I am interested in the elements of the Galois group of $\mathbb Q(\sqrt{2+\sqrt{2}})/\mathbb Q$.

Let $\alpha:=\sqrt{2+\sqrt{2}}$, then the minimal polynomial $m_{\alpha,\mathbb Q}(X)=X^4-4X^2+2$ has roots $$\pm\sqrt{2+\sqrt{2}}=\pm\alpha\\\pm\sqrt{2-\sqrt{2}}=\pm\beta$$ where $\beta=\frac{\alpha^2-2}{\alpha}\in\mathbb Q(\alpha)$ so the field extension is normal and we can extend the identity $id:\mathbb Q\to\mathbb Q$ to an automorphism $\phi:\mathbb Q(\alpha)\to\mathbb Q(\alpha)$ by permuting the $4$ roots:$$\phi_\alpha=\begin{cases}\alpha\mapsto\alpha\\\beta\mapsto\beta\end{cases}\\\phi_\beta=\begin{cases}\alpha\mapsto\beta\\\beta\mapsto-\alpha\end{cases}\\\phi_{-\alpha}=\begin{cases}\alpha\mapsto-\alpha\\\beta\mapsto-\beta\end{cases}\\ \phi_{-\beta}=\begin{cases}\alpha\mapsto-\beta\\\beta\mapsto\alpha\end{cases} $$ as the images of $\beta$ are determined by the images of $\alpha$ already. In total we have that $\phi_\alpha$ acts as the identity on $\mathbb Q(\alpha)$ and $$\text{Gal}(\mathbb Q(\sqrt{2+\sqrt{2}})/\mathbb Q)=\langle\phi_\beta\rangle=\langle\phi_{-\beta}\rangle\cong\mathbb Z_4$$.

  • 2
    $\begingroup$ is there a question ? $\endgroup$
    – mercio
    Commented May 15, 2018 at 8:52
  • $\begingroup$ Once you know that the extension is normal, you know it is Galois; since clearly the degree is $4$, the Galois group has four elements. Find an automorphism of order $4$ and you're done proving the group is cyclic. $\endgroup$
    – egreg
    Commented May 15, 2018 at 8:54
  • 1
    $\begingroup$ I don't see a question either, but see math.stackexchange.com/questions/575171/… $\endgroup$
    – Arnaud D.
    Commented May 15, 2018 at 8:58
  • $\begingroup$ Woops. My question is whether my thoughts above are correct. Specifically the automorphisms and their actions on the set of roots. $\endgroup$
    – Buh
    Commented May 15, 2018 at 9:27
  • $\begingroup$ Does this answer your question? Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$ $\endgroup$ Commented Jan 15, 2020 at 15:36

1 Answer 1


Everything looks correct. I would add argumentation that the polynomial you got is really irreducible (Eisenstein, for example) and I'd add some calculation to support the claim that, for example, $\alpha\mapsto \beta \implies \beta\mapsto -\alpha$, but I'm guessing you've already done that.

Let me just write this one example, in case you wasn't sure about it. Let's say we have $\alpha\mapsto \beta$. We know that automorphisms act as permutations on the set of roots of the minimal polynomial. We also know that $\beta$ is determined by action on $\alpha$, so it is enough to check that $\beta\mapsto -\alpha$ is indeed permutation consistent with $\beta = \frac{\alpha^2-2}\alpha$. Maybe it involves some case checking, but once you've found the correct thing, you immediately know it's the only possibility.

However, this is not really a viable strategy for bigger extensions. What I would do is write $\beta$ as a polynomial in $\alpha$. We have \begin{align}\alpha^4-4\alpha^2 = -2 &\implies \alpha(\alpha^3 - 4\alpha) = -2\\ & \implies \frac 1\alpha = \frac{4\alpha-\alpha^3}{2} \\ &\implies \beta = \alpha - \frac 2\alpha = \alpha - (4\alpha-\alpha^3) = \alpha^3-3\alpha\end{align}

which can simplify your calculations for automorphisms.

  • $\begingroup$ Alright, thanks! $\endgroup$
    – Buh
    Commented May 15, 2018 at 10:36
  • $\begingroup$ @Buh, you are welcome. Also notice that once you figured out $\beta^3 - 3\beta = -\alpha$, all the other calculations become trivial, no need to use radical expressions any more. $\endgroup$
    – Ennar
    Commented May 15, 2018 at 10:42

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