I'm trying to find the subfields of $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$. This is easily seen to be a degree $14$ extension of $\mathbb{Q}$.

I found that there is a unique subfield of degree $2$ over $\mathbb{Q}$. If $E$ and $F$ are distinct subfields of degree $2$ over $\mathbb{Q}$, then $$ [FE:\mathbb{Q}]=[FE:E][E:\mathbb{Q}]\leq[F:\mathbb{Q}][E:\mathbb{Q}]=4. $$ In particular, $[FE:E]\leq 2$. Since $FE\neq E$, I must get $[FE:E]=2$. But then this implies $[FE:\mathbb{Q}]=4$, a contradiction since $4\nmid 14$. So there is at most one subfield of degree $2$over $\mathbb{Q}$, and it is $\mathbb{Q}(\sqrt{3})$.

The only part left is to find the subfields of degree $7$ over $\mathbb{Q}$. Certainly $\mathbb{Q}(\sqrt[7]{5})$ is one, but I'm having trouble finding if there are others. I know the Galois closure of is $\mathbb{Q}(\sqrt{3},\sqrt[7]{5},\zeta_7)$, as this is a splitting field of the separable polynomial $(X^2-3)(X^7-5)$. Then the subfields will be the fixed fields of the subgroups in the Galois group containing the subgroup fixing $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$, but this seems computationally difficult.

Is there a way to do it without having to look at the Galois group?

  • $\begingroup$ I do not think there is any other way to go for all sub fields except for taking its algebraic closure and see its Galois group and see its sub fields.... :O $\endgroup$
    – user87543
    Mar 18, 2014 at 11:08

1 Answer 1


Let $E=\mathbb{Q}(\sqrt[7]{5})$ and suppose that you have another field $F$ of degree 7. Then $[F(\sqrt[7]{5}):F][F:\mathbb{Q}]\leq 14$ so we must have $[F(\sqrt[7]{5}):F]=2$. It follows that $\alpha=\sqrt[7]{5}$ satisfies an irreducible polynomial of degree 2 over $F$. There is now a general fact that states that if a polynomial $x^p-a$ is reducible over some field F for some $p$ prime, then it has a root in F. More precisely, this polynomial is of the form $(x-\zeta^i \alpha)(x-\zeta^j \alpha)=x^2 -(\zeta^i+\zeta^j)\alpha +\zeta^{i+j}\alpha^2$ where $\zeta$ is a primitive 7 root of unity and $0\leq i<j<7$ - this is because it divides $x^7-5=\prod_{i=1}^7 (x-\zeta^i \alpha)$.

You now have that $\zeta^{i+j}\alpha^2 \in F$ so by taking the 4th power you also have that $\zeta^{4i+4j}\alpha^8=5\zeta^{4i+4j}\alpha \in F$ and therefore $\zeta^{4i+4j}\alpha \in F$. We now have two options - either $\zeta^{4i+4j}=1$ but then $E=F$, or it is some primitive 7 root of unity, which then must be in $F(\alpha)$. But this means that $6\mid [F(\alpha):\mathbb{Q}]$ - contradiction.

  • $\begingroup$ Absolutely brilliant: +1 of course. As a variant you may also use that $F(\alpha)=\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$ only contains real numbers. $\endgroup$ Mar 18, 2014 at 15:27
  • $\begingroup$ Thanks. I follow most of this, but how do you know the irreducible quadratic has form $(x-\zeta^i\alpha)(x-\zeta^j\alpha)$? Where do the primitive $7$th roots of unity come in? I know how to prove the fact about $x^p-a$ preceding that, but why is that relevant if we're dealing with an irreducible quadratic, and not a reducible one? $\endgroup$ Mar 18, 2014 at 20:00
  • 1
    $\begingroup$ @Hana Bailey : this is because it is the minimal polynomial of $\alpha$ over $F$, and therefore must divide the polynomial $x^7 - 5$. Also added it to my answer. $\endgroup$
    – Ofir
    Mar 18, 2014 at 21:16
  • $\begingroup$ @Prometheus That clears it up, thank you. $\endgroup$ Mar 18, 2014 at 21:20
  • $\begingroup$ @Prometheus I was able to follow this until the very last part. Why must $6$ divide $[F(\alpha):\mathbb{Q}]$ given $F(\alpha)$ contains a primitive $7$th root of unity? $\endgroup$
    – pretzelman
    Apr 29, 2015 at 2:02

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