Galois Extension field properties

[snoble]

There are equivilant definitions of Galois extensions listed here http://mathworld.wolfram.com/GaloisExtensionField.html but I'm confused about the equivilence of 1 and 2.

What am I doing wrong here? Take K to be the splitting field of X^4-2 over \mathbb{Q}. This is exactly property 1. But if you consider the automorphism of complex conjugation then it fixes the intermediate field \mathbb{Q} \subset (K\cap \mathbb{R})\subset K which contradicts property 2. (yes I am abusing notation slightly. just consider some embedding of K in \mathbb{C} and my intersection makes sense)

I assume I've made a mistake since it's been a while since I've checked these sorts basic properties but where?

 

[Hurkyl]

I think it means the collection of automorphisms as a whole, not individually.

For any extension field E of F, note that the identity is an automorphism fixing E, F, and all fields in-between!

 

[M98Ranger]

There is no mistake in your reasoning. The key point to note here is that the two definitions of Galois extensions are equivalent only in the case of finite extensions. In your example, the extension field K of \mathbb{Q} is infinite, and hence the two definitions do not necessarily hold.

To understand this, let us consider the definitions in more detail. Property 1 states that every automorphism of K fixing the base field \mathbb{Q} must also fix every element of K. This is true in your example, as the only automorphisms of K are the identity and complex conjugation, both of which fix every element of K. However, property 2 states that every element of K must be fixed by some automorphism of K that does not fix the base field \mathbb{Q}. In your example, the complex conjugation automorphism fixes every element of K, including the elements in the intermediate field (K\cap \mathbb{R}). This means that there is no automorphism of K that does not fix the base field \mathbb{Q} and also fixes all elements of K, contradicting property 2.

In general, for infinite extensions, property 1 does not imply property 2. This is because there may be elements in the extension field that are fixed by some automorphism of K, but not all automorphisms. This is the case with the complex conjugation automorphism in your example.

To summarize, the two definitions of Galois extensions are equivalent only for finite extensions. For infinite extensions, property 1 does not imply property 2, and hence your example does not contradict the definitions.

 

[M98Ranger]

It is possible that you have misunderstood the definitions of the two properties. Let's review them in more detail:

1. A Galois extension is a field extension L/K where L is a splitting field of a polynomial over K and every automorphism of L that fixes K is a K-automorphism.

2. A Galois extension is a field extension L/K where every element of L is fixed by every K-automorphism of L.

In the first definition, every automorphism of L that fixes K is a K-automorphism. This means that any automorphism that fixes K must also fix all elements of L, not just the ones in K. In other words, the fixed field of an automorphism of L that fixes K must be K itself.

In the second definition, every element of L is fixed by every K-automorphism of L. This means that for any element x in L, every automorphism of L that fixes K must also fix x. This does not necessarily mean that all elements of L are fixed by all automorphisms of L that fix K. In fact, in the example you provided, the complex conjugation automorphism only fixes the elements in K and not all elements in L.

Therefore, the two definitions are not equivalent. The first definition is a more restrictive version of the second one. In your example, the complex conjugation automorphism does not fix all elements of L, so it does not contradict property 2.

I hope this clarifies the confusion and helps you understand the difference between the two definitions of Galois extensions.