Why is $ \mathbb{Q}(\sqrt[5]3 , e^{\frac{2\pi i}{5}} ) \cap \mathbb{Q}(e^{\frac{2\pi i}{7}} ) = \mathbb{Q}$? Say $\zeta_1 = e^{\frac{2\pi i}{5}} $, $\zeta_2 = e^{\frac{2\pi i}{7}} $ and $\zeta_3 = e^{\frac{2\pi i}{35}} $ . Define $E_1 = \mathbb{Q}(\sqrt[5]3 , \zeta_1)$ and $E_2 = \mathbb{Q}( \zeta_2)$
Show $E_1 \cap E_2 = \mathbb{Q}$ by Galois theorem.
(Hints : The splitting field, $K=\mathbb{Q}(\sqrt[5]3 , \zeta_3)$ for $f(x) = (x^5 -3)(x^7-1)$)
I'm trying to prove above considering "any $\sigma$" in $H_1 \cap H_2$, $\sigma(\sqrt[5]3) =\sqrt[5]3 $ and $\sigma(\zeta_i) =\zeta_i$. (Here $H_i= G(K /E_i)$ and $i \in \{1,2\}$)
But I've stuck to proceed the next step. Please help me. Plus another method also welcomed.
Best regards.
 A: I'll finish @Jyrki_Lahtonen's comment. As $[E_1:\mathbb Q]=20$, by Sylow's theorem, the Galois group $\text{Gal}(E_1/\mathbb Q)$ has a unique (normal) subgroup of order $5$, which fixes a unique subfield of order $4$. We happen to know that $\mathbb Q(\zeta_5)$ is an intermediate extension of degree $4$, hence must be the unique one. Any intermediate extension $F$ of degree $2$ must be fixed by a subgroup of order $10$ which must contain the unique Sylow $5$-subgroup of $\text{Gal}(E_1/\mathbb Q)$, and hence must be a subfield of $\mathbb Q(\zeta_5)$.
As $\text{Gal}(\mathbb Q(\zeta_5)/\mathbb Q)\simeq (\mathbb Z/5\mathbb Z)^{\times}$ which is cyclic, it has a unique subgroup of any fixed order. In particular, it can only have one subfield of degree $2$ which is fixed by the subgroup $\{\pm1\}\subset(\mathbb Z/5\mathbb Z)^{\times}$, hence the element $x=\zeta_5+\zeta_5^{-1}$ is in the subfield, and $x^2=\zeta_5^2+\zeta_5^3+2=(-1-x)+2$, $x^2+x-1=0$, $x=\frac{-1\pm\sqrt{5}}{2}$.
Hence $\mathbb Q(\sqrt 5)$ is the unique subfield of degree $2$ in $E_1$.
Similarly, we can find that $x:=\zeta_7+\zeta_7^2+\zeta_7^4$ satisfies $x^2=x+2(-1-x)$, $x^2+x+2=0$, $x=\frac{-1\pm\sqrt{-7}}{2}$, and $\mathbb Q(\sqrt {-7})$ is the unique subfield of degree $2$ in $E_2$. However, while $\sqrt 5$ is real, $\sqrt{-7}$ is imaginary, hence $E_1\cap E_2=\mathbb Q$.
A: The Galois group $G(K/\Bbb Q)$ has order $5\cdot\varphi (35)=120$.
Since the extensions $E_i/\Bbb Q$ are of degrees $5\cdot \varphi (5)=20$ and $\varphi (7)=6$, the corresponding subgroups $H_i$ of $G(K/\Bbb Q)$ have index $6$ and $20$ respectively.  That means their orders are $20$ and $6$.
Under the correspondence,  the intersection $E_1\cap E_2$ is the field corresponding to the join $\langle H_1,H_2\rangle $ of the subgroups.
We need that $H_1\cap H_2=\{e\}$.  One way to see that is to check that $E_1E_2=K$.
So that join has order $120$.  Thus $\langle H_1,H_2\rangle =G(K/\Bbb Q)$.
The whole Galois group corresponds to $\Bbb Q$.
