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Consider, for $\zeta_5$ the fifth root of unity, $\alpha := \zeta_5^2 + \zeta_5^3 ∈ ℂ$.

To demonstrate: $[ℚ(α): ℚ] = 2$.

We know that $\zeta_5$ is a root of the cyclotomic polynomial $f := \frac{X^5-1}{X-1} = X^4 + X^3 + X^2 + X +1 ∈ ℤ[X]$. As it is monic and irreducible, we must in fact have that $f = f^{ζ_5}_ℚ$ (the minimum polynomial of $ζ_5$ over ℚ) and hence $[ℚ(ζ_5): ℚ] = \deg f = 4$.

How then can $[ℚ(α): ℚ]$ be 2? Or have I already concluded something wrongly?

Edit

Without resorting to trigonometric identities (because the end game here is to derive them), does this allow one to prove that $ℚ(\alpha) = ℚ(√{5})$?

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    $\begingroup$ Why is it inconsistent to have $[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$? $\endgroup$ Mar 15 '20 at 21:58
  • $\begingroup$ To convince yourself that $\alpha$ is in fact an algebraic number of second degree, try computing $\alpha^2+\alpha-1$. As I have showed in my answer, the result is $0$. $\endgroup$
    – Caffeine
    Mar 15 '20 at 22:19
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As you note $[\Bbb{Q}(\zeta_5):\Bbb{Q}]=4$ and clearly $\alpha\in\Bbb{Q}(\zeta_5)$, so you have a tower of fields $$\Bbb{Q}\subset\Bbb{Q}(\alpha)\subset\Bbb{Q}(\zeta_5).$$ The degree is multiplicative over towers of fields so $[\Bbb{Q}(\alpha):\Bbb{Q}]$ must divide $[\Bbb{Q}(\zeta_5):\Bbb{Q}]=4$.

As you know, the first four powers of $\zeta_5$ form a basis for $\Bbb{Q}(\zeta_5)$ as a vector space over $\Bbb{Q}$. From this it is immediate that $\alpha\notin\Bbb{Q}$, which shows that $[\Bbb{Q}(\alpha):\Bbb{Q}]\neq1$. To determine whether the degree equals $2$ or $4$ you can compute a few powers of of $\alpha$: If the degree is $2$ then $\alpha^0,\alpha^1,\alpha^2\in\Bbb{Q}(\alpha)$ must be linearly dependent over $\Bbb{Q}$. A few simple computations show that \begin{eqnarray*} \alpha^0&=&1=\zeta_5^0,\\ \alpha^1&=&\zeta_5^2+\zeta_5^3,\\ \alpha^2&=&(\zeta_5^2+\zeta_5^3)^2=\zeta_5^4+2\zeta_5^5+\zeta_5^6=2+\zeta_5+\zeta_5^4, \end{eqnarray*} where we used the fact that $\zeta_5^5=1$. Now keep in mind that $1+\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4=0$, so $$\alpha^2+\alpha-1=0.$$ This shows that $\Bbb{Q}(\alpha)$ is a quadratic extension of $\Bbb{Q}$.

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    $\begingroup$ Great answer! As a very small observation: $f_\alpha=X^2+bX+c$, since the minimal polynomial is monic $\endgroup$
    – Caffeine
    Mar 15 '20 at 22:29
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    $\begingroup$ @Caffeine Thank you! And I agree, though I wasn't sure whether to assume this given the basic level of the question. I rephrased the argument to work without mentioning minimal polynomials. $\endgroup$
    – Servaes
    Mar 15 '20 at 22:38
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    $\begingroup$ To correct the minorest slip in this perfect answer: it’s the first four powers of $\zeta_5$ that form a basis, since $\zeta_5^4$ is a linear combination of the lower powers. $\endgroup$
    – Lubin
    Mar 16 '20 at 0:00
  • $\begingroup$ @Lubin Thanks for spotting that, it's fixed now. $\endgroup$
    – Servaes
    Mar 16 '20 at 1:00
  • $\begingroup$ All magnificent answers! I wish I could 'accept' them all, but @Caffeine was first. If anyone cares to go further, check the edit $\endgroup$ Mar 16 '20 at 22:36
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Fast method: (even though not very intuitive)

$$H(X):=X^2+X-1\\ Q(X):=H(X^2+X^3)=(X^2+X^3)^2+X^2+X^3-1=X^6+2X^5+X^4+X^3+X^2-1\\ Q(\zeta_5)=\zeta_5+2+\zeta^4+\zeta^3+\zeta^2=\zeta^4+\zeta^3+\zeta^2+\zeta+1=0\\ H(\alpha)=0$$

Actually, with a little bit more computations, one is able to prove

$$\alpha=-\frac{1}{2}-\frac{\sqrt{5}}{2}$$ Intuitive method:

First note that, since $\alpha\in \mathbb{R}$ (because $\overline{\zeta_5^2}=\frac{1}{\zeta_5^2}=\zeta_5^3$), $\mathbb{Q}(\alpha)\subsetneq\mathbb{Q}(\zeta_5)$

We then have $$4=[\mathbb{Q}(\zeta_5):\mathbb{Q}]=[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]\\ P(X):=X^3+X^2-\alpha\in \mathbb{Q}(\alpha)[X]$$ So \begin{cases}1<[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]\le \text{deg}(P)=3\\ [\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]|4\end{cases}

And thus

$$[\mathbb{Q}(\zeta_5):\mathbb{Q}(\alpha)]=2\\ [\mathbb{Q}(\alpha):\mathbb{Q}]=\frac{[\mathbb{Q}(\zeta):\mathbb{Q}]}{[\mathbb{Q}(\alpha):\mathbb{Q}]}=2$$

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A lot of nice solutions have already been posted, but I thought I'd add one that was a bit more elementary. The solution below exploits that palindromic property of the coefficients of $x^4+x^3+x^2+x+1$.

First, let $\omega=\zeta_5^2$, and note that

$$\alpha=\zeta_5^2+\zeta_5^3=\omega+\frac{1}{\omega}.$$

Since the minimal polynomial of $\omega$ over $\Bbb{Q}$ is $x^4+x^3+x^2+x+1$, we have that

$$\omega^4+\omega^3+\omega^2+\omega+1=0.$$

And since $\omega\ne0$, we can divide by $\omega^2$ to obtain

$$\omega^2+\omega+1+\frac{1}{\omega}+\frac{1}{\omega^2}=0,$$

which we can rearrange to obtain

$$\left(\omega^2+\frac{1}{\omega^2}\right)+\left(\omega+\frac{1}{\omega}\right)+1=0.$$

We can then complete the square:

$$\left(\omega^2+2+\frac{1}{\omega^2}\right)+\left(\omega+\frac{1}{\omega}\right)-1=0.$$

Hence:

$$\left(\omega+\frac{1}{\omega}\right)^2+\left(\omega+\frac{1}{\omega}\right)-1=0.$$

So $\alpha=\omega+\dfrac{1}{\omega}$ is a root of $x^2+x-1$. Since this is a quadratic polynomial with no rational roots, it is irreducible over $\Bbb{Q}$, and is therefore the minimal polynomial of $\alpha$ over $\Bbb{Q}$. Thus, $\left[\Bbb{Q}(\alpha):\Bbb{Q}\right]=2$.

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  • $\begingroup$ Let $\beta=\zeta_5+\zeta_5^4$. If we had instead defined $\omega=\zeta_5$, then the above proof can be used to show that $\left[\Bbb{Q}(\beta):\Bbb{Q}\right]=2$, as well. $\endgroup$
    – user729424
    Mar 16 '20 at 3:15

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