MHB What Determines the Galois Group of a Polynomial's Splitting Field?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Group
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

We consider the polynomial $f(x)=x^3+x^2-2x-1 \in \mathbb{Q}[x]$ and let $E$ be its splitting field.

How can we find the group $Gal(E/\mathbb{Q})$ ?? (Wondering)
 
Physics news on Phys.org
Hi,

The automorphisms will be well defined with the image of the roots of $f$, and are also permutations over the roots, so you only have to check when a so defined automorphism is in the Galois group.
 
Fallen Angel said:
Hi,

The automorphisms will be well defined with the image of the roots of $f$, and are also permutations over the roots, so you only have to check when a so defined automorphism is in the Galois group.

Could you explain it further to me?? (Wondering)
 
Compute the discriminant of the cubic polynomial (it is irreducible). Then check if the discriminant is a square or not, in the square field. If it is a square the group is $A_3$, if it is a non-square then it is $S_3$.
 
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 3 ·
Replies
3
Views
2K
  • · Replies 26 ·
Replies
26
Views
678
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
6
Views
3K
Replies
0
Views
2K
Replies
6
Views
1K