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I am having some trouble calculating the Galois group (over $\mathbb{Q}$) of $(x^3-5)(x^2-3)$. I can see the splitting field is $F:=\mathbb{Q}(\sqrt[3]{5},\omega,\sqrt{3})=\mathbb{Q}(\sqrt[3]{5},i,\sqrt{3})$, where $\omega$ is a primitive 3rd root of unity, and it has degree 12 over $\mathbb{Q}$. Since the extension is Galois (it is a splitting field extension for a separable polynomial), this makes $|\operatorname{Gal}(F/\mathbb{Q})|=12$ so it is either $S_3 \times S_2$ or $A_4$ as these are the only subgroups of of $S_5$ of order 12. It has a transposition (complex conjugation) so I conclude that it is $S_3 \times S_2$. I found all the subgroups of $S_3 \times S_2$, (all 16 of them), but I can't see to match them up with the intermediate fields. For example, there are three subgroups of order 6 in $S_3 \times S_2$ which should correspond to three field extensions of degree 2. One of them should be $\mathbb{Q}(i)$ since this is an intermediate field with a degree 2 minimal polynomial, but none of the subgroups of order 6 I found have every element fix $i$, so this is a problem. (A subgroup of order 6 has an element of order 3, which should be permuting the roots of $x^3-5$, no?)

Have I gone wrong somewhere? I would appreciate the help.

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    $\begingroup$ How do you deduce that complex conjugation must correspond to a transposition? It's an element of order $2$, true, but $A_4$ has elements of order $2$ as well... $\endgroup$ Commented May 11, 2012 at 4:58
  • $\begingroup$ By transposition I mean "element of order 2", i.e. it is an involution. Yes... this seems silly but I was looking only at the generators of the groups and totally forgot about the other elements. I guess this means it should be $A_4$ then? $\endgroup$
    – nullUser
    Commented May 11, 2012 at 5:02
  • $\begingroup$ Is there a better way to deduce it should be $A_4$ other than brute forcing the wrong option and then finding it didn't work? $\endgroup$
    – nullUser
    Commented May 11, 2012 at 5:07
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    $\begingroup$ It is not $A_4$, because $A_4$ can't have orbits of size 2! $\endgroup$
    – user641
    Commented May 11, 2012 at 5:09
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    $\begingroup$ The splitting field of $x^3-5$ has a Galois group $S_3$ (the discriminant is not a square), and is contained in your field, so your field must have Galois group that has a quotient isomorphic to $S_3$; since $A_4$ does not have $S_3$ as a quotient, the Galois group is not $A_4$. The three subextensions of degree $2$ should be $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\omega)$, and $\mathbb{Q}(i)$. So perhaps you are misidentifying the actions? $\endgroup$ Commented May 11, 2012 at 5:25

3 Answers 3

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Let's use the following numbering on the roots of your polynomial: $$\{z_1=\root3\of5,z_2=ωz_1,z_3=ω^2z_1,z_4=\sqrt3,z_5=−\sqrt3\}.$$ Any automorphism $\sigma$ of the splitting field must permute these numbers as they are the zeros of your polynomial. As the splitting field $F$ is generated by them, the automorphism $\sigma$ is fully determined once we know $\sigma(z_j),j=1,2,3,4,5.$ This gives us the usual way of identifying $\sigma$ with an element of $S_5=\operatorname{Sym}(\{z_1,z_2,z_3,z_4,z_5\})$.

But there are further constraints. The numbers $z_1,z_2,z_3$ are zeros of the factor $x^3-5$. As that factor has rational coefficients, all automorphisms must permute these three roots among themselves. Similarly for the remaining pair $z_4,z_5$ as they are the zeros of $x^2-3$. Therefore in the identification of an automorphism with a permutation in $S_5$ of the previous paragraph only the permutations in $\textrm{Sym}(\{z_1,z_2,z_3\})\times \textrm{Sym}(\{z_4,z_5\})$ are allowed. There are 12 such permutations forming a group isomorphic to $S_3\times S_2$. As you had established by other means that $[F:\mathbb{Q}]=12$, we can conclude that all such permutations come from actual automorphisms, and thus the Galois group is isomorphic to $G=S_3\times S_2$.

Let us first calculate the fixed field of the subgroup $H_1=\langle\sigma\rangle$, where $\sigma=(123)(45)$. Here we are given that $\sigma(z_1)=z_2$ and that $\sigma(z_2)=z_3$. As $\sigma$ is an automorphism of fields we get $$ \sigma(\omega)=\sigma\left(\frac{z_2}{z_1}\right)=\frac{\sigma(z_2)}{\sigma(z_1)}=\frac{\omega^2z_1}{\omega z_1}=\omega. $$ Therefore $\omega$ is fixed by $\sigma$, and therefore also by any power of $\sigma$. Thus $\mathbb{Q}(\omega)\subseteq Inv(H_1)$. From Galois theory we know that $[Inv(H_1):\mathbb{Q}]=|G|/|H|=12/6=2$. As $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ it follows that $\mathbb{Q}(\omega)=Inv(H_1)$.

It is fairly clear that the fixed field of the subgroup $H_2\simeq S_3$ that keeps both $z_4$ and $z_5$ fixed is $Inv(H_2)=\mathbb{Q}(\sqrt3)$.

Let's try the last subgroup $H_3$ of order 6. I describe it as follows. We have a homomorphism $f$ from $S_3=Sym(\{z_1,z_2,z_3\})$ to $S_2=\textrm{Sym}(\{z_4,z_5\})$ that maps a permutation $\alpha\in S_3$ to the identity element $(4)(5)$ (resp. to the 2-cycle $(45)$) according to whether $\alpha$ is an even (resp. odd) permutation. Then $$ H_3=\{(\alpha,f(\alpha))\in S_3\times S_2\mid \alpha\in S_3\}. $$ We easily see that $H_3$ is generated by $\beta=(123)$ and $\gamma=(12)(45)$. Because $\beta$ acts on the set $\{z_1,z_2,z_3\}$ the same way as $\sigma$ above, we see that $\beta(\omega)=\omega$ and also that $$\beta(\sqrt{-3})=\beta(2\omega+1)=2\omega+1=\sqrt{-3}.$$ Because $\beta(z_4)=z_4$, we obviously also have $\beta(\sqrt3)=\sqrt3$. What about $\gamma$? First we get $$ \gamma(\omega)=\gamma(\frac{z_2}{z_1})=\frac{\gamma(z_2)}{\gamma(z_1)}=\frac{z_1}{z_2}=\omega^2. $$ As $2\omega=-1+\sqrt{-3}$ and $2\omega^2=-1-\sqrt{-3}$, this implies that $\gamma(\sqrt{-3})=-\sqrt{-3}$. But we also have $$ \gamma(\sqrt3)=\gamma(z_4)=z_5=-\sqrt3. $$ Putting all these bits together we see that $$ \gamma(i)=\gamma\left(\frac{\sqrt{-3}}{\sqrt3}\right)=\frac{-\sqrt{-3}}{-\sqrt3}=i. $$ Similarly we see that $\beta(i)=i$. We can then conclude that $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.

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Jyrki has already typed up a great answer showing how to compute the fixed fields. In my post I will concentrate on how to compute the Galois group. Let us write $E = \Bbb{Q}(\sqrt{3},i,\sqrt[3]{5}) = \Bbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_4)$ and $F = \Bbb{Q}$ where

$$\begin{eqnarray*} \alpha_1 &=& \sqrt[3]{5} \\ \alpha_2 &=& \omega\sqrt[3]{5} \\ \alpha_3 &=& \omega^2 \sqrt[3]{5} \\ \alpha_4 &=& \sqrt{3} \\ \alpha_5 &=& -\sqrt{3}\\ \end{eqnarray*}$$

and $\omega = e^{e\pi i/3}$. Now what you said about "complex conjugation" and concluding from there that the Galois group was $D_{12} \cong S_3 \times S_2$ and not $A_4$ is not right. I think you meant by complex conjugation the automorphism that exchanges the roots of $x^2- 2$? In any case what I will do now is use my bare hands to compute the Galois group. Then you will be $10^6$ percent sure that it must be $D_{12}$ of order 12. I will now bash out the possibilities for the action of a $\sigma \in \operatorname{Gal}(E/F)$ on the roots (in other words, bash out all possibilities for automorphisms). We will write for these automorphisms cycles of $S_5$ since we can view $\operatorname{Gal}(E/F)$ as a subgroup of $S_5$. For example, the cycle $(123)$ sends $\alpha_1 \mapsto \alpha_2, \alpha_2 \mapsto \alpha_3$ and $\alpha_3 \mapsto \alpha_1$. $\alpha_4$ and $\alpha_5$ are fixed. The elements are:

$$ e = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_2 \\ \alpha_3 \mapsto \alpha_3 \\ \alpha_4 \mapsto \alpha_4 \\ \alpha_5 \mapsto \alpha_5 \end{cases} \hspace{4mm} (45) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_2 \\ \alpha_3 \mapsto \alpha_3 \\ \alpha_4 \mapsto \alpha_5 \\ \alpha_5 \mapsto \alpha_4 \end{cases} \hspace{4mm} (23)(45) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_3 \\ \alpha_3 \mapsto \alpha_2 \\ \alpha_4 \mapsto \alpha_5 \\ \alpha_5 \mapsto \alpha_4 \end{cases} \hspace{4mm} (23) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_3 \\ \alpha_3 \mapsto \alpha_2 \\ \alpha_4 \mapsto \alpha_4 \\ \alpha_5 \mapsto \alpha_5 \end{cases}$$

and the rest of the cycles are $(123),(123)(45),(12),(12)(45),(132)(45),(132),(13),(13)(45)$. Now I explain a bit more on how these were calculated. You start say with $\alpha_1$. Then if it's sent to itself, then you have only two choices for where $\alpha_2$ is sent to. Either itself, or $\alpha_3$. We don't need to do this $\alpha_3$ as where $\alpha_2$ is sent to already determines this. Then you deal with $\alpha_4$ and $\alpha_5$. Keep going and eventually get all the elements I talked about above. Each of those is a valid automorphism of $E/F$ and since the Galois group has 12 elements, this must be all of them.

If we set $x = (123)(45)$ and $y = (12)$ we see that we have the elements of $\operatorname{Gal}(E/F)$ being

$$\{e,x,x^2 ,x^3,x^4, x^5, y,xy, x^2y ,x^3y,x^4, x^5y\}$$

that satisfy the relations $x^6 = y^2 = 1$ and as you can check we have $yx = x^5y = x^{-1}y$. Hence we conclude immediately from here that $\operatorname{Gal}(E/F)$ must be isomorphic to $D_{12}$ because the structure of the elements and the relations that they satisfy are exactly those of $D_{12}$.

Now I see you have some trouble finding the subgroup that fixes $\Bbb{Q}(i)$. Here is how you can do it. Your three subgroups of order 6 are:

$$\langle x \rangle, \{e,x^2,x^4,y,x^2y,x^4, \}, \{e,x^2,x^4, xy,x^3y,x^5y\}.$$

Now I don't think the first subgroup will fix $i$ because it kinda "moves everything around. My guess is that the last one in the right is the one that fixes $i$. We check this. Recall that

$$x^2 = (132),x^4= (123), xy = (13)(45),x^3y = (12)(45),x^5y = (23)(45)$$

and from the fact that $\alpha_2 = \omega \sqrt[3]{5} = \left(\frac{-1 + i\sqrt{3}}{2}\right)\sqrt[3]{5}$ we get that $i = \frac{1}{\alpha_4}\left(\frac{2\alpha_2}{\alpha_1}+ 1\right)$. Now if you apply each element of the subgroup to this and noting that $\omega^2 = \omega^{-1}$ you will easily see that this fixes $i$. For example, if you let $xy = (13)(45)$ act on $i$ we have that

$$\begin{eqnarray*} (13)(45) \Bigg( \frac{1}{\alpha_4}\left( \frac{2\alpha_2}{\alpha_1} +1 \right) \Bigg) &=& \frac{1}{\alpha_5}\left( \frac{2\alpha_2}{\alpha_3} +1 \right) \\ &=& \frac{1}{\alpha_5}(2\omega^2 + 1)\\ &=& \frac{1}{\alpha_5}(-i\sqrt{3})\\ &=& i. \end{eqnarray*}$$

You should be able to bash out the rest. If you do this the smart way like what Jyrki has done you just need to show that $i$ is fixed by just the generators of this subgroup, namely $(123)$ and $(12)(45)$. Does this help you?

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It may help you to have a general remark on when to recognize that a Galois group splits as a direct product.

Suppose that $E/F$ is Galois with Galois group $G$, and that $G$ factors as a product $G = G_1 \times G_2$. Then by the Galois correspondence we have Galois subextensions $E_i/F$ contained in $E/F$, where $E_i = E^{G_i}$. The fact that $G_1$ and $G_2$ generate $G$ says that $E_1 \cap E_2 = E^G = F.$ The fact that $G_1 \cap G_2 = \{1\}$ says that $E_1$ and $E_2$ together generate $E$ (in standard, if slightly flowery, terminology, we say that $E$ is the compositum of $E_1$ and $E_2$).

So to show that $Gal(E/F)$ is a product, we have to write $E$ as the compositum of two subfields $E_1$ and $E_2$, each Galois over $F$, such that $E_1\cap E_2 = F$.

In your case $E = \mathbb Q(5^{1/3}, \omega,\sqrt{3})$ can be written as the compositum of $E_1 = \mathbb Q(5^{1/3},\omega)$ and $E_2 = \mathbb Q(\sqrt{3})$. This gives the product struture on $G$.

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  • $\begingroup$ The general case is pretty much the same. The compositum is a pushout, and via the Galois correspondence, the Galois group of the compositum is a pullback. $\endgroup$
    – user641
    Commented May 13, 2012 at 21:10

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