# Thread: Galois group, cyclic real roots :)

1. ## Galois group, cyclic real roots :)

Prove that if the Galois group of the splitting field of a cubic over $\mathbb{Q}$ is the cyclic group of order 3 then all the roots of the cubic are real.

So here is what I know

$G(\mathbb{F}/ \mathbb{Q})=\{1,\sigma,\sigma^2 \}$ since the group is cyclic it is Abelian( and so are all of its subgroups)

The Galois group has only the trivial subgroups and they are normal.

This implies that each field extention is Galois and that there is a 1-1 corrispondance to the subgroups of G and the subfields of $\mathbb{F}$ ie Since there are no non trivial sub groups there are not any non trivial subfields???

I think I need to argue the degree of the extentions, but I don't see how to make the connection.

Thanks for any help

TES

2. Originally Posted by TheEmptySet
Prove that if the Galois group of the splitting field of a cubic over $\mathbb{Q}$ is the cyclic group of order 3 then all the roots of the cubic are real.

So here is what I know

$G(\mathbb{F}/ \mathbb{Q})=\{1,\sigma,\sigma^2 \}$ since the group is cyclic it is Abelian( and so are all of its subgroups)

The Galois group has only the trivial subgroups and they are normal.

This implies that each field extention is Galois and that there is a 1-1 corrispondance to the subgroups of G and the subfields of $\mathbb{F}$ ie Since there are no non trivial sub groups there are not any non trivial subfields???

I think I need to argue the degree of the extentions, but I don't see how to make the connection.

Thanks for any help

TES
If the polynomial had a complex zero then it would have a complex conjugate as well, if $\tau$ is complex conjugation then $\tau$ belong to the Galois group but it has order 2. However, the Galois group is of order three and not divisible by 2. Can you see the contradiction?

3. Originally Posted by ThePerfectHacker
If the polynomial had a complex zero then it would have a complex conjugate as well, if $\tau$ is complex conjugation then $\tau$ belong to the Galois group but it has order 2. However, the Galois group is of order three and not divisible by 2. Can you see the contradiction?
This would imply that $[\mathbb{Q}([i\tau]):\mathbb{Q}]=2$ and since it is Galois there would need to be a subgroup of order 2, but since the Galois group is cyclic the order of any subgroup must divide the order of the group therefore $2|3$ contradiction

4. Originally Posted by TheEmptySet
This would imply that $[\mathbb{Q}([i\tau]):\mathbb{Q}]=2$ and since it is Galois there would need to be a subgroup of order 2, but since the Galois group is cyclic the order of any subgroup must divide the order of the group therefore $2|3$ contradiction
Complex conjugation $\tau$ is an automorphism of this field. Therefore, $\tau \in G$. By Lagrange's theorem the order of $\tau$ must divide $|G| = 3$ however $\tau$ has order $2$ which is impossible. Therefore, there can be no complex roots. [The mapping $\tau$ is an automorphism if and only if the field has complex roots, otherwise it is just the identity].