Galois group

  • Thread starter math_grl
  • Start date
  • #1
49
0
This statement was made in my class and I'm trying still to piece together the details of it...

We say that some rational polynomial, [tex]f[/tex] has a Galois group isomorphic to the quaternions. We can then conclude that the polynomial has degree [tex]n \geq 8[/tex].

I have a few thoughts on this and I might be overlooking something simple...but letting [tex]K[/tex] be the splitting field, then [tex][K:\mathbb{Q}][/tex] divides [tex]n![/tex] so [tex]n \geq 4[/tex].

My other thought is that since the Galois group has finite number of subgroups then between K and the rational numbers are a finite number of intermediate fields, thus K is simple. With [tex]K \cong \mathbb{Q}(a)[/tex] for some root of a minimal polynomial whose degree must be 8 since [tex]\mathbb{Q}(a) \cong \mathbb{Q}[x]/\langle m_a \rangle [/tex].

Does [tex]m_a[/tex] divide [tex]f[/tex] or something?
 

Answers and Replies

  • #2
430
3
(notation: In this post I shall use [itex](\sigma f)[/itex] to denote f with [itex]\sigma[/itex] applied to all the coefficients of f; it's well-known that the new polynomial has the same degree and if x is a root of f, then [itex]\sigma x[/itex] is a root of [itex]\sigma f[/itex]).

Does [tex]m_a[/tex] divide [tex]f[/tex] or something?
Almost, let [itex]\beta \in K[/itex] be a root of f. Let [itex]\sigma : K \to K[/itex] be an automorphism fixing k and sending [itex]\beta \mapsto \alpha[/itex]. Now [itex](\sigma f)(\alpha) = 0[/itex] so [itex]m_\alpha | \sigma f[/itex] and since [itex]\deg(\sigma f) = \deg(f)[/itex] this lets you conclude [itex]8=\deg(m_\alpha) \leq \deg(f) = n[/itex].
 
  • #3
49
0
Let [itex]\sigma : K \to K[/itex] be an automorphism fixing k and sending [itex]\beta \mapsto \alpha[/itex]. Now [itex](\sigma f)(\alpha) = 0[/itex] so [itex]m_\alpha | \sigma f[/itex] and since [tex]\deg(\sigma f) = \deg(f)[/tex] this lets you conclude [tex]8=\deg(m_\alpha) \leq \deg(f) = n[/tex].
Isn't an automorphism from K to K fixing k the identity? or do you mean fixing [tex]\mathbb{Q}[/tex], the base field?

And just so I understand what you are trying to say is that the roots of [tex]m_a[/tex] are roots of f, since K is the splitting field for both [tex]m_a[/tex] and [tex]f[/tex]?
 
  • #4
97
0
As a permutation group, the quaternions are a subgroup of [tex]S_8[/tex] generated by the permutations [tex]\{(1234)(5678), (1537)(2846)\}[/tex]. Therefore [tex]f[/tex] must have 8 roots in [tex]K/\mathbb{Q}[/tex]. So the degree of [tex]f[/tex] must be [tex]\geq 8[/tex] (with equality if [tex]f[/tex] is irreducible).
 
  • #5
397
0
I shouldn't mess with this question but I'm curious: I once did an algebra problem where one of the roots of the polynomial was

sqrt(2) + sqrt(3) + sqrt(5)

and of course there are eight roots for all possible choices of the signs, so there's an eighth degree polynomial. I wonder if that's an example of the quaternion group?
 
  • #6
49
0
As a permutation group, the quaternions are a subgroup of [tex]S_8[/tex] generated by the permutations [tex]\{(1234)(5678), (1537)(2846)\}[/tex]. .
Is S8 the only permutation group the quaternions are a subgroup of?
 
  • #7
97
0
I'm pretty sure there are a few subgroups of [tex]S_8[/tex] containg the quaternions, although I don't know off-hand. For the purposes of this question, all you need to know is that as a permutation group they act on 8 objects - in this case: 8 non-rational roots.
 

Related Threads on Galois group

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
16
Views
7K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
20
Views
4K
Top