# Intermediate fields of splitting field

I'm trying to find the intermediate fields of the extension $\mathbb Q\big /\mathbb Q(\alpha)$, where $\alpha = \sqrt{7+\sqrt{13}}$. To do so I've tried to use the Galois correspondence. I've already found that $\rm{Gal}\left(\mathbb Q\big /\mathbb Q(\alpha)\right)$ has order $4$ and that is isomorphic to $\mathbb Z\big/2\mathbb Z\times \mathbb Z\big/2\mathbb Z$, which has three subgroups of order 2. Therefore, there must be three normal intermediate fields in the extension.

$E:=\mathbb Q(\alpha)$ is the splitting field of $f(x)=x^4-14x^2+36$, which has $4$ roots numbered as:

$$\left\{\alpha_1 = \alpha, \alpha_2 = -\sqrt{7+\sqrt{13}}, \alpha_3 = \sqrt{7-\sqrt{13}}, \alpha_4 = -\sqrt{7-\sqrt{13}}\right\}.$$

Then, the automorphisms in $\rm{Gal}\left(\mathbb Q\big /\mathbb Q(\alpha)\right)$ are

$$\sigma_1(\alpha) = \alpha_1$$ $$\sigma_2(\alpha) = \alpha_2$$ $$\sigma_3(\alpha) = \alpha_3$$ $$\sigma_4(\alpha) = \alpha_4$$

EDIT: corrected subgroups

If we see them as elements of $S_4$, they are $id, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)$, respectively. Thus, the subgroups are $H_1:=\langle(1,2)(3,4)\rangle, H_2:=\langle(1,3)(2,4)\rangle$ and $H_3:=\langle(1,4)(2,3)\rangle$ (all isomorphic to $\mathbb Z\big/2\mathbb Z$).

Then, to find the intermediate fields I'm looking for $E^{H_i}=\{x\in E\mid \sigma(x), \forall\sigma\in H_i\}$

However, when I try with $H_2$, for example, it gets very nasty. In this case we'd have to impose that $\sigma_2(\gamma)=\gamma$ for all $\gamma\in E$. On the one hand, since $\{1,\alpha,\alpha^2,\alpha^3\}$ is a $\mathbb Q-$basis for $E$,

$$\gamma = a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3,\qquad a_i\in\mathbb Q$$

On the other hand, we have

$$\sigma_2(\gamma) = a_0+a_1\sigma_2(\alpha)+a_2\sigma_2(\alpha^2)+a_3\sigma_3(\alpha^3)$$

$$=a_0-a_1\alpha+a_2\alpha^2-a_3\alpha^3$$

And then, coefficients of both expressions should be equal.

EDIT: added manipulations of these expressions

From these expressions we get that $\alpha_1=\alpha_3=0$ and $\alpha_0,\alpha_2$ are free (so it has 2 degrees of freedom as expected).

$$\Rightarrow E^{H_2}=\{a+b(7+\sqrt{13})\mid a, b\in \mathbb Q\}=\mathbb Q(7+\sqrt{13})=\mathbb Q(\sqrt{13}).$$

So, $\mathbb Q(\sqrt{13})$ is the intermediate field that corresponds to the group $H_2$.

But I don't know how to do it with the other subgroups..

With $\sigma_3(\alpha) = \alpha_3 = \frac{6}{\alpha}$,

$$\sigma_3(\gamma) = a_0+a_1\frac{6}{\alpha}+a_2\left(\frac{6}{\alpha}\right)^2+a_3\left(\frac{6}{\alpha}\right)^3$$

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Something is wrong here. The permutations: $id, (12), (13), (14)$ do not form a group. For example $(12)(13)=(132)$. If the group has only four elements (I think that is correct), then the non-identity elements cannot have fixed points among the four roots. –  Jyrki Lahtonen Jan 27 '13 at 12:52
You're right. However, $\mathbb Q(\alpha)\big / \mathbb Q$ is simple and $\alpha$ is algebraic over $\mathbb Q$, so the automorphisms are defined by the mappings of $\alpha$ to the roots of the polynomial $f=\rm{Irr}\left(\alpha,\mathbb Q\right)$. I don't see where is my mistake –  synack Jan 27 '13 at 14:08
That is true, but if $\sigma_2$ is defined as the automorphism that maps $\alpha\mapsto\alpha_2$, then (exactly because the automorphism is already fully determined) you cannot conclude that $\sigma_2(\alpha_3)=\alpha_3$. It turns out that instead we are forced to have $\sigma_2(\alpha_3)=\alpha_4$. This shouldn't be a surprise, because $$\alpha_1\alpha_3=\sqrt{(7+\sqrt{13})(7-\sqrt{13})}=\sqrt{7^2-13}=6.$$ So we must have $\sigma(\alpha_3)=6/\sigma(\alpha_1)$ for all the automorphisms $\sigma$. If $\sigma$ moves $\alpha_1$, it must also move $\alpha_3$ and vice versa. –  Jyrki Lahtonen Jan 27 '13 at 15:17
Okay, now I see it, but isn't there any easier way to determine where $\sigma_2$ sends the other roots? Is there any general method? –  synack Jan 27 '13 at 16:00
Well, if you know that the other roots are in the field generated by one of the roots, here $\alpha$, doesn't that imply that you already know how to write the other roots in terms of $\alpha$? The automorphisms must respect such relations. How did you conclude that $\alpha_3\in\mathbb{Q}(\alpha)$ here? –  Jyrki Lahtonen Jan 27 '13 at 16:07

There is a general method to deal with these sorts of extensions. Suppose we have $F$ a field of characteristic not equal to $2$ and suppose we adjoin $\alpha = \sqrt{a + b\sqrt{c}}$ Suppose that $b\neq0$, $a^2 - b^2c = h^2$ for some $h\in F$. Let $f(x)$ be the minimal polynomial of $\sqrt{a + b\sqrt{c}}$ (put some nice conditions on $a,b,c$ so that $F(\alpha)$ is an extension of degree $4$).

Let us call $\alpha_1 = \alpha$ , $\alpha_2 = \sqrt{a - b\sqrt{c}}$, $-\alpha_1= \alpha_3$ and $-\alpha_2 = \alpha_4$. We choose the following presentation for $V_4$ (the Klein 4- group isomorphic to $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$): $$V_4 = \{e, (12)(34), (13)(24),(14)(23)\}.$$ Note that $(14)(23) = \big((13)(24)\big)\big((12)(34)\big)$. Then we can see that each cycle in $V_4$ acts on the roots according to the numbering convention: For example the cycle $(12)(34)$ sends $\alpha_1 \mapsto \alpha_2$, $\alpha_3\mapsto \alpha_4$. Now for the sake of simplicity let us call

$$\theta = (12)(34), \hspace{5mm} \gamma = (13)(24), \hspace{5mm} \gamma\theta = (14)(23).$$

Then it is easy to see that $\theta(\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2, \gamma\theta(\alpha_1 - \alpha_2) = \alpha_1 - \alpha_2$. Now because $\gamma(\alpha_1) = \alpha_3$, squaring both sides we see that $\gamma(\sqrt{c} ) = \sqrt{c}$. It follows from these observations that

$$F(\alpha_1 + \alpha_2) \subset E^{\langle \theta \rangle}, \hspace{5mm} F(\sqrt{c}) \subset E^{\langle \gamma \rangle}, \hspace{5mm} F(\alpha_1 - \alpha_2) \subset E^{\langle \gamma\theta\rangle }.$$

By computing minimal polynomials/degrees we get that these subset inclusions are actually equalities and so you are done.

Edit: As Jyrki points out, the only presentation for $V_4$ that we can choose is the one given in my answer as that presentation is the only one which gives a transitive subgroup of $S_4$.

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+1: But it is not always clear that you can choose the permutation presentation of the Galois group just from knowing its isomorphism type. In this case it is ok, because that is the only transitive copy of $V_4$ inside $S_4$. Transitivity coming from irreducibility of the polynomial we started with. In general you have to calculate the action of the automorphisms on the roots from their definitions. –  Jyrki Lahtonen Jan 27 '13 at 12:59
@JyrkiLahtonen I agree. That's why when I studied Galois theory I was completely stunned to find that we can have two - copies of $V_4$ inside of $S_4$, one normal and one not! –  user38268 Jan 27 '13 at 13:01
@JyrkiLahtonen We had such a problem as an assignment to do; when I was doing this assignment I was trying so hard to understand as to why the copy of $V_4 = \{e,(12),(34),(12)(34)\}$ just wouldn't work and your comment above says exactly why. –  user38268 Jan 27 '13 at 13:02
The last part is not very clear to me. I don't see how you determine so easily that those fields are in fact the intermediate fields. Is there a way to prove it using the fixed elements and a basis as I proposed in my question? –  synack Jan 27 '13 at 16:40
@Kits89 This should be clear from the definition of $\theta$ and $\gamma$. For example I have already chosen a numbering of the roots. If you look at the automorphism $\theta = (12)(34)$ what do you think it most obviously fixes? Well it sends $1$ to $2$ and $2$ to $1$ so if you take their sum this will be fixed by $\theta$ yes? –  user38268 Jan 28 '13 at 2:44

Perhaps it is worth mentioning that $\mathbf{Q}(\alpha) = \mathbf{Q}\left(\sqrt{\dfrac{13}{2}}, \sqrt{\dfrac{1}{2}}\right)$. In fact, as in the post of @BenjaLim, if $a^2 - b^2 c$ is a square (in $\mathbf{Q}$, in this case), you can solve $$\sqrt{a + b \sqrt{c}} = \sqrt{u} + \sqrt{v}$$ for $u, v \in \mathbf{Q}$, as they are the solutions of $x^2 - a x + \dfrac{b² c}{4} = 0$. Rewriting the extension this way makes the computations of the Galois group and of the intermediate fields slightly easier.

Thus, we are regarding the extension $\mathbf{Q}\left(\sqrt{\dfrac{13}{2}}, \sqrt{\dfrac{1}{2}}\right) / \mathbf{Q}$ as the splitting field of the reducible polynomial $$\left(x^2 - \dfrac{13}{2}\right) \cdot \left(x^2 - \dfrac{1}{2}\right),$$ whose roots are $$\beta_{1} = \sqrt{\frac{13}{2}}, \quad \beta_{2} = - \sqrt{\frac{13}{2}}, \quad \beta_{3} = \sqrt{\frac{1}{2}}, \quad \beta_{4} = - \sqrt{\frac{1}{2}}.$$ If you regard the Galois group as a permutation group acting on the indices of the $\beta_i$, it takes the form $\{1,(12),(34),(12)(34)\}$. Not a transitive group, but the roots are those of a reducible polynomial here.

The intermediate subfields of course are $$\mathbf{Q}\left(\sqrt{\dfrac{13}{2}}\right), \quad \mathbf{Q}\left(\sqrt{\dfrac{1}{2}}\right), \quad \mathbf{Q}\left(\sqrt{\dfrac{13}{2}} \cdot \sqrt{\dfrac{1}{2}}\right) = \mathbf{Q}\left(\sqrt{13}\right) .$$

EDIT In a previous version I had made an incorrect claim. Thanks to @Kits89 for letting me know in his comment below.

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Your basis is not $\mathbb Q$-linearly independent, $\beta_1 = -\beta_2$ –  synack Jan 27 '13 at 18:14
So, by your answer I see that the extension is actually: $\mathbb Q(\sqrt{13},\sqrt{2})=\mathbb Q(\sqrt{13}+\sqrt{2})$. –  synack Jan 27 '13 at 18:19
And could you explain the rationale of that factorization, please? Why the roots of the polynomial $x^2-ax+\frac{b^2c}{4}$ give it? –  synack Jan 27 '13 at 18:54
@Kits89 Thanks for your first comment. Let me fix this, and I will come back to the third one. –  Andreas Caranti Jan 27 '13 at 20:45
As to your third comment, square both sides of $\sqrt{a + b \sqrt{c}} = \sqrt{u}+\sqrt{v}$ to get $a+b \sqrt{c} = u+v + 2 \sqrt{u v}$. Comparing, you obtain $u+v=a$ and $uv = b^2 c /4$, so that $u,v$ are the roots of that quadratic equation. The rationale of the factorization, though, is best seen by starting from $\sqrt{u} + \sqrt{v}$ (where $u, v, uv$ are not squares in $\mathbf{Q}$), finding the polynomial over $\mathbf{Q}$ of degree 4 of which this is the root, noticing that it is a biquadratic, and now solving with the usual substitution $y = x^2$ to get the form $\sqrt{a + b \sqrt{c}}$. –  Andreas Caranti Jan 27 '13 at 20:54