# Galois extensions over $\mathbb{Q}$.

Determine whether the following fields are Galois over $\mathbb{Q}$.

(a) $\mathbb{Q}(\omega )$, where $\omega = exp(2\pi i/3)$.

(b) $\mathbb{Q}(\sqrt[4]{2})$.

(c) $\mathbb{Q}(\sqrt{5}, \sqrt{7})$.

I have already shown that (a) is Galois over $\mathbb{Q}$, and (b) is not. For (c), I know that $\mathbb{Q}(\sqrt{5},\sqrt{7}) = \mathbb{Q}(\sqrt{5} + \sqrt{7})$, but I couldn't conclude. Any hint?

PS: Use only results of automorphism and Galois extensions.

Definition. Let $K$ be an algebraic extension of $F$. Then $K$ is Galois over $F$ if $F = \mathcal{F}(Gal(K/F))$.

$\mathcal{F}$ is the fixed field.

Proposition. Let $K$ be a field extension of $F$. Then $K/F$ is Galois if and only if $|Gal(K/F)|=[K:F]$.

• Are you sure (a) is not Galois? Looks cyclotomic (and of degree 2) to me. – Torsten Schoeneberg Apr 29 '18 at 2:39
• @TorstenSchoeneberg, Sorry, I should not have been clear... (a) is Galois over $\mathbb{Q}$. The automorphism are identity and complex conjugation. – Lucas Corrêa Apr 29 '18 at 2:42

Hint: The minimal polynomial of $\sqrt{5} + \sqrt{7}$ is $x^4-24x^2+4$. Now show that $\mathbb{Q}(\sqrt{5} + \sqrt{7})$ is the splitting field of this polynomial.

Edit: Since you can't use splitting fields, here is another way to do it. We can check pretty easily that $K=\mathbb{Q}(\sqrt{5}, \sqrt{7})$ has degree $4$ over $\mathbb{Q}$ with basis $(1, \sqrt{5}, \sqrt{7}, \sqrt{35} )$.

Let $\sigma \in Aut(K/\mathbb{Q})$. Then $\sigma$ is uniquely determined by its action on $\sqrt{5}$ and $\sqrt{7}$.

If $\sigma(\sqrt{5})=a+b\sqrt{5}+c\sqrt{7} +d\sqrt{35}$, then we have $$5=\sigma(\sqrt{5})^2=(a+b\sqrt{5}+c\sqrt{7} +d\sqrt{35})^2$$ By equating coefficients we see that $b^2=1$, $a=c=d=0$ and thus $\sigma(\sqrt{5})=\pm \sqrt{5}$.

By the same reasoning, $\sigma(\sqrt{7})=\pm \sqrt{7}$.

We then get $4$ distinct automorphism and thus $K$ is Galois.

• Thanks for the help! But, I still cannot use splitting field... this is only defined in the next chapter. – Lucas Corrêa Apr 29 '18 at 2:45
• Nice proof! Thanks a lot! – Lucas Corrêa Apr 29 '18 at 3:08

$c)$ is the splitting field of $X^2-5$ and $X^2-7$ so it is a normal extension, since $\mathbb{Q}$ is perfect, it is separable and henceforth Galois.

Let $\sigma$ be a homomorphism of $\mathbb Q(\sqrt5,\sqrt7)$ into $\mathbb C$. Clearly $\sigma$ sends $\sqrt5$ to $\pm\sqrt5$ and $\sqrt7$ to $\pm\sqrt7$, so it sends a linear combination $a+b\sqrt5+c\sqrt7$ to another linear combination $a'+b'\sqrt5+c'\sqrt7$. This shows that $\sigma$ is an automorphism of $\mathbb Q(\sqrt5,\sqrt7)$ over $\mathbb Q$. Thus $\mathbb Q(\sqrt5,\sqrt7)/\mathbb Q$ is a normal field extension.

Hope this helps.

• Yes, very good solution. But my professor wants to be solved using only results of automorphisms and Galois extensions. Normal and separable extensions were taught later. Anyway, thank you very much! – Lucas Corrêa Apr 29 '18 at 2:59
• @LucasCorrêa May I suggest you to put your definition of a Galois extension here, so that we know what we shall prove? I am using the definition that a Galois extension is a normal and separable extension. Thanks. – awllower Apr 29 '18 at 3:01
• You are completely right. I updated. – Lucas Corrêa Apr 29 '18 at 3:07