# Extension of $\zeta_5^2 + \zeta_5^3$ over $ℚ$ is quadratic.

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 $$ℚ(\alpha) = ℚ(√{5})$$?

• Why is it inconsistent to have $[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$? Mar 15 '20 at 21:58
• 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 $0$. Mar 15 '20 at 22:19

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 $$\Bbb{Q}$$.

• Great answer! As a very small observation: $f_\alpha=X^2+bX+c$, since the minimal polynomial is monic Mar 15 '20 at 22:29
• @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. Mar 15 '20 at 22:38
• 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. Mar 16 '20 at 0:00
• @Lubin Thanks for spotting that, it's fixed now. Mar 16 '20 at 1:00
• All magnificent answers! I wish I could 'accept' them all, but @Caffeine was first. If anyone cares to go further, check the edit Mar 16 '20 at 22:36

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

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

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, $$\left[\Bbb{Q}(\alpha):\Bbb{Q}\right]=2$$.

• 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. Mar 16 '20 at 3:15