# Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$

Let $$\alpha = \sqrt[3]{4+\sqrt{5}}$$. I would like to prove that $$\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$$. We have $$\alpha^3 = 4 + \sqrt{5}$$, and so $$(\alpha^3 - 4)^2 = 5$$, hence $$\alpha$$ is a root of the polynomial $$f(x)=x^6 - 8 x^3 + 11$$. I tried to prove with various approaches that $$f(x)$$ is irreducible over $$\mathbb{Q}$$ without success, so I devised the following strategy.

Since $$x^2 - 5$$ is irreducible over $$\mathbb{Q}$$, we have $$\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$$. Now from $$\alpha^3 = 4 + \sqrt{5}$$ we get $$\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$$, so that $$\mathbb{Q} \left( \sqrt{5} \right )$$ is a subfield of $$\mathbb{Q} \left( \alpha \right )$$, $$\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$$, and we have $$$$\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .$$$$ Now $$\alpha$$ is a root of the polynomial $$g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$$ given by $$g(x) = x^3 - 4 - \sqrt{5}$$. So to prove our thesis it is enough to prove that this polynomial is irreducible in $$\mathbb{Q} \left (\sqrt{5} \right) [ x ]$$. Being $$g(x)$$ of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that $$g(x)$$ would have some root in $$\mathbb{Q} \left (\sqrt{5} \right)$$. Hence our problem boils down to show that there are no integers $$m_0, m_1, n$$, with $$n \neq 0$$, such that $$$$\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},$$$$ which gives $$$$m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1 + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,$$$$ or $$$$m_0^3 + 15 m_0 m_1^2 - 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1 - n^3 \right)=0,$$$$ which implies, being $$\sqrt{5}$$ irrational, $$\begin{cases} m_0^3 + 15 m_0 m_1^2 - 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1 - n^3 = 0. \end{cases}$$ At this point I am stuck, because I do not know how to prove that this system admits the only integer solution $$m_0 = m_1 = n = 0$$.

Any help is welcome!

• A possible alernative hint: If $\beta ^3 = 4 + \sqrt{5}$ then the norm $Nm(\beta^3) = Nm(4 + \sqrt{5})$ - can you derive a contradiction. Sep 18, 2020 at 11:35
• Here is a silly but cool check that you can use: $n = 16$ gives $16777099$, which is prime. Sep 18, 2020 at 11:38
• @EdwardEvans Thanks for quoting my answer! I knew it would be helpful to somebody. Sep 18, 2020 at 11:39
• @TeresaLisbon You'll be happy, then, to know that I've never since used the criterion! Hahaha, it's very cool though and it always pops into my head on such questions Sep 18, 2020 at 11:41
• $x^6 - 8 x^3 + 11$ is irreducible mod $7$. See WA
– lhf
Sep 18, 2020 at 12:17

As requested by OP I am rewriting my comment as an answer. We will show that $$[\mathbb{Q}(\sqrt[3]{4+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 3$$ by showing that $$f(x) = x^3 - (4+\sqrt{5})$$ has no solution in $$\mathbb{Q}(\sqrt{5})$$.

Rather than the approach in the question we notice that $$\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) = 11$$. In particular suppose that $$\alpha$$ is a root of $$f(x)$$ in $$\mathbb{Q}(\sqrt{5})$$, then \begin{align*} \operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha)^3 &= \operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha^3) \\ & =\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) =11 \end{align*} a contradiction.

What's really going on under the hood here is that $$f(x)$$ is Eisenstein for the prime ideal $$\mathfrak{p} = (4 + \sqrt{5})$$.

• @Mummythetureky Thank you very very much for having written down the answer! I only did not understand your last observation. What does it mean that $f(x)$ is Eisenstein for the prime ideal $\mathfrak{p} = (4+\sqrt{5})$? Sep 18, 2020 at 16:06
• @MaurizioBarbato Eisenstein's criterion has a generalisation to domains (e.g., the ring of integers of $\mathbb{Q}(\sqrt{5})$, see the wikipedia page Sep 18, 2020 at 16:08
• Ops I did not know it: I am just a beginner in algebra! Thank you very very much again for your very very brilliant answer. I would never have got it by myself! Sep 18, 2020 at 16:12

There's of course, the wacky way as suggested by Edward above, thanks to him!

But there's a criterion by Osada , which fits the bill perfectly.

Let $$f(x) =x^n + a_{n-1}x^{n-1} + ... + a_1x \pm p$$ be a monic polynomial with integer coefficients, such that $$p$$ is a prime with $$p > 1 + |a_{n-1}| + ... + |a_1|$$, then $$f(x)$$ is irreducible over the rationals.

Here the criterion applies, and we are done, because the polynomial $$x^6 - 8x^3+11$$ is then irreducible but also one that $$\alpha$$ satisfies, hence must be the minimal polynomial. Therefore the extension by $$\alpha$$ has degree $$6$$ as desired.

We should still see how this works, because looking at the location of complex roots is actually quite a nice way of showing irreducibility of integer polynomials : then the roots are linked to the coefficients via Vieta and something goes wrong in the event of a factorization. This is somewhat different from Eisenstein and mod $$p$$ reduction, so it is nice!

I will give you a sketch of this proof, with spoilers. Let $$f$$ be a polynomial satisfying the premise of Osada's criterion.

• Suppose $$f = gh$$ as polynomials in $$\mathbb Z[x]$$ with $$g,h$$ non-constant. Why should one of $$g$$ or $$h$$ have constant coefficient $$\pm 1$$?

This is because $$|f(0)| = |g(0)h(0)| = p$$, but $$g(0),h(0)$$ are integers so one of them has coefficient $$\pm 1$$.

• WLOG let $$h$$ have constant coefficient $$\pm 1$$. Why is there a root $$\beta$$ of $$h$$ such that $$|\beta| \leq 1$$?

Elsewise all the roots of $$h$$ would be greater than $$1$$ in modulus. By Vieta's formula, $$|h(0)|$$ is the product of the modulus of all the roots, but this is equal to $$1$$, which can't happen if all roots had moduli $$>1$$.

• We actually then have $$f(\beta) \neq 0$$. (HINT : Triangle inequality)

Well, $$f(\beta) = 0$$ implies $$|\beta^n +a_{n-1}\beta^{n-1} + ... + a_1\beta| = p$$, but then using the triangle inequality, the LHS is atmost $$1 + |a_{n-1}| + |a_{n-2}| + ... + |a_1|$$, so can't be equal to $$p$$.

• But $$\beta$$ cannot be a root of $$h$$, and not a root of $$f$$, because $$h$$ divides $$f$$! This completes the proof.

I should add that these techniques for proving irreducibility come under the category "Polynomials with dominant coefficient", where one coefficient is much larger than the others. Indeed, this allows us to locate roots of factor polynomials under their existence , which could not be roots of the original polynomial!

The theorems of Ram Murty and Cohn don't come under this category but come under the category of "polynomials taking prime values". There are others, like "polynomials taking small values", and the most difficult but rewarding theory of "Newton polygons".

As a bonus, I would like to direct you to "Polynomials" by Viktor Prasolov, which is one of the most rewarding books to read if you like to prove irreducibility of polynomials (which you will see a lot in Galois theory) and other estimates and computations regarding polynomials (like orthonormal bases, approximation, inequalities etc.)

• Dear Teresa, I do not know how to thank you for having shared with us another wonderful result about the irreducibility of polynomials in $\mathbb{Z}[x]$. It is really a wonderful world, and I am so happy that people can share these little gems of mathematical beauty! Thank you very very ... much again with all my heart! Sep 18, 2020 at 16:02
• @MaurizioBarbato Thank you for giving me the opportunity to help you. I hope we will meet on this site in the future! Sep 18, 2020 at 16:03
• Dear Teresa, I am really moved by how mathematics can unite people in this broken world! Thanks G-d for giving all of us the desire to know the truth and to share this knowledge with other people! Thank you again, Teresa! Sep 18, 2020 at 16:17

As $$\Bbb Q(\sqrt 5)$$ is a subfield, its automorphism $$\sqrt 5\to-\sqrt 5$$ can be extended, which means that $$\sqrt[3]{4-\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5})$$ and also $$\sqrt[3]{11}=\sqrt[3]{4-\sqrt 5}\sqrt[3]{4+\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5}).$$ Now, $$[\Bbb Q(\sqrt[3]{11}):\Bbb Q]$$ is clearly $$3$$. With a subfield of degree 2 and another of degree 3 over $$\Bbb Q$$, ouf field must be at least of degree $$6$$, hence exactly degree $$6$$.

• Why the automorphism of $\mathbb{Q}(\sqrt{5})$ which sends $\sqrt{5}$ into $-\sqrt{5}$ can be extended? Forgive my ignorance, but I am just a beginner in algebra. Could give me some reference, please? Thank you very very much for your help! Sep 18, 2020 at 16:29
• I don't think this line of attack works. For the reason @MaurizioBarbato described. The automorphism could be extended with certainty if the bigger field were Galois over the smallest. But $\Bbb{Q}(\root3\of{4+\sqrt5})$ is not Galois over $\Bbb{Q}$. For a smaller example consider $F=\Bbb{Q}(\sqrt{2+i})$. Surely $[F:\Bbb{Q}]=4$. The automorphism $\sigma:i\mapsto -i$ does not extend to an automorphism of $F$. For if it did, then $\sqrt{(2+i)(2-i)}=\sqrt5$ would be an element of $F$. But then we would have $F=\Bbb{Q}(i,\sqrt5)$ etc. Sep 18, 2020 at 16:49
• In other words, you can conclude that $\root3\of{11}$ belongs to the normal closure of $\Bbb{Q}(\root3\of{4+\sqrt5})$ but not to that field itself. Sep 18, 2020 at 16:51
• @JyrkiLahtonen Thank you very very very much Jyrki for having clarified this intrigate and delicate issue! I would never have the skills to do it by myself! Sep 20, 2020 at 9:13
• @MaurizioBarbato In that example the contradiction is that $\sqrt{2+i}$ is not an element of $\Bbb{Q}(i,\sqrt5)$. The other is about $\root3\of{4-\sqrt5}$ being an element of the normal closure of $\Bbb{Q}(\root3\of{4+\sqrt5})$. Then so is the product of the two conjugates, $\root3\of{11}$. Nov 25, 2020 at 17:43