Consider the splitting field $K$ over $\mathbb Q$ of the cyclotomic polynomial $f(x)=1+x+x^2 +x^3 +x^4 +x^5 +x^6$. Find the lattice of subfields of K and for each subfield $F$ find polynomial $g(x) \in \mathbb Z[x]$ such that $F$ is the splitting field of $g(x)$ over $\mathbb Q$.

My attempt: We know the Galois group to be the cyclic group of order 6. It has two proper subgroups of order 2 and 3 and hence we are looking for only two intermediate field extensions of degree 3 and 2.

$\mathbb Q[\zeta_7+\zeta_7^{-1}]$ is a real subfield.

$\mathbb Q[\zeta_7-\zeta_7^{-1}]$ is also a subfield.

How do I calculate the degree and minimal polynomial?

  • 4
    $\begingroup$ Find a generator $\sigma$ of the Galois group (such a $\sigma$ is given by $\zeta_7 \mapsto \zeta_7^a$ where $a$ generates $(\mathbb{Z}/7\mathbb{Z})^*$), then see which power of $\sigma$ is preserved by your subfields (you only need to look at what happens to the generator). $\endgroup$
    – Joel Cohen
    Jun 3, 2012 at 19:32
  • 1
    $\begingroup$ Note that $\zeta_7^{-1}=\frac{1}{2}((\zeta_7-\zeta_7^{-1})-(\zeta_7-\zeta_7^{-1}))$, and so $\mathbb{Q}(\zeta_7-\zeta_7^{-1})=\mathbb{Q}(\zeta_7)$. $\endgroup$
    – M Turgeon
    Jun 3, 2012 at 20:24
  • 3
    $\begingroup$ When you work on Joel's suggestion, you should find (for full credit you must justify this yourself) that the element $$G=\zeta_7+\zeta_7^2+\zeta_7^4$$ becomes interesting. Calculate $G^2$, and try to express it as a linear combination of $G$ and $1$. In the end you should be able to write $G$ in terms of the square root of a certain rational integer. I'm off to bed now, so wont' write any more details. Good luck! $\endgroup$ Jun 3, 2012 at 21:41

1 Answer 1


Somehow, the theme of symmetrization often doesn't come across very clearly in many expositions of Galois theory. Here is a basic definition:

Definition. Let $F$ be a field, and let $G$ be a finite group of automorphisms of $F$. The symmetrization function $\phi_G\colon F\to F$ associated to $G$ is defined by the formula $$ \phi_G(x) \;=\; \sum_{g\in G} g(x). $$

Example. Let $\mathbb{C}$ be the field of complex numbers, and let $G\leq \mathrm{Aut}(\mathbb{C})$ be the group $\{\mathrm{id},c\}$, where $\mathrm{id}$ is the identity automorphism, and $c$ is complex conjugation. Then $\phi_G\colon\mathbb{C}\to\mathbb{C}$ is defined by the formula $$ \phi_G(z) \;=\; \mathrm{id}(z) + c(z) \;=\; z+\overline{z} \;=\; 2\,\mathrm{Re}(z). $$ Note that the image of $\phi$ is the field of real numbers, which is precisely the fixed field of $G$. This example generalizes:

Theorem. Let $F$ be a field, let $G$ be a finite group of automorphisms of $F$, and let $\phi_G\colon F\to F$ be the associated symmetrization function. Then the image of $\phi_G$ is contained in the fixed field $F^G$. Moreover, if $F$ has characteristic zero, then $\mathrm{im}(\phi_G) = F^G$.

Of course, since $\phi_G$ isn't a homomorphism, it's not always obvious how to compute a nice set of generators for its image. However, in small examples the goal is usually just to produce a few elements of $F^G$, and then prove that they generate.

Let's apply symmetrization to the present example. You are interested in the field $\mathbb{Q}(\zeta_7)$, whose Galois group is cyclic of order $6$. There are two subgroups of the Galois group to consider:

The subgroup of order two: This is the group $\{\mathrm{id},c\}$, where $c$ is complex conjugation. You have already used your intuition to guess that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is the corresponding fixed field. The basic reason that this works is that $\zeta_7+\zeta_7^{-1}$ is the symmetrization of $\zeta_7$ with respect to this group.

The subgroup of order three: This is the group $\{\mathrm{id},\alpha,\alpha^2\}$, where $\alpha\colon\mathbb{Q}(\zeta_7)\to\mathbb{Q}(\zeta_7)$ is the automorphism defined by $\alpha(\zeta_7) = \zeta_7^2$. (Note that this indeed has order three, since $\alpha^3(\zeta_7) = \zeta_7^8 = \zeta_7$.) The resulting symmetrization of $\zeta_7$ is $$ \mathrm{id}(\zeta_7) + \alpha(\zeta_7) + \alpha^2(\zeta_7) \;=\; \zeta_7 + \zeta_7^2 + \zeta_7^4. $$ Therefore, the corresponding fixed field is presumably $\mathbb{Q}(\zeta_7 + \zeta_7^2 + \zeta_7^4)$.

All that remains is to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7 + \zeta_7^2 + \zeta_7^4$. This is just a matter of computing powers until we find some that are linearly dependent. Using the basis $\{1,\zeta_7,\zeta_7^2,\zeta_7^3,\zeta_7^4,\zeta_7^5\}$, we have $$ \begin{align*} \zeta_7 + \zeta_7^{-1} \;&=\; -1 - \zeta_7^2 - \zeta_7^3 - \zeta_7^4 - \zeta_7^5 \\ (\zeta_7 + \zeta_7^{-1})^2 \;&=\; 2 + \zeta_7^2 + \zeta_7^5 \\ (\zeta_7 + \zeta_7^{-1})^3 \;&=\; -3 - 3\zeta_7^2 - 2\zeta_7^3 - 2\zeta_7^4 - 3\zeta_7^5 \end{align*} $$ In particular, $(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 = 0$, so the minimal polynomial for $\zeta_7+\zeta_7^{-1}$ is $x^3 + x^2 - 2x - 1$. Similarly, we find that $$ (\zeta_7 + \zeta_7^2 + \zeta_7^4)^2 \;=\; -2 - \zeta_7 - \zeta_7^2 - \zeta_7^4 $$ so the minimal polynomial for $\zeta_7 + \zeta_7^2 + \zeta_7^4$ is $x^2+x+2$.

  • 1
    $\begingroup$ That this "doesn't come across very clearly", I think, is because to explain this result involves the normal basis theorem, which is not usually part of a first course discussing Galois theory. $\endgroup$
    – KCd
    Jun 3, 2012 at 23:47
  • $\begingroup$ @KCd I'm not sure I agree -- the result isn't too hard to prove on its own. In particular, it's easy to see that anything in $\mathrm{im}(\phi_G)$ will be fixed by the action of $G$, and it's also easy to see that $\phi_G$ acts as multiplication by $|G|$ on the fixed field. $\endgroup$
    – Jim Belk
    Jun 3, 2012 at 23:50
  • 1
    $\begingroup$ You are right. I was thinking about the use of an element generating a normal basis as a single number whose image under (to use your notation) $\phi_H$ for any subgroup $H \subset G$ will provide you with a primitive element of the subfield fixed by $H$. $\endgroup$
    – KCd
    Jun 4, 2012 at 1:16
  • $\begingroup$ @JimBelk I know this answer was posted quite some time ago, but I have found it so useful I have to try. What can you do when the result of the symmetrization is zero? That case came up when studying cyclotomic polynomials and I can't make sense of it. Thank you! $\endgroup$
    – Zanzag
    Nov 13, 2021 at 16:36

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