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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 $\mathbb{Q}(\epsilon_{5})$.

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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.

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