Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?

Question

I was wondering about the minimal polynomial of real number $$u=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$$ over field $\mathbb{Q}$.

As you can see here, I worked out that $u$ is a root of monic rational polynomial $x^3+3x-4$. This is not irreducible: $$x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)$$ and the second, quadratic, factor has complex roots $\frac{-1\pm i\sqrt{15}}{2}$.

Can I claim that the minimal polynomial of $u$ over $\mathbb{Q}$ is $x-1$?

In other words: does this prove that $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?

In case it does, how can we demonstrate $u=1$ working on the roots in a way different from how I did in the link?

Edit:

$\sqrt[3]{2-\sqrt{5}}$ is meant to be the real cube root of $2-\sqrt{5}$.

According to Mathematica, the minimal polynomial of that number is $t^4-t^3-3 t^2-4 t+16$. — Mariano Suárez-Álvarez, Apr 18 '14 at 21:36
It depends on which cube root of the negative number $2-\sqrt{5}$ you take. Mathematica uses the principal branch. If you use the real cube root $-(-2+\sqrt{5})^{1/3}$ you do get $1$. — Robert Israel, Apr 18 '14 at 21:38
@Mariano: I had seen it. This would mean $u$ of degree $4$ over $\mathbb{Q}$, what you think about it? — MattAllegro, Apr 18 '14 at 21:38

score 7 · Accepted Answer · answered Apr 18 '14 at 21:38

7

A calculation shows that $(1+\sqrt{5})^3=16+8\sqrt{5}$, so the real cube root of $2+\sqrt{5}$ is $\frac{1}{2}(1+\sqrt{5})$.

Similarly, or by using conjugation, the real cube root of $2-\sqrt{5}$ is $\frac{1}{2}(1-\sqrt{5})$.

Add.

answered Apr 18 '14 at 21:38

André Nicolas

507,029

Mark Bennet · Answer 2 · 2014-04-18T22:33:21.357

4

Here's another way of looking at this through a reverse lens:

Let's solve $u^3+3u-4=0$ by Cardano's method, putting $u=x+y$.

Then $(x+y)^3-3xy(x+y)-(x^3+y^3)=0$ and we require:

$$x^3+y^3=4$$ and $$-3xy = 3 \text { so that }xy=x^3y^3=-1$$

Then we note that $x^3$ and $y^3$ are roots of the quadratic $$z^2-4z-1=0$$So that $$z=\frac{4\pm\sqrt{16+4}}{2}=2\pm \sqrt 5$$

Whence $$u=\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}$$ where the cube roots have to be compatible with the constraint $xy=-1$. There are three possibilities and three roots of the cubic. Since $u=1$ is a root of the cubic, one choice will give the root $u=1$.

Since the cubic has only one real root, choosing the real cube roots will give $u=1$.

edited Apr 18 '14 at 22:33

answered Apr 18 '14 at 21:57

Mark Bennet

100,194

Ok, great. I only don't get why the quadratic $z^2-4z-1$, everything else is clear! – MattAllegro Apr 18 '14 at 23:38
1

@MattAllegro If $p+q=4$ and $pq=-1$ then the quadratic $(z-p)(z-q)=z^2-(p+q)z+pq=z^2-4z-1=0$ has roots $p$ and $q$. Put $p=x^3, q=y^3$. – Mark Bennet Apr 19 '14 at 06:04

Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?

Edit:

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