# Find $\mathrm{Gal}(L, \mathbb{Q})$ and prove that $\sqrt{3} \in L$

I am really struggling on these consecutive questions on Galois Theory:

Question 1: Suppose $L$ is a normal closure of $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ over $\mathbb{Q}$. Find $\mathrm{Gal}(L/\mathbb{Q})$.

Would this be the cyclic product $\mathbb{Z_2} \times \mathbb{Z_3}$ or $\mathbb{Z_6}$?

Question 2: Apply Galois theory to prove that $\sqrt{3} \not\in L$.

$\sqrt{3}$ clearly does not belong to $\mathbb{Q}$ and cannot be obtained using a composition of typical operations (addition, subtraction, multiplication etc) of $\sqrt{2}$ and $\sqrt{5}$ but how can I show this using Galois theory?

Many thanks for your help :-)

The possible values by an $L$-automorphism for $\sqrt{2}$ are $\sqrt{2}$, $j\sqrt{2}$ and $j^2\sqrt{2}$. And for $\sqrt{5}$ it is $\sqrt{5}$, $j\sqrt{5}$ and $j^2\sqrt{5}$. But the complex $j$ appears in our expressions, so the closure of $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ is $\mathbb{Q}(\sqrt{2}, \sqrt{5},j)$. And the possible images of $j$ by an $L$-automorphism are $j$ and $j^2$. So by considering all the combinations possible, there is a total of 18 possible $L$-automorphisms, and 17 of these automorphisms are of order 3, 17 are of order 6, one is of order 2 and there is $\mathrm{Id}$, so $\mathrm{Gal}(L/\mathbb{Q})$ is isomorphic with $S_3 \times \mathbb{Z_3}$.
• I'm assuming $j$ is a primitive cube root of unity? What about the map $j\mapsto j^2$? – carmichael561 Apr 8 '16 at 17:00
• $j^2$ is also a primitive root of the unity. Indeed $(j^2)^2=j$ and $(j^2)^3=1$ – Jennifer Apr 8 '16 at 17:05
• You're missing the map $j\mapsto j^2$. The Galois closure of $\mathbb{Q}(2^{\frac{1}{3}},5^{\frac{1}{3}})$ is $\mathbb{Q}(2^{\frac{1}{3}},5^{\frac{1}{3}},j)$, which has degree 18 over $\mathbb{Q}$. – carmichael561 Apr 8 '16 at 17:13