# Galois group

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?

## Answers and Replies

(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 lets you conclude $8=\deg(m_\alpha) \leq \deg(f) = n$.

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 lets 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 $$f$$?

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).

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?

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?

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.