The splitting field of $x^4-11$ over $\mathbb{Q}$ is $\mathbb{Q}(i, \sqrt[4]{11})$, so $\mathbb{Q}(i, \sqrt[4]{11})/\mathbb{Q}$ is a Galois extension. I managed to prove that the Galois group of this extension is isomorphic to $D_8$, the dihedral group of order $8$. Write this group as $\{ \langle \sigma, \varphi \rangle \, | \, \sigma^4=\varphi^2=1, \sigma\varphi\sigma=\sigma\}$, where $\sigma$ is the map sending $i$ to itself and $\sqrt[4]{11}$ to $i\sqrt[4]{11}$ and $\varphi$ maps $i$ to $-i$ and keeps $\sqrt[4]{11}$ fixed.

I am trying to show an example of the fundamental theorem of Galois theorem, but I seem to be making a mistake again and again: Consider the subgroups $\langle \sigma\varphi, \sigma^2 \rangle$ and $\langle \sigma^3\varphi \rangle$. To find the intermediate fields corresponding to these subgroups, I look at the fixed fields of the generators.

The map $\sigma\varphi$ sends:

$$ \sqrt[4]{11} \mapsto i\sqrt[4]{11} \\ \sqrt[4]{11^2} \mapsto -\sqrt[4]{11^2} \\ \sqrt[4]{11^3} \mapsto -i\sqrt[4]{11^3} \\ i \mapsto -i \\ i\sqrt[4]{11} \mapsto \sqrt[4]{11} \\ i\sqrt[4]{11^2} \mapsto i\sqrt[4]{11^2} \\ i\sqrt[4]{11^3} \mapsto -\sqrt[4]{11^3}, $$

so $\sigma\varphi$ only fixes $\mathbb{Q}(i\sqrt{11})$. By the same method, I showed that $\sigma^2$ fixes $\mathbb{Q}(\sqrt{11}, i\sqrt{11})$ and $\sigma^3\varphi$ fixes $\mathbb{Q}(i\sqrt{11})$. Then the subgroup $\langle \sigma^3\varphi \rangle$ fixes $\mathbb{Q}(i\sqrt{11})$. But the subgroup $\langle \sigma\varphi, \sigma^2 \rangle$ also fixes this field since $\mathbb{Q}(i\sqrt{11}) \cap \mathbb{Q}(\sqrt{11}, i\sqrt{11}) = \mathbb{Q}(i\sqrt{11})$. I know that different subgroups give different fields, so I must be doing something wrong.


1 Answer 1


One nice feature of the Galois correspondence is that if a subgroup $H < G$ corresponds to an intermediate field $L = K^H$, then we have $|H| = [K : L]$. In your case, the subgroup $\langle \sigma \varphi, \sigma^2 \rangle$ has order 4, so it should correspond to a subfield $L \subset K = \mathbb Q(i, \sqrt[4]{11})$ with $[K : L] = 4$ and thus $[L : \mathbb Q] = 2$. This agrees with your calculation that the fixed field is the quadratic field $\mathbb Q(i\sqrt{11}) = \mathbb Q(\sqrt{-11})$. However, the subgroups $\langle \sigma \varphi \rangle$ and $\langle \sigma^3 \varphi \rangle$ each have order $2$, so they should correspond to quartic extensions of $\mathbb Q$. That is, there must be more elements fixed by these automorphisms than you found.

To summarize your calculation of the fixed fields: you chose a basis of $K$ as a $\mathbb Q$-vector space, and for each element of that basis (except 1, which is of course fixed by every automorphism) you calculated its image under $\sigma \varphi$. You then asserted that the field generated by the fixed basis vectors is the fixed field. But this is not generally true: an automorphism can fix some linear combinations of the basis vectors without fixing the individual basis vectors involved in those combinations. Indeed, because \begin{align*} \sigma \varphi(\sqrt[4]{11}) & = i \sqrt[4]{11} \text{ and} \\ \sigma \varphi(i \sqrt[4]{11}) & = \sqrt[4]{11}, \end{align*} it follows that $\sigma \varphi$ fixes the element $(1+i)\sqrt[4]{11}$. You can similarly check that $\sigma^3 \varphi$ fixes the element $(1-i)\sqrt[4]{11}$. Each of these elements generates a quartic extension of $\mathbb Q$ (containing $\mathbb Q(i \sqrt{11})$), so these quartic extensions must be the true fixed fields. Finally, intersecting $K^{\sigma \varphi} = \mathbb Q((1+i)\sqrt[4]{11})$ with $K^{\sigma^2} = \mathbb Q(\sqrt{11}, i\sqrt{11})$ confirms your claim that the fixed field of the order-4 subgroup $\langle \sigma \varphi, \sigma^2 \rangle$ is $\mathbb Q(i\sqrt{11})$.

If you want to calculate the fixed field without relying on dimension considerations to know when you've got everything, one approach would be to "linearize" the calculation you've already done. Namely, write a general element of $K$ as $a_1 + a_2 \sqrt[4]{11} + \dots + a_8 i \sqrt[4]{11^3}$ with each $a_i \in \mathbb Q$, write down its image under $\sigma \varphi$ (or whatever automorphism you're interested in), and solve the $8$-by-$8$ system of linear equations given by asserting that it's a fixed point.

  • 1
    $\begingroup$ That makes sense. The automorphism may not fix either of the elements by swapping them, while still fixing their linear combinations. For a second, I thought "Then I'll check if the automorphism swaps any two elements of the basis." but of course, it may map $a$ to $b$ to $c$ to a, thus preserving their linear combinations. This is precisely the cycle $(1 2 3)$ in $S_n$, so there are a lot of ways an automorphism can fail to fix the elements of the basis while fixing a larger subfield. Galois theory might just become my favorite topic! Thank you for writing such a detailed explanation! $\endgroup$
    – rmdnusr
    Commented Jan 21, 2022 at 9:26

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