MHB The extension is Galois iff H_i is a normal subgroup of H_{i-1}

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Let $E/F$ be a finite Galois extension and let the chain of extensions $F =
K_0 \leq K_1 \leq \dots \leq K_n = E$.

Let $G = Gal(E/F)$ and, for $i = 0, 1, \dots , n$, let $H_i$ be the subgroup of $G$, that corresponds to $K_i$ through the Galois mapping.

I want to show that, for any $i \in \{1, \dots, n\}$ it holds that the extension $K_i/K_{i−1}$ is Galois iff $H_i \triangleleft H_{i−1}$.
In my notes there is the following proposition:

$E/F$ is finite Galois
View attachment 6156
where $G=\text{Gal}(E/F)$

$K/F$ is normal (and so Galois) iff $H\triangleleft G$ (normal subgroup).
In this case we have the following:
View attachment 6157
right? (Wondering)

Can we just apply the above proposition for each $K_i/K_{i-1}$ ? (Wondering)
 

Attachments

  • diag (2).PNG
    diag (2).PNG
    1.7 KB · Views: 103
  • diagg.png
    diagg.png
    1.9 KB · Views: 107
Physics news on Phys.org
mathmari said:
Can we just apply the above proposition for each $K_i/K_{i-1}$ ? (Wondering)
Or can we not just apply this proposition and we have to do something else? (Wondering)
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K