Let $L/K$ be a Galois extension. I would like to understand how to compute the fixed field of a subgroup $H \leq Gal(L/K)$ as explicitly as possible. The Fundamental theorem of Galois theory often gives us extensions of $K$ in the form $L^H$, which is not very illuminating.

Specifically, the question was inspired be the following construction:

Let $L$ be the splitting field of $X^{24}-1$ over $\mathbb{Q}$. Then $L=\mathbb{Q}(\xi)$ is a cyclotomic extension where $\xi$ is a primitive $24$th root of unity. The reciprocity map $\chi : Gal(L/\mathbb{Q}) \to (\mathbb{Z}/24\mathbb{Z})^*$ given via $\sigma \mapsto a$ where $\sigma \in Gal(L/\mathbb{Q})$ is such that $\sigma(\xi)=\xi^a$ is an injective group homomorphism in general. It is an isomorphism here because cyclotomic polynomials are irreducible over $\mathbb{Q}$.

Therefore $Gal(L/\mathbb{Q})\cong (\mathbb{Z}/24 \mathbb{Z})^* = \{1,5,7,11,13,17,19,23\} $ is abstractly isomorphic to $(\mathbb{Z}/2 \mathbb{Z})^3$ and it is relatively easy to write down the lattice of its subgroups. The Fundamental theorem of Galois theory now gives us a bijection with the lattice of intermediate fields $\mathbb{Q} \subset M \subset L$ via $H \mapsto L^H$. How can we express the fixed fields $L^H$ as $\mathbb{Q}(\alpha)$ or as $\mathbb{Q}(\alpha, \beta)$?

For example, take $H=\{1,11,17,19 \}$. All of these automorphisms fix $\eta = \xi+\xi^{11}+\xi^{17}+\xi^{19}$ and so $\mathbb{Q}(\eta) \subset L^H$. Now $\eta \notin \mathbb{Q}$ so this is indeed a proper extension of $\mathbb{Q}$ and $L^H=\mathbb{Q}(\eta)$.

There are two reasons why I am not entirely happy with the process above:

  1. It does not always work: For $H=\{1, 13\}$ we get $\mathbb{Q}(\xi + \xi^{13})$, but actually $\xi + \xi^{13}=0$ so the inclusion is strict.
  2. Even if it does work, it is not a priori clear that some sum of powers of $\xi$ is not fixed by any other automorphism.

I would be interested in both a specific $\alpha$ such that $\mathbb{Q}(\alpha)=L^{\{1,11,13,23\}}$ and a more general framework about how to go about finding generators for fixed fields given by the Galois correspondence.

  • $\begingroup$ How about $\xi^2+\xi^{22}$? $\endgroup$ Apr 18, 2020 at 14:59
  • $\begingroup$ This works, but it's very random. What is the motivation behind choosing this? $\endgroup$
    – mathmo
    Apr 18, 2020 at 17:05
  • $\begingroup$ If the trace of $\xi$ doesn't work, try the trace of $\xi^2$.... $\endgroup$ Apr 18, 2020 at 18:15

1 Answer 1


This is actually fairly straightforward in the case that you know a primitive element for your Galois extension.

If $H$ is the subgroup and $\alpha$ is the primitive element, simply take the field $E$ generated by the coefficients of the following polynomial $$f(x) = \prod_{\sigma \in H} (x - \sigma(\alpha))$$ Then $L^H = E$.

By construction, the coefficients are all invariant under $H$, $E\subseteq L^H$. On the other hand, $\alpha$ is a root of $f(x) \in E[x]$ which has degree $|H|$ and so $[L:E] \leq |H|$. Comparing the containment with this degree, we can only have $E=L^H$ (if the containment were proper, the degree $[L:E]$ would be strictly greater than than $[L:L^H] = |H|$, contradiction).

In the cases you have, sometimes not all of the coefficients are necessary, as for instance the 2nd coefficient of this polynomial for your first subgroup is exactly the element you already named (with a minus sign). In general these "coincidences" won't happen.

If we try it for your subgroup $H= \{1,13\}$ we get $$f(x) = (x-\zeta)(x-\zeta^{13}) = x^2 - (\zeta + \zeta^{13})x + \zeta\zeta^{13}$$ So generators over $K$ for the fixed field would be $\zeta + \zeta^{13}$ and $\zeta^{14}$. As you observed, the first is actually zero, so the only generator you need is $\zeta^{14}$ (which we can readily see is fixed by $H$ and has the right degree, since it is a primitive $12$th root of unity).

  • $\begingroup$ Thank you. In general though, this polynomial may have a large degree and we get an extension of $\mathbb{Q}$ generated by a lot of elements. Is there a way to see which of them are unnecessary? I would like to avoid using $\mathbb{Q}(\xi+\xi^{13},\xi^{14})$ for $\mathbb{Q}(\xi^{14})$. $\endgroup$
    – mathmo
    Apr 19, 2020 at 8:29
  • $\begingroup$ Not that I know of. There are surely example analogous to this where neither of those two generators alone is sufficient and I’d expect in general one can’t dispose of them all very easy. You could use the more constructive forms of the primitive element theorem to reduce to a single generator in a straightforward algorithmic fashion but the computation would get messy. $\endgroup$
    – user208649
    Apr 19, 2020 at 15:19
  • $\begingroup$ Thank you. Then is seems this is as good as possible in the general case. $\endgroup$
    – mathmo
    Apr 19, 2020 at 15:46

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