# What are the proper subfields of the cyclotomic field extension generated by a primitive fifth root of unity?

I'm currently studying up for an abstract algebra qualifier exam (and final exam). In my review, I've found this problem. I think it's a good practice problem; it touches upon a few good aspects of Galois theory and field theory that feel relevant in my course. So, here's the problem statement:

Let $$\epsilon_{5}$$ be a primitive 5th root of unity, and denote $$\theta = \epsilon_{5} + \epsilon_{5}^{-1}$$ as an element of the cyclotomic field $$\mathbb{Q}(\epsilon_{5})$$. Show that the minimal polynomial of $$\theta$$ over $$\mathbb{Q}$$ is $$m_{\theta, \mathbb{Q}}(x) = x^{2} + x - 1$$, and show that $$\mathbb{Q}$$ and $$\mathbb{Q}(\theta)$$ are the only proper subfields of $$\mathbb{Q}(\epsilon_{5})$$.

Here's my solution. If we plug $$\theta$$ into the (suspected) minimal polynomial, we get $$m_{\theta, \mathbb{Q}}(\theta) = (\epsilon_{5} + \epsilon_{5}^{-1})^{2} + (\epsilon_{5} + \epsilon_{5}^{-1}) - 1$$ $$= (\epsilon_{5}^{2} + 2 + \epsilon_{5}^{-2}) + (\epsilon_{5} + \epsilon_{5}^{-1}) - 1$$ $$= 1 + \epsilon_{5} + \epsilon_{5}^{2} + \epsilon_{5}^{3} + \epsilon_{5}^{4} = 0,$$ because $$\epsilon_{5}^{-1} = \epsilon_{5}^{4}$$ and $$\epsilon_{5}^{-2} = \epsilon_{5}^{3},$$ and because $$\epsilon_{5}$$ is a root of the $$5$$th cyclotomic polynomial. So $$\theta$$ is a root of $$m_{\theta, \mathbb{Q}}(x)$$.

Now, we can find the roots of $$m_{\theta, \mathbb{Q}}(x)$$ by the quadratic formula: $$x = \frac{-1 \pm \sqrt{5}}{2}.$$ Neither of those are in $$\mathbb{Q}$$, so $$m_{\theta, \mathbb{Q}}(x)$$ has no roots in $$\mathbb{Q}$$; since it's a quadratic, it's irreducible over $$\mathbb{Q}$$, too. So $$m_{\theta, \mathbb{Q}}(x)$$ is definitely the minimal polynomial of $$\theta$$ over $$\mathbb{Q}$$.

Now, we aim to show that $$\mathbb{Q}$$ and $$\mathbb{Q}(\theta)$$ are the only proper subfields of $$\mathbb{Q}(\epsilon_{5})$$. First, we want to show $$\mathbb{Q}(\theta) \neq \mathbb{Q}(\epsilon_{5}).$$ Certainly, $$\mathbb{Q}(\theta) \subseteq \mathbb{Q}(\epsilon_{5})$$, so we can write $$[\mathbb{Q}(\epsilon_{5}) : \mathbb{Q}] = [\mathbb{Q}(\epsilon_{5}) : \mathbb{Q}(\theta)] \times [\mathbb{Q}(\theta) : \mathbb{Q}].$$

Notice that $$\mathbb{Q}(\epsilon_{5})/Q$$ is a cyclotomic (and hence, Galois) extension over $$\mathbb{Q}$$. The Galois group of a cyclotomic extension is well-known; in this case, it's $$\text{Gal}(\mathbb{Q}(\epsilon_{5})/\mathbb{Q}) \cong (\mathbb{Z}/5\mathbb{Z})^{\times} =$$ {$$1, 2, 3, 4$$}.

In fact, since it's Galois, we have $$[\mathbb{Q}(\epsilon_{5}) : \mathbb{Q}] = |\text{Gal}(\mathbb{Q}(\epsilon_{5})/\mathbb{Q})| = |(\mathbb{Z}/5\mathbb{Z})^{\times}| = 4.$$

And as an aside that will be useful later, it's not too hard to check that $$(\mathbb{Z}/5\mathbb{Z})^{\times}$$ has exactly three subgroups: the trivial subgroup, the whole group, and the non-trivial subgroup {$$1,4$$}.

Now, since $$\mathbb{Q}(\theta)$$ is certainly a subfield of $$\mathbb{Q}(\epsilon_{5})$$ which is Galois over $$\mathbb{Q}$$, $$\mathbb{Q}(\theta)$$ is also Galois over $$\mathbb{Q}$$. Since $$\theta$$ has minimal polynomial $$x^{2}+x-1$$ (of degree $$2$$) over $$\mathbb{Q}$$, and $$\mathbb{Q}(\theta)/\mathbb{Q}$$ is Galois, we have $$\mathbb{Q}(\theta) \cong \mathbb{Q}[x]/(x^{2}+x-1)$$ $$\implies [\mathbb{Q}(\theta) : \mathbb{Q}] = \text{deg}(x^{2}+x-1) = 2.$$

Plugging these values into our equation above, we have $$4 = [\mathbb{Q}(\epsilon_{5}) : \mathbb{Q}(\theta)] \times 2$$ $$\implies [\mathbb{Q}(\epsilon_{5}) : \mathbb{Q}(\theta)] = 2.$$ In particular, then, $$\mathbb{Q}(\theta) \neq \mathbb{Q}(\epsilon_{5})$$, so $$\mathbb{Q}(\theta)$$ is a proper subfield of $$\mathbb{Q}(\epsilon_{5})$$.

Finally, recall that $$\text{Gal}(\mathbb{Q}(\epsilon_{5})/\mathbb{Q}) \cong (\mathbb{Z}/5\mathbb{Z})^{\times}$$ has three subgroups. So by the Fundamental Theorem of Galois Theory, $$\mathbb{Q}(\epsilon_{5})$$ has three subfields; excluding itself, that's two proper subfields.

Certainly, $$\mathbb{Q}$$ is a proper subfield of $$\mathbb{Q}(\epsilon_{5})$$. And we've already shown that $$\mathbb{Q}(\theta)$$ is a proper subfield. Since there are only two such subfields, these must be the only proper subfields, and we're done.