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

Question

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 \begin{equation} \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] . \end{equation} 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 \begin{equation} \left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5}, \end{equation} which gives \begin{equation} 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, \end{equation} or \begin{equation} 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, \end{equation} 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!

Answer 1

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 $\mathfrak{p} = (4 + \sqrt{5})$.

Answer 2

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.

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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!

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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.

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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".

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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.)

Answer 3

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 $6$.