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In my book, Galois group is defined to mean the set of automorphisms on $E/F$ that "leave alone" the elements in $F$.

On Wikipedia it says:

"If $E/F$ is a Galois extension, then $Aut(E/F)$ is called the Galois group of (the extension) $E$ over $F$, $\dots$"

And Wikipedia's definition of Galois:

"An algebraic field extension $E/F$ is Galois if it is normal and separable. Equivalently, the extension $E/F$ is Galois if and only if it is algebraic, and the field fixed by the automorphism group $Aut(E/F)$ is precisely the base field $F$."

So in one case, Wikipedia, the extension is restricted to be algebraic. So the set of automorphisms on $\mathbb{Q}(\pi) / \mathbb{Q}$ is not a Galois group.

My question: How is it possible to have two different definitions of what a Galois group is? Do these not conflict? Or what am I missing here?

Many thanks for your help.

Edit:

I'm using J. Gallian, Contemporary Abstract Algebra and Allan Clark, Elements of Abstract Algebra. Both use the same terminology, not the same as Wikipedia.

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I'm not seeing the conflict here. –  Qiaochu Yuan Feb 16 '11 at 10:27
    
@Qiaochu: well, in one definition the automorphisms on $\mathbb{Q}(\pi) / \mathbb{Q}$ is a Galois group and in one it isn't. Is this not a conflict? –  Matt N. Feb 16 '11 at 10:45
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@Matt: are you sure you have read your book correctly? It may have said something at the beginning like "in this chapter E/F always denotes a Galois extension" or something like that. In any case, it is okay for books to disagree on definitions. Sometimes a particular definition is more suitable for the author's purposes if it is tweaked slightly. –  Qiaochu Yuan Feb 16 '11 at 10:49
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@Qiaochu, @Matt: The (potential) conflict is that some books call $\mathrm{Aut}(E/F)$ the "Galois group of the extension", whether or not the extension is Galois, while others use "Automorphism group" for the general case, and reserve the term "Galois group" exclusively for the case where the extension is Galois. For example, Hungerford always calls $\mathrm{Aut}(E/F)$ "the Galois group of the extension", and then defines the extension to be Galois if and only if the fixed field of the Galois group is $F$. –  Arturo Magidin Feb 16 '11 at 17:56
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@Arturo, @Matt Fraleigh does the same thing as Hungerford except he uses $G(E/F)$. –  PEV Feb 17 '11 at 6:13

3 Answers 3

up vote 13 down vote accepted

There is a slight divergence of nomenclature. Everyone agrees on what $\mathrm{Aut}(E/F)$ is. The question is what to call it.

  1. Some books (e.g., Hungerford, Rotman's Galois Theory), always refer to $\mathrm{Aut}(E/F)$ as the "Galois group" of $E$ over $F$ (or of the extension), whether or not the extension is a Galois extension.

  2. Other books (e.g., Lang), use the generic term "automorphism group" to refer to $\mathrm{Aut}(E/F)$ in the general case, and reserve the term Galois group exclusively for the situation in which $E$ is a Galois extension of $F$.

So, in Lang, even just saying "Galois group" already implies that the extension must be a Galois extension, that is, normal and separable. In Hungerford, just saying "Galois group" does not imply anything beyond the fact that we are looking at the automorphism of the extension.

Wikipedia is following Convention 2; your book is following convention 1.

There is also the question of whether to admit infinite extensions or not. A lot of introductory books only consider only finite extensions when dealing with Galois Theory, and define an extension to be Galois if and only if $|\mathrm{Aut}(E/F)| = [E:F]$. This definition does not extend to the infinite extension, so the definitions are restricted to finite (algebraic) extensions, with infinite extensions not considered at all. Other characterizations of an extension being Galois (e.g., normal and separable) generalize naturally to infinite extensions, so no restriction is placed. Likewise, some books explicitly restrict to algebraic extensions, others do not; but note that most define "normal" to require algebraicity, because it is defined in terms of embeddings into the algebraic closure of the base field, so even if you don't explicitly require the extension to be algebraic in order to be Galois, in reality this restriction is (almost) always in place.

This is not such a big deal as it might appear, because one can show that an arbitrary (possibly infinite) Galois extension $E/F$ is completely characterized in a very precise sense by the finite Galois extensions $K/F$ with $F\subseteq K\subset E$ with $[K:F]\lt\infty$, as the automorphism group $\mathrm{Aut}(E/F)$ is the inverse limit of the corresponding finite automorphism groups.

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1  
Calling $Aut(E/F)$ the Galois group when $E/F$ is not a Galois extension is confusing to say the least. –  Makoto Kato Nov 17 '12 at 19:27

I don't know which book you are using or what the precise statement therein is, but here is how the terminology is used (at least within the English speaking mathematical world) [Edit: This may be too categorical a statement; see Arturo Magidin's answer]:

If $E/F$ is any field extension, then $Aut(E/F)$ denotes the field automorphisms of $E$ that leave each element of $F$ fixed.

If $E/F$ is a Galois extension (i.e. finite, separable, and normal) then one writes $Gal(E/F)$ for $Aut(E/F)$ and calls it the Galois group of $E$ over $F$. So Galois groups are a special case of automorphism groups. The reason for introducing this new terminology (i.e. $Gal(E/F)$) for an existing more general concept (i.e. $Aut(E/F)$) is in part historical, and also to serve as a reminder that in the particular case when $E/F$ is Galois, there is a rich theory (the Galois correspondence) relating $Gal(E/F)$ and the field theoretic structure of the extension $E/F$, which does not hold in more general contexts.

Note that there is also a notion of an infinite degree Galois extension $E/F$ (if $E/F$ is an infinite extension of $F$, one says that it is Galois if $E$ is the union of finite subextensions of $F$; in particular, $E$ is still necessarily algebraic over $F$), and in this context one also uses the notation $Gal(E/F)$ for $Aut(E/F)$. But in this case one also equips $Gal(E/F)$ with extra structure (one makes it not just a group, but a topological group). Also, this case is typically not treated in undergraduate textbooks and courses (at least in the U.S.), so if you are learning this material for the first time, you won't want to worry about it.

Finally, I want to say that studying $Aut(\mathbb Q(\pi)/\mathbb Q)$ (just to give one non-Galois theoretic example) is interesting and important; it just isn't part of Galois theory.

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To further complicate matters, some books use $\mathrm{Gal}(E/F)$ or $G(E/F)$ for the automorphism group only when the extension is Galois, others use it for the generic case (though all examples I know of the latter follow Hungerford's convention of always calling the automorphism group "the Galois group"). –  Arturo Magidin Feb 17 '11 at 6:23
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Like the old joke goes, "The nice thing about standards is that there are so many to choose from..." –  Arturo Magidin Feb 17 '11 at 6:24
    
@Matt: Infinite degree Galois extensions and $\mathbb{Q}(\pi) / \mathbb{Q}$ are not part of the course but I find it interesting to think about. In fact, I would like to read about it. Are there any books you can recommend that treat either or both of the subjects? –  Matt N. Feb 17 '11 at 9:54
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@Matt: Dear Matt, To learn about infinite degree Galois exetnsions, type "Galois group" and "Krull topology" into google. ("Krull topology" is the name for the profinite topology that people put on infinite Galois groups.) This should lead you to some introductory material and references. (I'm sure Lang's algebra book will discuss this, and there are going to be many other sources too, but probably googling is the easiest way to find one that suits you.) As for the automorphisms of $k(X)/k$ (for a field $k$ and an indeterminante $X$), these are the same as the ... –  Matt E Feb 17 '11 at 14:24
    
... the automorphism of the curve $\mathbb P^1$ over $k$, and are given by the Mobius transformations $X \mapsto (a X + b)/(c X + d)$ (for some invertible matrix $(a b ; c d)$). This is part of the theory of algebraic geometry of curves, and is probably harder to learn about with your current background; but you could try to prove the claim I just made about $Aut(k(X)/k)$ by hand, and see how you get on! Regards, –  Matt E Feb 17 '11 at 14:26

It is supposed to be proved in your Galois theory book as an important theorem, if not, firstly, consider the degree of L|F, and secondly, consider the fixed field, or to take a look at Van Der Wareden's book Algebra is to be of some help.
Edit:Here I mean that the fixed field of Aut(E|F) is exactly equal to F when E|F is
(1)Finite
(2)Separable
(3)Normal
and the converse is also true, if this isn't your question, I will delete my post immediately, thanks.

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Sorry when you write 'it' do you mean that these two definitions are equivalent? –  Matt N. Feb 16 '11 at 10:43
    
And I cannot get the point why you said Q($\pi$)|Q is Galois, since there is no Galois group notion attached to non-Galois extensions which might be properly my false, please inform me, thanks very much. –  awllower Feb 16 '11 at 13:32

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