What Is the Galois Group of x^5 - 1 Over Q?

  • Thread starter Thread starter Zoe-b
  • Start date Start date
  • Tags Tags
    Confused Group
Click For Summary
SUMMARY

The Galois group of the polynomial x^5 - 1 over Q is isomorphic to the cyclic group C4, which corresponds to the multiplicative group of integers mod 5. The roots of the polynomial, specifically w = exp(2*I*PI/5), include w, w^2, w^3, and w^4, with Q(sqrt(5)) identified as the intermediate field fixed by the subgroup {e,t}, where e is the identity and t represents complex conjugation. The discussion clarifies that the Galois group is determined by the splitting field, and two polynomials can share a Galois group if they have the same splitting field, regardless of their roots.

PREREQUISITES
  • Understanding of Galois theory and its terminology
  • Familiarity with polynomial roots and splitting fields
  • Knowledge of cyclic groups and their properties
  • Basic complex number theory, particularly exponential forms
NEXT STEPS
  • Study the properties of Galois groups and their relation to field extensions
  • Explore the concept of splitting fields in greater detail
  • Learn about the relationship between roots of polynomials and their Galois groups
  • Investigate examples of polynomials with the same splitting field and their Galois groups
USEFUL FOR

Mathematics students, particularly those studying abstract algebra and Galois theory, as well as educators seeking to deepen their understanding of polynomial Galois groups and field theory.

Zoe-b
Messages
91
Reaction score
0

Homework Statement


I'm trying to find the galois group of x^5 - 1 over Q, and then for each subgroup of the galois group identify which subfield is fixed.


Homework Equations





The Attempt at a Solution


If w = exp(2*I*PI/5), then the roots not in Q are w, w^2, w^3, w^4. Its fairly easy to see by checking possible automorphisms that the Galois group is isomorphic to the multiplicative group of integers mod 5, which is in turn isomorphic to C4.

There may be a mistake there already.. but I'm not sure.

Here comes the confusion- C4 has one non-trivial proper subgroup, in this case found by multiplying only the elements w and w^4 together. But there are no intermediate fields as far as I can tell, since each root is primitive and so generates the other roots.

Please help! I've been teaching myself this course so I think I've probably just misunderstood something somewhere but not sure what..
 
Physics news on Phys.org
An intermediate field doesn't need to contain any root.
 
Ok... true. Will an intermediate field be one over which x^4 + x^3 + x^2 + x + 1 splits into two quadratics? or is the polynomial irrelevant for this...?
 
Think I've got it now- the intermediate field is Q(sqrt(5)) which is fixed by the subgroup {e,t} where e is the identity and t sends w to w^4, w^2 to w^3, that is, t is equivalent to complex conjugation.

Thank you!
 
Zoe-b said:
Think I've got it now- the intermediate field is Q(sqrt(5)) which is fixed by the subgroup {e,t} where e is the identity and t sends w to w^4, w^2 to w^3, that is, t is equivalent to complex conjugation.

Thank you!

This sounds right.

Zoe-b said:
Ok... true. Will an intermediate field be one over which x^4 + x^3 + x^2 + x + 1 splits into two quadratics? or is the polynomial irrelevant for this...?
I expect this to be true as well.
 
Fantastic- I have a more general question which as yet I've been unable to find the answer to in a textbook..
Does the galois group of a polynomial depend purely on its splitting field? Or is it in some way connected to the polynomial itself? For example, if two polynomials have different roots but the same splitting field, are their galois groups the same?
 
The definition I know of the Galois group of a polynomial is literally that it is the Galois group of the splitting field (over the relevant base field).
 
Ok thank you I thought that was the case but then got confused by questions where the splitting field seemed to be the same for different examples :)
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 28 ·
Replies
28
Views
5K
Replies
5
Views
2K
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K