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

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

I start with the following observation:

  • 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.

Best Answer

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.

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