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.


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 $3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.