I can compute minimal polynomials over $\mathbb Q$ without issue, but this question has me stumped (it's a practice problem for my competency exam).
Determine the minimal polynomial of $\alpha = \sqrt[5] {2}$ over the field $\mathbb Q (\sqrt{3}).$
I know a few things about this problem. First, I know an upper bound for the degree is $5$, since I can find a polynomial over $\mathbb Q$ with degree $5$ of which this is a root in the obvious way. I also know that since we have extended $\mathbb Q$, the minimal polynomial could possibly have lower degree, but need not.
The problem is that I don't know how to determine if $x^5-2$ can be improved upon by the addition of the new field element. I had a suspicion before using multiplication of degrees of extensions, but as pointed out in the comments, this does not work.
I appreciate any hints or advice on this question.
Often, the thing to do is to jot down a hierarchy of all possibly relevant fields, and tally up their relationship (particularly the degree of the extensions). Often you can just solve or easily reason out the unknown links.
$$\begin{matrix} & & \mathbb{Q}(\sqrt{3}, \sqrt[5]{2}) \\ & \nearrow x\!\!\!\!\!\! & & \nwarrow y\!\!\!\!\!\! \\ \mathbb{Q}(\sqrt{3}) & & & & \mathbb{Q}(\sqrt[5]{2}) \\ &\nwarrow 2\!\!\!\!\!\! & & \nearrow 5\!\!\!\!\!\! \\ & & \mathbb{Q} \end{matrix} $$
For example, we know that $y \leq 2$ and $x \leq 5$, and there are a number of ways to proceed with that information.