Showing that $\sqrt{5} \in \mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$

Question

I am attempting to show that $\sqrt{5} \in \mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$, where $p > 2$ is prime. I have already shown that $[\mathbb{Q}(\sqrt[p]{2}, \sqrt{5}) : \mathbb{Q}] = 2p$.

If needs be, I can understand that this might constitute proving that $\mathbb{Q}(\sqrt[p]{2} + \sqrt{5}) = \mathbb{Q}(\sqrt[p]{2}, \sqrt{5})$, which I know intuitively but am unsure how to prove. In that regard, I am aware of questions such as this and this, but all answers provided either

• do not seem to generalize easily to cases where not both of the roots are square.
• are beyond the scope of my current course.

Any help towards a (preferably low-level) proof of either the inclusion of $\sqrt{5}$ or the equality of $\mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$ and $\mathbb{Q}(\sqrt[p]{2}, \sqrt{5})$ is much appreciated.

Answer 1

So let $\alpha = \sqrt [p]2+\sqrt 5$ then $$\left(\alpha -\sqrt 5\right)^p=2$$

Expand the left-hand side using the binomial theorem to obtain $$p(\alpha)-q(\alpha)\sqrt 5=2$$ where $q(\alpha)\gt 0$ since all the terms involving $\sqrt 5$ have the same sign. Also $p(\alpha), q(\alpha)$ are polynomials in $\alpha$ and belong to $\mathbb Q(\alpha)$ Finally $$\sqrt 5=\frac {p(\alpha)-2}{q(\alpha)}$$

Answer 2

Here is an answer which uses some elementary properties of field automorphisms (essentially the statement that degree $2$ extension is Galois, but this is something which can be rather easily proven directly using just field theory).

Let $K=\mathbb Q(\sqrt[p]{2}+\sqrt{5})$. Suppose $\sqrt{5}\not\in K$. Then $K(\sqrt{5})$ is a degree $2$ extension of $K$, which implies there exists an automorphism $\sigma$ of $K(\sqrt{5})/K$ such that $\sigma(\sqrt{5})=-\sqrt{5}$. Since $\sqrt[p]{2}+\sqrt{5}\in K$, it is fixed by $\sigma$, so we have $$\sigma(\sqrt[p]{2})=\sigma(\sqrt[p]{2}+\sqrt{5}-\sqrt{5})=\sigma(\sqrt[p]{2}+\sqrt{5})-\sigma(\sqrt{5})=\sqrt[p]{2}+\sqrt{5}+\sqrt{5}.$$

However, $\sqrt[p]{2}+2\sqrt{5}$ is not a conjugate of $\sqrt[p]{2}$. Indeed, since $p>2$, the nontrivial conjugates of $\sqrt[p]{2}$ are not real numbers, and clearly $\sqrt[p]{2}+2\sqrt{5}\neq\sqrt[p]{2}$. This is a contradiction, so we conclude $\sqrt{5}\in K$ after all.

Answer 3

If you can't use Galois theory, we can just rule out the possibilities $[\mathbb{Q}(\sqrt[p]2+\sqrt5):\mathbb{Q}]=1,2,p$ by more elementary means (I assume you can use results such as tower law).

Clearly $\sqrt{5}+\sqrt[p]{2}\notin\mathbb{Q}$, or else $\mathbb{Q}(\sqrt[p]2)\ni\sqrt5$, contradiction.

Start with $$ (\sqrt[p]2+\sqrt5)^n=\sum\binom{n}{k}\sqrt[p]2^{n-k}\sqrt5^k $$ Since $p>2$, we immediately get, by considering the $\sqrt[p]2^2$ coefficient in any possible linear dependence, that $[\mathbb{Q}(\sqrt[p]2+\sqrt5):\mathbb{Q}]\neq 2$ (recall you have $\{\sqrt[p]2^j\sqrt5^k\mid 0\leq j<p; k=0,1\}$ as a basis of $\mathbb{Q}(\sqrt[p]2,\sqrt5)$).

If $[\mathbb{Q}(\sqrt[p]2+\sqrt5):\mathbb{Q}]=p$, then $x=\sqrt[p]{2}+\sqrt5$ satisfies both $(x-\sqrt{5})^p-2=0$ and another monic polynomial (over $\mathbb{Q}$) $m(x)=0$ of degree $p$. Expanding, \begin{align*} x^p-\sqrt{5}px^{p-1}+\binom{p}25x^{p-2}-\dots-5^{(p-1)/2}\sqrt5&=0\\ x^p+m_1x^{p-1}+m_2x^{p-2}+\dots+m_p &=0 \end{align*} therefore we have $$ \underbrace{(m_1+\sqrt{5}p)}_{\neq 0}x^{p-1}+\left[m_2-\binom{p}25\right]x^{p-2}+\dots+(m_p+5^{(p-1)/2}\sqrt5)=0 $$ Thus $x$ satisfies a nontrivial polynomial equation of degree $p-1$ over $\mathbb{Q}(\sqrt5)$, so $[\mathbb{Q}(\sqrt[p]2+\sqrt5,\sqrt{5}):\mathbb{Q}(\sqrt5)]\leq p-1$, which contradicts $[\mathbb{Q}(\sqrt[p]2,\sqrt5):\mathbb{Q}]=2p$.