# Galois group of $(x^5-3)(x^5-7)$ over $\Bbb Q$

I am having some trouble with the following question:

Let $f(x)=(x^5-3)(x^5-7)\in\Bbb Q[x]$. Find the degree of the splitting field of $f$ over $\Bbb Q$, and determine the Galois group of $f$ over $\Bbb Q$.

I noticed that $x^5-3$ and $x^5-7$ are irreducible in $\Bbb Q[x]$, and so their Galois groups are isomorphic to a transitive subgroup of $S_5$. I also found that the splitting fields of these two quintics are $\Bbb Q(\sqrt[5] 3,\alpha)$ and $\Bbb Q(\sqrt[5] 7,\alpha)$ where $\alpha^5=1$. Since $\alpha$ has degree $4$ over $\Bbb Q$ and $\sqrt[5] 3,\sqrt[5] 7$ have degree $5$ over $\Bbb Q$, we have that the degrees of these splitting fields are $20$. Hence, we have $$\operatorname{Gal}(x^5-3/\Bbb Q)\cong F_{20}\cong \operatorname{Gal}(x^5-7/\Bbb Q)$$ But now I'm supposed to use a theorem we proved that $\operatorname{Gal}(f/\Bbb Q)$ is a subgroup of the direct product of the above Galois groups. I am not sure how to do this: i.e., $\color{red}{\text{how to determine which subgroup$\operatorname{Gal}(f/\Bbb Q)$corresponds to}}$.

For the degree of the splitting field of $f$ over $\Bbb Q$, I argued that it is $100$ since the compositum $\Bbb Q(\sqrt[5] 7,\alpha)\Bbb Q(\sqrt[5] 3,\alpha)=\Bbb Q(\sqrt[5] 3,\sqrt[5] 7,\alpha)$ has degree $5\cdot 5\cdot 4=100$ over $\Bbb Q$. $\color{red}{\text{Does this make sense}}$?

I also had another question to determine the Galois group of $(x^3-2)(x^4-2)$ over $\Bbb Q$. I showed that $x^3-2$ has Galois group $S_3$ and $x^4-2$ has Galois group $D_4$. Then I argued that the intersection of the two splitting fields, $\Bbb Q(\sqrt[3]2,\omega)\cap\Bbb Q(\sqrt[4]2,i)$ is just $\Bbb Q$, so we have $\operatorname{Gal}((x^3-2)(x^4-2)/\Bbb Q)\cong S_3\times D_4$. $\color{red}{\text{Does this make sense}}$? In order to show $\Bbb Q(\sqrt[3]2,\omega)\cap\Bbb Q(\sqrt[4]2,i)=\Bbb Q$ I just argued that $\sqrt[4]2,i\notin\Bbb Q(\sqrt[3]2,\omega)$, $\color{red}{\text{is this sufficient}}$?

• Can't you use the theorem you just used in the last part to solve the first one? Once that $\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)\cap \mathbb{Q}(\alpha)=\mathbb{Q}$, $$Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)\mathbb{Q}(\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)/\mathbb{Q})\times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]3)/\mathbb{Q}) \times Gal(\mathbb{Q}(\sqrt[5]7)/\mathbb{Q}) \times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong \mathbb{Z}_5 \times \mathbb{Z}_5 \times Gal(\mathbb{Q}(\alpha)/\mathbb{Q}).$$ – TPace Feb 15 '18 at 22:34
• Ah! I think that makes sense, so in this case $\operatorname{Gal}(\Bbb Q(\alpha)/\Bbb Q)\cong C_4$? – Dave Feb 15 '18 at 22:40
• It may be $V_4$ too! – TPace Feb 15 '18 at 22:41
• But if $\alpha$ is a 5th root of unity, wouldn't its galois group just permute the roots of $x^4+x^3+x^2+x+1$ like a 4-cycle, or am I wrong? – Dave Feb 15 '18 at 22:43
• Then you need to write down the effect of the order four generator of $F_{20}$, i.e. the automorphism determined by $\alpha\mapsto\alpha^2,\root5\of3\mapsto\root5\of3, \root5\of7\mapsto\root5\of7$. I didn't fully justify everything, but it looks like you shoud get a semi-direct product $(C_5\times C_5)\rtimes C_4$ as the answer. – Jyrki Lahtonen Feb 16 '18 at 6:05

On the first "Does this make sense?": Yes, it pretty much does. Using the tower law, we get that $$[\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha):\mathbb{Q}(\sqrt[5]7,\alpha)][\mathbb{Q}(\sqrt[5]7,\alpha):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=5*5*4=100.$$ On your second "Does this make sense?": Yes, it also does. If $K_1$ and $K_2$ are Galois extensions and $K_1 \cap K_2=F$, $$Gal(K_1K_2/F)\cong Gal(K_1/F)\times Gal(K_2/F).$$ So, if $K_1=\mathbb{Q}(\sqrt[3]2,\omega)$ and $K_2=\mathbb{Q}(\sqrt[4]2,i)$ and $Gal(K_1/\mathbb{Q})\cong S_3$ and $Gal(K_2/\mathbb{Q})\cong D_4$, we get the result you got. However, I think that you could be more complete when arguing that $K_1\cap K_2=\mathbb{Q}$, explaining it a bit more- even though you're not wrong.
On the first part, as we argued in the comment section, $Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha)/\mathbb{Q})\cong C_5^2\times C_4$. This is due to the fact that, once $\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)\cap \mathbb{Q}(\alpha)=\mathbb{Q},$
$$Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)/\mathbb{Q})\times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]7)/\mathbb{Q})\times Gal(\mathbb{Q}(\sqrt[5]3)/\mathbb{Q})\times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong C_5\times C_5 \times C_4=C_5^2\times C_4$$
• Really appreciate the help here. How would I show more concretely that $\Bbb Q(\alpha)\cap \Bbb Q(\sqrt[5] 3,\sqrt[5] 7)=\Bbb Q$? – Dave Feb 15 '18 at 23:16
• Let $x$ be in the intersection of both splitting fields. If $1,b_1,...,b_n$ are basis elements of $\mathbb{Q}(\sqrt[3]2,\omega)$ and $1,c_1,...,c_m$ of $\mathbb{Q}(\sqrt[4]2,i)$, then $x=a_0+a_1b_1+...+a_nb_n=d_0+d_1c_1+...+d_mc_m$, where $a_i$ and $b_i$ are rationals. Suppose $a_i,b_i\neq0$. As a consequence, $b_1=\frac{-(a_0+a_2b_2+...+a_nb_n)+d_0+d_1c_1+...+d_mc_m}{a_1}$. This, however, isn't true in this case, so that $a_i=b_i=0, i>0$. This proves what we wanted, in the sense that $x$ must be strictly a member of the rationals. – TPace Feb 15 '18 at 23:50
• I am going through my notes and I just realized that we can't really define the Galois group of $\Bbb Q(\sqrt[5] 3)$ or other similar ones as they are not splitting fields of separable polynomials. I understand that we can still talk about their automorphism groups, but in my lectures we cannot classify them as Galois groups. I believe I have made a decent argument to get the result above though. Thanks for the help. – Dave Feb 16 '18 at 0:57
• The Galois groups of $x^5-3$ and $x^5-7$ are isomorphic to $F_{20}$. They are definitely not abelian. Hence the sought after Galois group cannot be abelian either. Your mistake is in writing $Gal(\Bbb{Q}(\root5\of3,\root5\of7)/\Bbb{Q})$. That extension is not Galois. The field $K=\Bbb{Q}(\root5\of3,\root5\of7)$ is real, so both $x^5-3$ and $x^5-7$ have a single zero in $K$ only. Therefore $K$ has only a single automorphism - the idenitity. – Jyrki Lahtonen Feb 16 '18 at 5:52