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

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Extension
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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)
 
Physics news on Phys.org
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)
 
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...
Back
Top