How to Prove Galois Extension Statements in a Normal and Separable Field?

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In summary, the conversation discusses the proof of two statements related to Galois extensions. The first statement shows that if $F\leq K_1\leq K_2\leq E$, then the fixed field of the Galois group of $E/K_1$ is larger than or equal to the fixed field of the Galois group of $E/K_2$. The second statement shows that if $H_1\leq H_2\leq \mathcal{G}(E/F)$, then the fixed field of $H_1$ is larger than or equal to the fixed field of $H_2$. The conversation also includes a summary of the proof for each statement. The speaker confirms
  • #1
mathmari
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Hey! :eek:

Let $E/F$ be a Galois extension. I want to show the following:
  1. $F\leq K\leq E \Rightarrow \mathcal{F}(\mathcal{G}(E/K))\geq K$
  2. $H\leq \mathcal{G}(E/F)\Rightarrow \mathcal{G}(E/\mathcal{F}(H))\geq H$

Since $E/F$ is a Galois extension, we have that the extension is normal and separable.

Could you give me some hints how we could show these statements? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

Let $E/F$ be a Galois extension. I want to show the following:
  1. $F\leq K\leq E \Rightarrow \mathcal{F}(\mathcal{G}(E/K))\geq K$
  2. $H\leq \mathcal{G}(E/F)\Rightarrow \mathcal{G}(E/\mathcal{F}(H))\geq H$

Since $E/F$ is a Galois extension, we have that the extension is normal and separable.

Could you give me some hints how we could show these statements? (Wondering)

Both statements follow easily from definitions.

(1). Let $x\in K$ be arbitrary. We will show that $x$ is in the fixed field of $\mathcal G(E:K)$. Pick $\sigma\in \mathcal G(E:K)$. Then $\sigma$ fixes $K$ point wise. Therefore $\sigma(x) = x$ and we have $x\in \mathcal F(\mathcal G(E:K))$.

(2). Let $h\in H$. Then $h$ fixes each element in $\mathcal F(H)$ (just by definition). Thus $h\in \mathcal G(E:\mathcal F(H))$ and we are done.
 
  • #3
I understand! (Nerd) I want to show the following statements:
  • $F\leq K_1\leq K_2\leq E\Rightarrow \mathcal{G}(E/K_1)\geq \mathcal{G}(E/K_2)$
  • $H_1\leq H_2\leq \mathcal{G}(E/F)\Rightarrow \mathcal{F}(H_1)\geq \mathcal{F}(H_2)$

For the first one I have done the following:
Let $\sigma \in \mathcal{G}(E/K_2)$ then $\sigma (k_2)=k_2,\forall k_2\in K_2$. Since $K_1\leq K_2$ we have that this is true for all $k_1\in K_1$, so $\sigma (k_1)=k_1,\forall k_1\in K_1$, therefore, $\sigma \in \mathcal{G}(E/K_1)$.

For the second one I have done the following:
Let $f\in \mathcal{F}(H_2)$ then $\sigma (f)=f,\forall \sigma \in H_2$. Since $H_1\leq H_2$ , we get that $\sigma (f)=f \forall \sigma \in H_1$ too, and so $f\in F(H_1)$.

Is everything correct? Could I improve something? (Wondering)
 
  • #4
mathmari said:
I understand! (Nerd) I want to show the following statements:
  • $F\leq K_1\leq K_2\leq E\Rightarrow \mathcal{G}(E/K_1)\geq \mathcal{G}(E/K_2)$
  • $H_1\leq H_2\leq \mathcal{G}(E/F)\Rightarrow \mathcal{F}(H_1)\geq \mathcal{F}(H_2)$

For the first one I have done the following:
Let $\sigma \in \mathcal{G}(E/K_2)$ then $\sigma (k_2)=k_2,\forall k_2\in K_2$. Since $K_1\leq K_2$ we have that this is true for all $k_1\in K_1$, so $\sigma (k_1)=k_1,\forall k_1\in K_1$, therefore, $\sigma \in \mathcal{G}(E/K_1)$.

For the second one I have done the following:
Let $f\in \mathcal{F}(H_2)$ then $\sigma (f)=f,\forall \sigma \in H_2$. Since $H_1\leq H_2$ , we get that $\sigma (f)=f \forall \sigma \in H_1$ too, and so $f\in F(H_1)$.

Is everything correct? Could I improve something? (Wondering)
This is fine.
 
  • #5
Thank you very much! (Happy)
 

FAQ: How to Prove Galois Extension Statements in a Normal and Separable Field?

What is a Galois extension?

A Galois extension is a type of field extension in abstract algebra that is named after the mathematician Évariste Galois. It is a field extension where the automorphisms of the extension field are in one-to-one correspondence with the automorphisms of the base field.

What is the significance of Galois extensions?

Galois extensions have many applications in algebraic number theory, algebraic geometry, and other areas of mathematics. They provide a way to understand the structure of fields and their extensions, and have been used to prove important results such as the insolvability of the quintic equation.

How is a Galois extension defined?

A Galois extension is defined as a finite field extension that is both normal and separable. Normality means that the extension field is a splitting field for a family of polynomials over the base field, and separability means that every irreducible polynomial over the base field has distinct roots in the extension field.

What is the Galois group of a Galois extension?

The Galois group of a Galois extension is the group of automorphisms of the extension field that fix the base field element-wise. This group plays a central role in studying the properties of the extension and its automorphisms.

Can a Galois extension be infinite?

No, a Galois extension must be finite. This is because the definition of a Galois extension requires it to be both normal and separable, and these properties are only satisfied for finite extensions. However, there are infinite extensions that have similar properties and are studied in the field of infinite Galois theory.

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