Possible $|F(\alpha, \beta):F|$ where $|F(\alpha):F|=6$ and $|F(\beta):F|=15$?

Question

Here is a past paper problem which I am struggling to solve currently.

Let $\alpha,\beta\in E$, where $E$ an extension of field $F$. We are given $|F(\alpha):F|=6$ and $|F(\beta):F|=15$.

What are the possible values of $|F(\alpha, \beta):F|$?

I know that $F(\alpha, \beta)$ is a field extension for both $F(\alpha)$ and $F(\beta)$. I tried to use the Tower Rule but could not conclude anything here. My best guess is that $|F(\alpha, \beta):F|$ must be a multiple of both $6$ and $15$.

Any suggestions? Thank you.

Answer 1

Your observation about $[F(\alpha, \beta) : F]$ being a multiple of both $6$ and $15$ is solid. Additionally, observe that $[F(\alpha, \beta) : F(\alpha)] \leq 15$, because the degree of $\beta$ over $F(\alpha)$ cannot be bigger than its degree over smaller field $F$. This means that $[F(\alpha, \beta) : F] \leq 6 \cdot 15 = 90$. I think that the upper bound is achievable for specific choices of $\alpha$ and $\beta$: I guess $\alpha = \sqrt[6]{2}, \beta = \sqrt[16]{3}$ will work, you'd have to check it for yourself

By similar, divisibility based arguments, you can that the only two other options are 30 and 60, and you can find specific examples with $\alpha$ and $\beta$ having appropriate algebraic dependency of low enough degree.

Answer 2

$F[\alpha,\beta]\ge F[\beta]\ge F$, so $[F[\alpha,\beta]:F]$ is divisible by $15$. Similarly, it is divisible by $6$, so it is divisible $30$. On the other hand, it does not exceed $15\cdot 6=90$ because $[F[\alpha,\beta]:F]=[F[\alpha,\beta]:F[\beta]]\cdot [F[\beta]:F]$ and $[F[\alpha,\beta]:F[\beta]]\le [F[\alpha]:F]$.

So $[F[\alpha,\beta]:F]$ is either $30, 60$ or $90$. You get $30$ if $\alpha=2^{1/6}.\beta=2^{1/15}$. You get $90$ if $\alpha=2^{1/6},\beta=3^{1/15}$ and you get $60$ if $\alpha=2^{1/6}, \beta=(e^{2\pi i/15})(2^{1/5})$.