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Consider the field $\mathbb{Q}(\alpha)$, where $\alpha$ is one of the (complex) roots of the polynomial $f(x) := x^3 + x + 1 \in \mathbb{Q}[x]$.

I now want to find out if $i \in \mathbb{Q}(\alpha)$ or not.

Now my hopes of having some "nicely looking" roots that can be easily verified as such were shattered when I typed the polynomial in Wolframalpha to see what they actually look like – unless there is a massive simplification for these complex numbers that Wolframalpha doesn't come up with, they don't look all that nice.

One of the roots is of course real-valued, so if $\alpha$ happens to be the real root, then we have $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, hence $i \notin \mathbb{Q}(\alpha)$. But for the other two roots, I don't really know how to approach them. The way this exercise is proposed to me makes me think that there's a more simple way of proving/disproving $i \in \mathbb{Q}(\alpha)$ that doesn't involve working with the explicit values of the roots, but so far, I haven't been able to find such a way.

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Our cubic is irreducible over the rationals, so $\mathbb{Q}(\alpha)$ has degree $3$ over the rationals. If $i$ were in $\mathbb{Q}(\alpha)$, then $\mathbb{Q}(\alpha)$ would have a subfield of degree $2$ over the rationals. This is impossible, since $2$ does not divide $3$.

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