Prove: Galois Extensions Homework - E is Isomorphic to B_N x B_H

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In summary, the first part deals with proving that a Galois extension of two fields is also a Galois extension of their tensor product. The second part involves using this result to show that a given Galois extension is isomorphic to a tensor product of two fields, under certain conditions.
  • #1
dabien
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Homework Statement



Prove
Suppose that [tex]B_1, B_2[/tex] are Galois extensions of F with respective Galois groups [tex]G_1, G_2[/tex], and that [tex]E=B_1 \otimes B_2[/tex] is a field. Then it is Galois over F with Galois group isomorphic to [tex]G_1 \times G_2[/tex].

#2
Suppose that E is a Galois extension of F with Galois group G and that G contains subgroups H and N with N normal in G, [tex]H \cap N=[/tex]{1} and [tex]HN=G[/tex]. Let [tex]B_N[/tex] be the fixed field of N (so [tex]B_N[/tex] is Galois over F) and [tex]B_H[/tex] be the fixed field of H. Prove that E is isomorphic to [tex]B_N \otimes B_H[/tex]. (If H is also normal in G then [tex]G\cong H \times N[/tex] giving a converse to the preceding)


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The Attempt at a Solution


I am trying to get started...
 
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  • #2
My progress:
For #1:
I note that one equivalent condition for an extension to be Galois is that the number of automorphisms which fix the base field equals the degree of the extension. It is clear that [tex]G_1 \times G_2[/tex] is contained in the group of automorphisms of [tex]B_1 \otimes B_2[/tex] which fix F. Now I need help trying to argue that any such automorphism is necessarily in [tex]G_1 \times G_2[/tex]. I also need help trying to check orders to confirm that the extension is Galois.

For 2) I have read a condition for when the tensor product of two fields is actually equal to the product of the fields (when the fields are linearly disjoint). I need help verifying that this holds here. I also need to show that this product has to be all of E.
I have a stated lemma: If H,K are two normal subgroups of a group G such that they intersect trivially and [tex]HK=G[/tex], then G is isomorphic to H times K. I need help to show that if [tex]E=B_N \otimes B_H[/tex] with H and K satisfying the above, then [tex]G=H \times K[/tex].
 

Related to Prove: Galois Extensions Homework - E is Isomorphic to B_N x B_H

1. What is a Galois extension?

A Galois extension is a type of field extension in algebra that has a particular symmetry property known as Galois symmetry. This means that any automorphism (a bijective map that preserves the algebraic structure) of the extension field must preserve the base field as well.

2. What is the significance of proving that E is isomorphic to B_N x B_H?

Proving that E is isomorphic to B_N x B_H means that the Galois group of E, denoted as Gal(E), is isomorphic to the direct product of the Galois groups of the subfields B_N and B_H. This provides a way to understand the structure and properties of the Galois group of E, which can then be used to study the Galois extensions and their applications.

3. What is the definition of isomorphism?

An isomorphism is a bijective map between two mathematical structures that preserves the structure and properties of the objects in the structures. In the context of Galois extensions, an isomorphism between two fields means that the two fields have the same algebraic structure and properties.

4. How do you prove that E is isomorphic to B_N x B_H?

To prove that E is isomorphic to B_N x B_H, you need to show that there exists a bijective map between the two fields that preserves their algebraic structure and properties. This can be done by constructing an isomorphism explicitly or by using known properties and theorems of Galois extensions to show the isomorphism.

5. What is the role of the Galois group in this proof?

The Galois group plays a crucial role in this proof as it is used to establish the isomorphism between E and B_N x B_H. The Galois group of E provides a way to understand the structure and properties of the Galois extension, and by showing that it is isomorphic to the direct product of the Galois groups of the subfields B_N and B_H, we can prove that E is isomorphic to B_N x B_H.

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