# Math Help - Galois Group, FTG

1. ## Galois Group, FTG

(a) Find a splitting field $K$ for the polynomial $f(x)=(x^2-3)(x^2-5) \in \mathbb{Q}[x]$.
(b) Determine the Galois Group of $K$ over $\mathbb{Q}$.
(c) Find all intermediate fields between $\mathbb{Q}$ and $K$, by refering to the Fundamental Theorem of Galois Theory [FTG] to show you have them all.

I'm pretty sure that (a) is $\mathbb{Q}(\sqrt(3), \sqrt(5))$, (b) is $\mathbb{Z}_2 \times \mathbb{Z}_2$ (c) $\mathbb{Q}, \mathbb{Q}(\sqrt(3), \sqrt(5)), \mathbb{Q}(\sqrt(3)), \mathbb{Q}(\sqrt(5)), \mathbb{Q}(\sqrt(15))$. I don't know how to justify (c) by using the Fundamental Theorem of Galois Theory. Also, I don't know how to determine the Galois Group for part (b). I know it is either $\mathbb{Z}_2 \times \mathbb{Z}_2$ or $\mathbb{Z}_4$. But, I don't know how to show this.

2. Originally Posted by diondex223
(a) Find a splitting field $K$ for the polynomial $f(x)=(x^2-3)(x^2-5) \in \mathbb{Q}[x]$.
We can write $(x -\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$ therefore the splitting field is $\mathbb{Q}(\sqrt{3},\sqrt{5})$.
(b) Determine the Galois Group of $K$ over $\mathbb{Q}$.
Since $[K:\mathbb{Q}]=4$ it tells us that $|G|=4$ where $G = \text{Gal}(K/\mathbb{Q})$. If $\theta$ is an automorphism of $K$ then it is determined by its value on $\theta (\sqrt{3}) \text{ and }\theta (\sqrt{5})$. Remember that automorphisms permute zeros of polynomials therefore $\theta (\sqrt{3})$ is a zero of $x^2 - 3$ and $\theta (\sqrt{5})$ is a zero of $x^2-5$. Thus, $\theta(\sqrt{3}) = \pm \sqrt{3} \text{ and }\theta (\sqrt{5}) = \pm \sqrt{5}$. We see that there are at most $4$ automorphisms but since $|G| = 4$ it means we must have an automorphism for each one of these possibilities. Define $\sigma: K\to K$ by $\sigma (\sqrt{3}) = -\sqrt{3} \text{ and }\sigma (\sqrt{5}) = \sqrt{5}$, define $\tau: K\to K$ by $\tau (\sqrt{3}) = \sqrt{3}\text{ and }\tau (\sqrt{5}) = -\sqrt{5}$. We see that $\sigma \tau$ is another automorphism of $K$. Therefore, $G = \left< \sigma,\tau \right>$ with $\sigma^2 = \tau^2 = \text{id}$. The non-trivial proper subgroups of $G$ are: $\left< \sigma\right> , \left<\tau\right>, \left< \sigma\tau \right>$ (just like $\mathbb{Z}_2\times \mathbb{Z}_2$).

(c) Find all intermediate fields between $\mathbb{Q}$ and $K$, by refering to the Fundamental Theorem of Galois Theory [FTG] to show you have them all.
The proper non-trivial intermediate are: $K^{\left< \sigma\right>}, K^{\left<\tau\right>}, K^{\left< \sigma\tau \right>}$. Notice that $\sqrt{5}$ is fixed by $\sigma$ therefore (fill in the minor details in this argument, do you see how or you want me to?) that $K^{\left< \sigma\right>} = \mathbb{Q}(\sqrt{5})$ likewise $K^{\left<\tau\right>} = \mathbb{Q}(\sqrt{3})$. For $\sigma\tau$ notice that $\sqrt{3} \sqrt{5}$ is fixed because the two minus signs go away, therefore (again you need to fill in a small detail) $K^{ \left< \sigma\tau \right>} = \mathbb{Q}(\sqrt{15})$.