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.

  • $\begingroup$ I think you should instead consider $[\Bbb{Q}(\alpha, \sqrt[5]{2}):\Bbb{Q}]=[\Bbb{Q}(\alpha, \sqrt[5]{2}):\Bbb{Q}(\alpha)][\Bbb{Q}(\alpha):\Bbb{Q}]$, where we can take $\alpha=\sqrt{3}$ $\endgroup$ – JasonM Aug 4 '16 at 4:53
  • $\begingroup$ @AlexWertheim I couldn't see it while typing the code until I posted it. I fixed it $\endgroup$ – JasonM Aug 4 '16 at 4:56
  • $\begingroup$ @AlexWertheim I completely forgot about that, you are right. In that case, I am even more stumped than before. I think I will have to think more in the morning, but I welcome any good advice for how to approach this question. $\endgroup$ – Alfred Yerger Aug 4 '16 at 4:56
  • $\begingroup$ @AlfredYerger: by a wonderful coincidence, the following recently posted question answers yours as well: math.stackexchange.com/questions/1880695/…. See Prof Lubin's excellent answer there. $\endgroup$ – Alex Wertheim Aug 4 '16 at 4:57
  • $\begingroup$ @AlexWertheim Sorry, I meant they should also consider it. $\endgroup$ – JasonM Aug 4 '16 at 4:59

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.

  • 2
    $\begingroup$ Nice. In fact, we need not use casework, since both $2$ and $5$ must divide the degree $n$ of $\mathbb{Q}(\sqrt{3}, \sqrt[5]{2})$ over $\mathbb{Q}$, and $n \leqslant 10$ certainly... $\endgroup$ – Alex Wertheim Aug 4 '16 at 5:10

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