implication is easy; the other one has been an open question on MathOverflow for more than a year.

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I just looked at the relevant exercises in Hungerford's book, and it turns out that the first example I give above appears in one of his exercises. (FWIW, I only came by my copy of the book recently and never saw those exercises before. This leads me to believe that this is "the most natural example", in some sense.) –  Pete L. Clark Mar 28 '11 at 9:56

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