0
$\begingroup$

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 $|Gal(K/F)|=[K:F]$.

$\endgroup$
  • $\begingroup$ Are you sure (a) is not Galois? Looks cyclotomic (and of degree 2) to me. $\endgroup$ – Torsten Schoeneberg Apr 29 '18 at 2:39
  • $\begingroup$ @TorstenSchoeneberg, Sorry, I should not have been clear... (a) is Galois over $\mathbb{Q}$. The automorphism are identity and complex conjugation. $\endgroup$ – Lucas Corrêa Apr 29 '18 at 2:42
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the help! But, I still cannot use splitting field... this is only defined in the next chapter. $\endgroup$ – Lucas Corrêa Apr 29 '18 at 2:45
  • $\begingroup$ Nice proof! Thanks a lot! $\endgroup$ – Lucas Corrêa Apr 29 '18 at 3:08
1
$\begingroup$

$c)$ 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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Lucas Corrêa Apr 29 '18 at 2:59
  • $\begingroup$ @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. $\endgroup$ – awllower Apr 29 '18 at 3:01
  • 1
    $\begingroup$ You are completely right. I updated. $\endgroup$ – Lucas Corrêa Apr 29 '18 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.