# Prove that $[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}] = 8$.

I'm trying to prove this result using elementary Field and Galois theory, but in an "efficient" way. It is desirable to avoid the use of powerful theorems of group theory or results about the structure of $$\operatorname{S}_4$$, as my professor suggested me.

Anyway, if a less elementary solution is posted, it will be welcome.

Here the statement required to be proved and my attempt of solution:

Prove that $$[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}] = 8$$.

First of all, I make the following observations:

• $$\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}})$$ is the splitting field of the polynomial $$p(x) = x^4-8x^2+11$$.

• The polynomial $$p(x)$$ is irreducible over $$\mathbb{Q}[x]$$. I can show this, first, proving that it has no rational root and, second, seeing (by hand) that it can't be factored into a product of quadratic polynomials.

• The field extension $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{5},\sqrt{11})$$ is a Galois extension of degree 4. Furthermore, $$\operatorname{Gal}(\mathbb{Q}(\sqrt{5},\sqrt{11})/\mathbb{Q})$$ is the group spanned by \begin{align*}\sigma:&\mathbb{Q} \overset{\operatorname{id}}{\mapsto} \mathbb{Q},\\ &\sqrt{5} \mapsto -\sqrt{5},\\ &\sqrt{11} \mapsto \sqrt{11},\end{align*} and \begin{align*}\tau:&\mathbb{Q} \overset{\operatorname{id}}{\mapsto} \mathbb{Q},\\ &\sqrt{5} \mapsto \sqrt{5},\\ &\sqrt{11} \mapsto -\sqrt{11}.\end{align*}

With the above observations, I will prove that $$\sqrt{4+\sqrt{5}} \not\in \mathbb{Q}(\sqrt{5},\sqrt{11}).$$ I proceed as follows:

Suppose, in order to get a contradiction, that $$\sqrt{4+\sqrt{5}} \in \mathbb{Q}(\sqrt{5},\sqrt{11})$$. Since $$\sqrt{4+\sqrt{5}}\sqrt{4-\sqrt{5}} = \sqrt{11},$$ we have that $$\sqrt{4-\sqrt{5}} \in \mathbb{Q}(\sqrt{5},\sqrt{11}).$$

Then, I compute the orbit of the action of the Galois group $$\operatorname{Gal}(\mathbb{Q}(\sqrt{5},\sqrt{11})/\mathbb{Q})$$ over $$\sqrt{4+\sqrt{5}}$$.

• $$id(\sqrt{4+\sqrt{5}}) = \sqrt{4+\sqrt{5}}.$$

• $$\sigma(\sqrt{4+\sqrt{5}})^2 = \sigma(4+\sqrt{5}) = 4-\sqrt{5}$$, hence $$\sigma(\sqrt{4+\sqrt{5}}) = \sqrt{4-\sqrt{5}}.$$

• $$\tau(\sqrt{4+\sqrt{5}})^2 = \sigma(4+\sqrt{5}) = 4+\sqrt{5}$$, hence $$\tau(\sqrt{4+\sqrt{5}}) = \sqrt{4+\sqrt{5}}.$$

• $$\sigma(\tau(\sqrt{4+\sqrt{5}}))^2 = \sigma(\sqrt{4+\sqrt{5}})^2 = 4-\sqrt{5}$$, hence $$\sigma(\tau(\sqrt{4+\sqrt{5}})) = \sqrt{4-\sqrt{5}}.$$

Therefore, the orbit of $$\sqrt{4+\sqrt{5}}$$ is $$\bigg \{ \sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}} \bigg \}.$$ In particular, $$\sqrt{4+\sqrt{5}}$$ has degree 2 over $$\mathbb{Q}(\sqrt{5},\sqrt{11})^{\operatorname{Gal}(\mathbb{Q}(\sqrt{5},\sqrt{11})/\mathbb{Q})}$$, but it is a Galois extension so $$\mathbb{Q}(\sqrt{5},\sqrt{11})^{\operatorname{Gal}(\mathbb{Q}(\sqrt{5},\sqrt{11})/\mathbb{Q})}=\mathbb{Q}$$. Then we get a contradiction since we know that $$\sqrt{4+\sqrt{5}}$$ has degree 4 over $$\mathbb{Q}$$ (is the root of an irreducible rational polynomial of degree 4).

At this point, we can solve the original problem.

Evidently $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{5},\sqrt{11}) \subset \mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}})$$. Applying the product formula to this chain of extensions: \begin{align*}[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}] &= [\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}(\sqrt{5},\sqrt{11})][\mathbb{Q}(\sqrt{5},\sqrt{11}):\mathbb{Q}]\\&= [\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}(\sqrt{5},\sqrt{11})]*4. \end{align*}

Since $$\sqrt{4+\sqrt{5}} \not\in \mathbb{Q}(\sqrt{5},\sqrt{11})$$ but $$\sqrt{4+\sqrt{5}}^2 \in \mathbb{Q}(\sqrt{5},\sqrt{11})$$, we conclude that $$[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}(\sqrt{5},\sqrt{11})] = 2.$$

This ends the proof.

I want to know if my solution is actually correct. If not, let me know if it can be corrected or if my attempts will not give an "efficient" solution.

Thank's everyone!

Reference:

• Michael Artin - Algebra, 2nd. Edition. Chapter 16, p. 494, Example 16.9.2(a).

Edit 25/01/19:

I realise that I can't conclude directly $$\sigma(\sqrt{4+\sqrt{5}})=\sqrt{4-\sqrt{5}}$$ and the same for the other elements of $$\operatorname{Gal}(\mathbb{Q}(\sqrt{5},\sqrt{11})/\mathbb{Q})$$, as Jyrki Lahtonen pointed out in the comments.

After this, I searched for a new correct solution, and the comment by eduard gives me an adequate idea for trying to prove it.

Let $$\mathbb{K}=\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}})$$.

• We have the chain of extensions $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{11}) \subset \mathbb{K}$$. By applying the product formula for finite extensions, $$[\mathbb{K}:\mathbb{Q}]=[\mathbb{K}:\mathbb{Q}(\sqrt{11})][\mathbb{Q}(\sqrt{11}):\mathbb{Q}]=[\mathbb{K}:\mathbb{Q}(\sqrt{11})]*2.$$

Now, $$\mathbb{K}=\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{11})$$, so we only have to compute the grade of $$\sqrt{4+\sqrt{5}}$$ over $$\mathbb{Q}(\sqrt{11}).$$

We have an irreducible polynomial for $$\sqrt{4+\sqrt{5}}$$ over $$\mathbb{Q}$$, $$p(x)$$, lets prove that this polynomial is still irreducible over $$\mathbb{Q}(\sqrt{11}).$$

• $$\sqrt{4+\sqrt{5}} \not\in \mathbb{Q}(\sqrt{11})$$, if not, $$\sqrt{4+\sqrt{5}}^2 \in \mathbb{Q}(\sqrt{11})$$ which is false.

• $$\sqrt{4-\sqrt{5}} \not\in \mathbb{Q}(\sqrt{11})$$ by the reasons given above.

• If $$p(x)$$ is not irreducible over $$\mathbb{Q}(\sqrt{11})$$, it must factor as a product of quadratics polynomials, but all the possibilities give quadratics polynomials with coefficients not in $$\mathbb{Q}(\sqrt{11})$$.

The possibilities mentioned above are the following:

1. $$(x-\sqrt{4+\sqrt{5}})(x+\sqrt{4+\sqrt{5}})=x^2-(4+\sqrt{5}) \notin (\mathbb{Q}(\sqrt{11}))[x].$$

2. $$(x-\sqrt{4+\sqrt{5}})(x-\sqrt{4-\sqrt{5}})=x^2-\sqrt{8+2\sqrt{11}}x+\sqrt{11} \notin (\mathbb{Q}(\sqrt{11}))[x].$$

3. $$(x-\sqrt{4+\sqrt{5}})(x+\sqrt{4-\sqrt{5}})=x^2-\sqrt{8-2\sqrt{11}}x+\sqrt{11} \notin (\mathbb{Q}(\sqrt{11}))[x].$$

Above we used that $$\sqrt{8\pm2\sqrt{11}} \notin \mathbb{Q}(\sqrt{11})$$. If not, $$\exists a,b \in \mathbb{Q} \quad | \quad \sqrt{8\pm2\sqrt{11}} = a + b\sqrt{11}.$$ Hence, $$8 \pm 2\sqrt{11} = a^2+ 11b^2 +2ab\sqrt{11},$$ which is equivalent to $$ab=\pm 1 \quad \text{and} \quad a^2+11b^2=8.$$ Then, $$a^2+\frac{11}{a^2} = 8 \Longleftrightarrow a^4 - 8a^2 +11 = p(a) = 0.$$ But $$p$$ has no roots in $$\mathbb{Q}$$ and this conclude the proof of the irreducibility of $$p$$ over $$(\mathbb{Q}(\sqrt{11}))[x].$$

With this, we conclude the original statement.

I'm sorry on publishing such a long question.

• From $(\sigma(\sqrt{4+\sqrt{5}}))^2=4-\sqrt{5}$ you can only infer that $\sigma(\sqrt{4+\sqrt{5}})=\pm\sqrt{4-\sqrt{5}}$. The sign ambiguity is still there. Same with the other stuff. – Jyrki Lahtonen Jan 25 '19 at 6:38
• You are correct @JyrkiLahtonen, I saw the mistake just 5 minutes after posting my question. I'm trying to fix the error, but actually I think that fixing it take me to an "non-efficient" solution. My last attempt is to do something similar as eduard did in his answer. I will edit my post updating all this stuff. I'm really grateful to your response. – DrinkingDonuts Jan 25 '19 at 18:34
• I've found an improvement of your strategy. I will post it later too ! – eduard Jan 25 '19 at 18:55
• Thanks for the time you have dedicated to my question @eduard. If in the following days my question doesn't have any new answers, I will mark your answer as the correct one. My best regards. – DrinkingDonuts Jan 25 '19 at 19:39
• You are welcome! – eduard Jan 26 '19 at 18:13

Let $$P= x^4- 8 x^2 + 11$$ and $$\pm\theta, \pm\theta'$$ the roots of $$P$$ in $$\mathbb R$$, $$\theta, \theta'$$ positive.

Since $$\theta \cdot \theta' =\sqrt{11}$$ then $$\mathbb Q(\theta, \theta')=\mathbb Q(\theta,\sqrt{11})$$ and $$[\mathbb Q(\theta, \sqrt{11}):\mathbb Q(\theta)]=:n$$ is either $$1$$ or $$2$$. Let's see that it is $$2$$.

A possible strategy using Galois theory: Assume that $$n=1$$, then $$[\mathbb Q(\theta, \sqrt{11}):\mathbb Q(\sqrt{11}]=2$$ so the minimal polynomial of $$\theta$$ over $$\mathbb Q(\sqrt{11})$$ is degree $$2$$. This is a divisor of $$P$$ thus it is one of the following. $$\begin{array}{c} P_1=(x-\theta)(x+\theta)\\ P_2=(x-\theta)(x-\theta')\\ P_3=(x-\theta)(x+\theta') \end{array}$$ It is easy to see that $$P_1\notin \mathbb Q(\sqrt{11})[x]$$. Indeed $$P_1(x) = x^2 - 4 - \sqrt{5}$$ and $$\sqrt{5}\notin \mathbb Q(\sqrt{11})$$. It is slightly harder to prove that $$P_i$$ is not in $$\mathbb Q(\sqrt{11})[x]$$ for every $$i=2,3$$. This leads to a contradiction.

Using $$29$$-adics: $$P$$ has a simple root $$14$$ in $$\mathbb F_{29}$$. By Hensel's Lemma, $$\mathbb Q(\theta)$$ admits an embedding in $$\mathbb Q_{29}$$, field of $$29$$-adic numbers.

($$29$$ is the smallest prime $$q$$ such that $$P$$ has a simple root in $$\mathbb F_{q}$$).

If $$n=1$$, that is if $$\sqrt{11}\in \mathbb Q(\theta)$$, then $$\mathbb Q_{29}$$ contains a root of $$X^2 -11$$. Hence $$11$$ is a square in $$\mathbb F_{29}$$. This is a contradiction.

Equivalently, since $$P$$ factors in irreducible factors as $$(X+ 14) (X+ 15)(X^2 +14)$$ in $$\mathbb F_{29}$$ then $$\mathbb Q_{29}$$ only contains $$2$$ out of $$4$$ roots of $$P$$. This is a $$p$$-adic version of the following argument over $$\mathbb C$$: $$\mathbb Q(\sqrt{1+\sqrt{7}})/\mathbb Q$$ is not Galois since its minimal polynomial has real and complex non-real roots.

Edit: Following OP's strategy assume that $$\theta \in \mathbb Q(\sqrt{11},\sqrt{5})$$. Since $$G=\mathbb Q(\sqrt{11},\sqrt{5})/\mathbb Q$$ is Galois and $$P$$ is irreducible then $$G$$ acts transitively on $$S=\{\pm\theta,\pm\theta'\}$$. In particular $$S=G(\alpha) :=\{ g(\theta):g\in G\}$$ for every $$\alpha \in S$$. Hence $$S=\{\theta,\sigma(\theta),\tau(\theta),\sigma\tau(\theta)\}$$. But by OP's argument $$\tau(\theta) =\pm \theta$$, hence $$\tau(\theta)=-\theta$$. Similarly one prove that $$\tau(\theta') = -\theta'$$. Hence $$\sqrt{11}=\theta \theta' = \tau(\theta\theta') = \tau(\sqrt{11}) = -\sqrt{11}.$$

If you aim only to show that $$[\mathbf Q (\sqrt{4 \pm \sqrt 5}):\mathbf Q]=8$$, it's much quicker to introduce the quadratic field $$k=\mathbf Q(\sqrt 5)$$ and consider the quadratic extensions $$k(\sqrt{4 \pm \sqrt 5})/k$$. Because $$\mathbf Q(\sqrt{4 \pm \sqrt 5})$$ contains $$\sqrt 5$$, it's obvious that $$k(\sqrt{4 \pm \sqrt 5})=\mathbf Q(\sqrt{4 \pm \sqrt 5})$$. Now, it's well known and straightforward that $$k(\sqrt x)=k(\sqrt y)$$ iff $$xy\in {k^*}^2$$ (this is a particular case of Kummer's theory). Here $$xy=16-5=11$$ cannot live in $$k^*$$ : this can be checked "by hand", or by using again the fact that $$\mathbf Q(\sqrt 5)=\mathbf Q(\sqrt 11)$$ iff $$55$$ is a square in $$\mathbf Q^*$$, which is impossible. So the two quadratic extensions $$k(\sqrt{4 \pm \sqrt 5})/k$$ are distinct, their compositum is biquadratic over $$k$$, and you're done.