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

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SUMMARY

This discussion focuses on proving Galois extension statements in the context of normal and separable fields. The participants demonstrate that for a Galois extension \(E/F\), if \(F \leq K \leq E\), then \(\mathcal{F}(\mathcal{G}(E/K)) \geq K\) holds true. Additionally, it is established that if \(H \leq \mathcal{G}(E/F)\), then \(\mathcal{G}(E/\mathcal{F}(H)) \geq H\). The proofs rely on the definitions of fixed fields and Galois groups, confirming the correctness of the statements presented.

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  • Knowledge of fixed fields and Galois groups
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mathmari
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Hey! :o

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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mathmari said:
Hey! :o

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.
 
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)
 
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.
 
Thank you very much! (Happy)
 

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