Exploring the Consequences of Galois Group Isomorphism to Quaternions

[math_grl]

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, f has a Galois group isomorphic to the quaternions. We can then conclude that the polynomial has degree n \geq 8.

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

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 K \cong \mathbb{Q}(a) for some root of a minimal polynomial whose degree must be 8 since \mathbb{Q}(a) \cong \mathbb{Q}[x]/\langle m_a \rangle.

Does m_a divide f or something?

 

[rasmhop]

(notation: In this post I shall use (\sigma f) to denote f with \sigma 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 \sigma x is a root of \sigma f).

Does m_a divide f or something?

Almost, let \beta \in K be a root of f. Let \sigma : K \to K be an automorphism fixing k and sending \beta \mapsto \alpha. Now (\sigma f)(\alpha) = 0 so m_\alpha | \sigma f and since \deg(\sigma f) = \deg(f) this let's you conclude 8=\deg(m_\alpha) \leq \deg(f) = n.

 

[math_grl]

Let \sigma : K \to K be an automorphism fixing k and sending \beta \mapsto \alpha. Now (\sigma f)(\alpha) = 0 so m_\alpha | \sigma f and since \deg(\sigma f) = \deg(f) this let's you conclude 8=\deg(m_\alpha) \leq \deg(f) = n.

Isn't an automorphism from K to K fixing k the identity? or do you mean fixing \mathbb{Q}, the base field?

And just so I understand what you are trying to say is that the roots of m_a are roots of f, since K is the splitting field for both m_a and f?

 

[mrbohn1]

As a permutation group, the quaternions are a subgroup of S_8 generated by the permutations \{(1234)(5678), (1537)(2846)\}. Therefore f must have 8 roots in K/\mathbb{Q}. So the degree of f must be \geq 8 (with equality if f is irreducible).

 

[conway]

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?

 

[math_grl]

As a permutation group, the quaternions are a subgroup of S_8 generated by the permutations \{(1234)(5678), (1537)(2846)\}. .

Is S8 the only permutation group the quaternions are a subgroup of?

 

[mrbohn1]

I'm pretty sure there are a few subgroups of S_8 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.