Let $E/\mathbb{Q}$ and $F/E$ be finite extensions of fields, let $u$ be an element of $Aut(F/E)$, and let $f$ be an element of $F$. Suppose that (i) $[F:E]=3$, (ii) $F=\mathbb{Q}(f)$ and (iii)$u(f)=f^2$. Prove that f is a primitive 7th root of 1, prove that $F/\mathbb{Q}$ is Galois, and describe the Galois group $Gal(F/\mathbb{Q})$ and describe the extension $E/\mathbb{Q}$ by adjoining radicals.

Attempt at solution: First, $f$ is not in $\mathbb{Q}$ hence $u(f)=f^2$ implies the automorphism is not trivial. Since $[F:E]=3$ is prime, there are no subfields between $F$ and $E$. Hence, the fixed field of $Aut(F/E)$ is either $E$ or $F$, but $u$ is an automorphism which does not fix $F$. Hence the fixed field of $Aut(F/E)$ is $E$, hence the extension is Galois, and hence $|Gal(F/E)|=[F:E]=3$.

Hence the Galois group is cyclic with 3 elements, hence generated by the non-identity automorphism $u$, hence $u^3(f)=f^8=f$, hence we have $f^7=1$ hence f is a primitive 7th root of unity.

The extension is Galois since it is given by the degree 6 irreducible (hence separable since we are over a field of characteristic 0) polynomial $x^6 + x^5+...+1$, hence the extension is Galois. By a theorem, the Galois group is isomorphic to $\mathbb{Z}_{\phi (n)}^x$ hence in this case it is the cyclic group on 6 elements.

Finally, we know the extension given is a quadratic extension. Here things are a bit murky. Intuitively, the extension should be $E=\mathbb{Q}(\zeta_3)$ for a primitive third root of unity, but is there a more precise way to say this?

  • $\begingroup$ What is $u$? An element of $Aut(F/E)$ or $Aut(F/\mathbb{Q})$? $\endgroup$
    – user10676
    Aug 15, 2014 at 23:01
  • $\begingroup$ @user10676 Aut(F/E), edited. $\endgroup$ Aug 15, 2014 at 23:15
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    $\begingroup$ How is that intuitive?? (that the extension $E/\mathbb Q$ should be $\mathbb Q(\zeta_3)$) $\endgroup$ Aug 15, 2014 at 23:37

1 Answer 1


The Galois group of $F/\mathbb Q$ is cyclic of order $6$ as you described above, and if $\sigma$ is a generator for this group, then since the unique subgroup of order $2$ is $\{1, \sigma^2, \sigma^4 \}$, we can assume without loss of generality that $\sigma^2 = u$. If you compute $\sigma$, you can compute $\sigma^2$ and explicit $E$ as the fixed field of $\sigma^2$.

Now the extension $F/\mathbb Q$ is simple, so $\sigma$ is characterized entirely by where it maps $f$ (let me write $f = \zeta_7$ for clarity). It suffices to find a primitive root of $(\mathbb Z/7\mathbb Z)^{\times}$ and we got it. After some testing you realize that $3^6 = 1$ but $3^2, 3^3 \neq 1 \pmod 7$, so $3$ is a primitive root. This means that $\sigma(\zeta_7) = \zeta_7^3$ is a possibility for $\sigma$ as a generator of the Galois group.

Now $\sigma^2(\zeta_7) = \zeta_7^9 = \zeta_7^2$ (as promised by $\sigma^2 = u$ and $u(f) = f^2$), so we're trying to find the fixed field corresponding to the group $\{ 1,\sigma^2,\sigma^4\}$.

Because of the simple relation $\zeta_7^7 = 1$, it is not hard to solve the equation $\sigma^2(x)-x = 0$ for $x \in F$ ; write $$ x = a_0 + a_1 \zeta_7 + \cdots + a_5 \zeta_7^5 $$ and apply $\sigma^2$ to it ; the equation $\sigma^2(x) = x$ will give you linear conditions on the coefficients. I'm afraid finding generators involves linear algebra. If you have the patience of writing down the equation by hand, you'll see the computations do not require any smart-ideas and you get $$ E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4). $$ The minimal polynomial can be computed, it is $$ (X - (\zeta_7 + \zeta_7^2 + \zeta_7^4))(X - (\zeta_7^3 + \zeta_7^5 + \zeta_7^6)) = X^2 + X + 2. $$ The roots of this polynomial are $\frac{-1 \pm \sqrt{-7}}2$, so you can write $$ E = \mathbb Q(\sqrt{-7}). $$ Note : The magical reason why I get only one canonical element (i.e. $\zeta_7 + \zeta_7^2 + \zeta_7^4$) as a generator for $E/\mathbb Q$ is because $[E:\mathbb Q] = 2$, so I know the extension can only be generated by one element and that a basis for $E/\mathbb Q$ will be given by $\{1,x\}$ for some $x \in E$. This means when I tried to compute the fixed field of $\sigma^2$, I know the term for $1$ would be irrelevant and the rest would only give me one term.

Interesting remark : You know that $\mathbb Z / 6 \mathbb Z$ has only one subgroup of index $2$ and that the extension $F/\mathbb Q$ is Galois, so by the Galois correspondence $F$ has only one subfield of order $2$ over $\mathbb Q$. The element $\zeta_7 + \sigma^2(\zeta_7) + \sigma^4(\zeta_7)$ is invariant by $\sigma^2$ and is not in $\mathbb Q$, so you could've skipped the linear algebra computations and know already that $E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. We don't escape the computation of the minimal polynomial to simplify this though. (These ideas of averaging are often used in invariant theory ; I averaged the orbit of $\zeta_7$ over the subgroup $\{1,\sigma^2,\sigma^4\}$ to find the invariant substructure I wanted.)

Hope that helps,

  • 1
    $\begingroup$ Very nice! Thanks for the work. $\endgroup$ Aug 15, 2014 at 23:51
  • $\begingroup$ @Bombyxmori : To be honest, OP did most of the work, I'm just a guy not afraid of computing stuff all the way through =) $\endgroup$ Aug 16, 2014 at 0:44

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