no quadratic extension of $\mathbb{Q}$ in $\mathbb{Q}[e^{\frac{2\pi i}{5}}]$?

Question

$\mathbb{Q}[e^{\frac{2\pi i}{5}}]$ is an extension of $\mathbb{Q}$ of degree 4, since $x^4+x^3+x^2+x+1$ is the irreducible polynomial of $\theta=e^{\frac{2\pi i}{5}}$ over $\mathbb{Q}$.

I'm asked if there is a quadratic extension $K$ of $\mathbb{Q}$ inside $\mathbb{Q}[e^{\frac{2\pi i}{5}}]$. I suspect that the answer is no.

Since a quadratic extension is always of the form $\mathbb{Q}[\sqrt{k}]$ for an integer $k$, a naive way is to show that the equality $$(a+b\theta+c\theta^2+d\theta^3)^2=k$$ for an integer $k$ is impossible. But that seems tedious.

Another approach would be that in such case, $\mathbb{Q}[e^{\frac{2\pi i}{5}}]$ is itself a quadratic extension, so we are expected to find an element that looks like $\sqrt{a+b\sqrt{k}}$ inside (for rational $a,b$). But then I'm stuck again.

Any ideas?

Answer 1

$\theta$ is a fifth root of unity; $\mathbb{Q}(\theta) / \mathbb{Q}$ is an abelian extension. That is, it is a Galois extension with abelian Galois group.

Every abelian group $G$ of order $n$ has, for every $m \mid n$, at least one subgroup $H$ of order $m$.

Consequently, the extension $\mathbb{Q}(\theta) / \mathbb{Q}$ has at least one subextension of every degree dividing $[\mathbb{Q}(\theta) : \mathbb{Q}]$.

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There are two ways to produce the quadratic subextension.

We can identify the quadratic extension by looking at the ramification in the ring of integers $\mathbb{Z}[\theta]$. For every $p$ except $5$, the algebraic closure of $\mathbb{F}_p$ has four distinct primitive roots of unity. However, over $\mathbb{F}_5$, every fifth root of unity is $1$.

Consequently, the extension ramifies only over the prime $5$. The extension must be either $\mathbb{Q}(\sqrt{5})$ or $\mathbb{Q}(\sqrt{-5})$. Studying how ramification over $2$ works implies that we must be taking the square root of a number that is $1 \bmod 4$, thus the extension is $\mathbb{Q}(\sqrt{5})$.

In general, for odd $p$, $\mathbb{Q}(\zeta_p)$ will contain either $\mathbb{Q}(\sqrt{p})$ or $\mathbb{Q}(\sqrt{-p})$ as a subfield; the correct square root is whichever $\pm p$ is $1 \bmod 4$.

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Another way to produce the quadratic subextension is to observe that $\mathbf{Q}(\theta)$ has complex embeddings, and that complex conjugation acts on the field.

Thus, it has subfield fixed by complex conjugation. We can even identify the subfield as:

$$\mathbb{Q}(\theta + \bar{\theta}) \subseteq \mathbb{Q}(\theta)$$

Since complex conjugation has order 2 (or simply by writing down the minimal polynomial of $\theta$ over the subfield), this field extension is order $2$, and thus $[\mathbb{Q}(\theta + \bar{\theta}) : \mathbb{Q}] = 2$.

Answer 2

We can get an answer in this simple case by solving the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$ Dividing by $x^{2}$ and putting $y=x+1/x$ we get $y^{2}+y-1=0$ so that $$y=\frac{-1\pm\sqrt{5}}{2}$$ and $$x=\frac{y\pm\sqrt{y^{2}-4}} {2}$$ and then clearly we can see that there is a tower of field extensions $$\mathbb{Q} \subset \mathbb{Q} (y) \subset \mathbb{Q} (x) $$ and the degree of each extension is $2$.

More generally Gauss proved that if $p=2^{2^{k}}+1$ is prime then the equation $x^{p} - 1=0$ can be solved by a series of $2^k$ quadratic equations so that the extension $\mathbb{Q} \subset \mathbb{Q} (\zeta_{p}) $ of degree $p-1=2^{2^{k}}$ can be realized as a tower of $2^{k}$ quadratic extensions of $\mathbb{Q}$. This development by Gauss is not so popular as compared to the more abstract work of Galois. See this and this for more details.