Having trouble big time.

I am asked to find the normal closure for the extension $\mathbb{Q}(a):\mathbb{Q}$ where $a$ is the fifth root of $3$ and real. Then I am asked to find Galois groups for the extension above and $N:\mathbb{Q}$, both.

Now, I found $N$ to be $\mathbb{Q}(a,\omega)$ where $\omega=\cos{\frac{2 \pi}{5}}+i \sin{\frac{2 \pi}{5}}$, the fifth root of unity. Problem is, for this $N:\mathbb{Q}=\mathbb{Q}(a,\omega):\mathbb{Q}$ extension, finding the Galois group, equivalently finding the automorphisms $Aut_{\mathbb{Q}}(\mathbb{Q}(a,\omega)$ is looking unbelievably tedious and difficult.

I mean, the elements of $N=\mathbb{Q}(a,\omega)$ will be something like $p+qa+ra^2+...+ta^4+w\omega+x\omega^2+...+ba\omega+ca^2\omega+...+ga^3\omega^3$ where coefficients are in $\mathbb{Q}$. Basically, the form of the element is massive. Everything except $\omega^4$ terms appear(since $\omega^4=-(\omega^3+\omega^2+\omega+1)$.

..Am I right so far? If so, finding automorphisms on such elements must be so tedious...say, conjugates(if I should be calling them that), $p-qa+ra^2+...+ta^4+w\omega+x\omega^2+...+ba\omega+ca^2\omega+...+ga^3\omega^3$,$p+qa-ra^2+...+ta^4+w\omega+x\omega^2+...+ba\omega+ca^2\omega+...+ga^3\omega^3$ etc would still be a $\mathbb{Q}$-automorphism so anything map as above would be in the Galois group, yes?

I mean, there's just loads of maps (automorphisms) of $\mathbb{Q}(a,\omega)$ that I doubt I have to write them all down for this Galois group.

Or is that really the answer? Some massive monstrous group?

  • $\begingroup$ I might write a complete answer later, but here are some tips. 1) What is the degree of the normal closure over $\mathbb{Q}$? This will be the order of your Galois group. $\endgroup$ Mar 1, 2016 at 23:20
  • 3
    $\begingroup$ 2) There are many ways of describing a group, and listing all the elements with a multiplication table is one of the worst, unless the group is of very small order. In this case, I would recommend trying to find a presentation for the group by finding generators and then computing relations they satisfy. Anyway, I'm pretty sure in the end the Galois group is the Frobenius group of order $20$. $\endgroup$ Mar 1, 2016 at 23:23

2 Answers 2


You correctly state that the normal closure of $F = \mathbb{Q}(\sqrt[5]{3})$ is $L = \mathbb{Q}(\sqrt[5]{3}, \zeta)$, where $\zeta$ is a primitive $5^\text{th}$ root of unity. $L$ has two important subfields, $F$ and $K = \mathbb{Q}(\zeta)$. Note that $K/\mathbb{Q}$ is Galois, since the other roots of its minimal polynomial $\frac{x^5 - 1}{x-1} = x^4 + x^3 + x^2 + x +1$ are just the powers of $\zeta$. Also note that $[\mathbb{Q}(\zeta): \mathbb{Q}] = \varphi(5) = 4$ and $[\mathbb{Q}(\sqrt[5]{3}) : \mathbb{Q}] = 5$ are relatively prime, so $20$ divides $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}]$. Since $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}] \leq 20$, this shows that $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}] = 20$. Thus we have the following field diagram

$\hspace 2.5cm$enter image description here

and the corresponding Galois group diagram.

$\hspace 2cm$enter image description here

Since $K/\mathbb{Q}$ is Galois, then $N = \text{Gal}(L/K)$ is normal in $G = \text{Gal}(L/\mathbb{Q})$. Moreover, since $N \cap H = 1$ by order considerations, then $|NH| = \frac{|N||H|}{|N \cap H|} = \frac{4 \cdot 5}{1} = 20$, so $NH = G$. Then $G \cong N \rtimes H$, i.e., $G$ is the semidirect product of $N$ and $H$. Since $K \cap F = \mathbb{Q}$ and $L = KF$, then $$ H = \text{Gal}(L/F) \cong \text{Gal}(K/\mathbb{Q}) = \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/5\mathbb{Z})^\times \cong \mathbb{Z}/4\mathbb{Z} $$ by results on composite fields and cyclotomic extensions. Now $|N| = 5$ so $N \cong \mathbb{Z}/5\mathbb{Z}$, hence we have (abstractly), $G \cong N \rtimes H \cong \mathbb{Z}/5\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$.

Maybe you're satisfied with that, but we can also get a more concrete description by examining the action of $H$ on $N$ in the semidirect product. Let \begin{align*} \sigma: L &\to L\\ \sqrt[5]{3} &\mapsto \sqrt[5]{3}\\ \zeta &\mapsto \zeta^2 \end{align*} \begin{align*} \tau: L &\to L\\ \sqrt[5]{3} &\mapsto \zeta\sqrt[5]{3}\\ \zeta &\mapsto \zeta \end{align*} Note that $\sigma$ fixes $F$ and $\tau$ fixes $K$, so $\sigma \in H$ and $\tau \in N$. Moreover, since $2$ generates $(\mathbb{Z}/5\mathbb{Z})^\times$, then $\sigma$ generates $H$, and since $\tau$ has order $5$, it generates $N$. Since $N \trianglelefteq G$, then $H$ acts on $N$ by conjugation. Note that $\sigma^{-1}$ is $\zeta \mapsto \zeta^3$ since $3 = 2^{-1} \pmod{5}$. Computing $\sigma \tau \sigma^{-1}$, we find \begin{align*} \sigma \tau \sigma^{-1}: \sqrt[5]{3} &\overset{\sigma^{-1}}{\longmapsto} \sqrt[5]{3} \overset{\tau}{\longmapsto} \zeta \sqrt[5]{3} \overset{\sigma}{\longmapsto} \zeta^2 \sqrt[5]{3}\\ \zeta &\overset{\sigma^{-1}}{\longmapsto} \zeta^3 \overset{\tau}{\longmapsto} \zeta^3 \overset{\sigma}{\longmapsto} \zeta \, . \end{align*} This is exactly the same action as $\tau^2$, so $\sigma \tau \sigma^{-1} = \tau^2$, i.e., $\sigma \tau = \tau^2 \sigma$, which provides us with a commutation relation. This allows us to write every element of $G$ as $\tau^i \sigma^j$ for some $i \in \{0, 1, 2, 3\}$ and $j \in \{0, 1, 2, 3, 4\}$, which accounts for all $20$ of the elements of $G$. Thus we have the presentation $$ G = \langle \sigma, \tau \mid \sigma^4 = \tau^5 = 1, \sigma \tau = \tau^2 \sigma \rangle \, . $$ As you can see here, this is a presentation for the Frobenius group of order $20$.


An automorphism of a Galois extension is uniquely defined be the images of the conjugate elements. So one way to describe the automorphism group is to describe how the conjugate elements are being mapped.

You are right in seeing that $N = \mathbb{Q}(\zeta, a)$, where $a = \sqrt[5]3$ and $\zeta = \exp(\frac{2\pi i}{5})$. As splitting field of $f(X) = X^5 - 3$ over $\mathbb{Q}$, $N/\mathbb{Q}$ is a Galois extension.
Next, in most cases it is very useful to determine the order of the Galois group.

$f$ is minimal polynomial of $a$ over $\mathbb{Q}$, so we have $[\mathbb{Q}(a):\mathbb{Q}] = 5$.

As cyclotomic field, $[\mathbb{Q}(\zeta):\mathbb{Q}] = \phi(5) = 4$, where $\phi$ is Euler's phi function.

$\gcd(5,4) = 1$, so we get $$[\mathbb{Q}(\zeta, a):\mathbb{Q}] = [\mathbb{Q}(\zeta):\mathbb{Q}] \cdot [\mathbb{Q}(a):\mathbb{Q}] = 20$$

Now, we want to find the $\mathbb{Q}$-automorphisms of $N$.
We know (well, in fact I'm assuming that you know ^^) that $$(\mathbb{Z}/5\mathbb{Z})^* \simeq Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$$$$ k \mapsto \zeta^k$$ A $\mathbb{Q}$-automorphism maps roots of an irreducible factor of a polynomial to another root of the same factor. So an element of the Galois group is only able to map $a$ to $\sqrt[5]3 \cdot \zeta^k$ for any $k \in (\mathbb{Z}/5\mathbb{Z})$.

Putting it together, a $\sigma \in Gal(N/\mathbb{Q})$ must fulfil $\sigma(a) = \sqrt[5]3 \cdot \zeta^g$ and $\sigma(\zeta) = \zeta^h$ for a $(g,h) \in (\mathbb{Z}/5\mathbb{Z}) \times (\mathbb{Z}/5\mathbb{Z})^*$.
Because the order of the Galois group is 20, there are no less elements. So these 20 elements form $Gal(N/\mathbb{Q})$

  • $\begingroup$ The end is a bit misleading. Indeed you have the 20 elements of $(\mathbb{Z}/5\mathbb{Z})×(\mathbb{Z}/5\mathbb{Z})^*$, but this is not a direct product, which is the cyclic group $\mathbb{Z}/20\mathbb{Z}$. There are more relations involved. $\endgroup$
    – Ur Ya'ar
    Jul 25, 2016 at 13:21
  • $\begingroup$ @UrBen-Ari-Tishler What would you say to simply "for $g \in \mathbb{Z}/5 \mathbb{Z}$ and $h \in (\mathbb{Z}/5 \mathbb{Z})^*$"? $\endgroup$
    – johnnycrab
    Jul 25, 2016 at 19:13
  • $\begingroup$ it would be better I think. But still, it doesn't give a complete answer, since a group is determined both by the elements and the group action, so this last part needs to be specified to really answer "what is the Galois group" $\endgroup$
    – Ur Ya'ar
    Jul 28, 2016 at 16:21

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