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

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.

• If the fields are, indeed, finite, then we can conclude that $[F(\alpha,\beta):F]=30$. If $F$ can be an infinite field (and you added the tag finite-fields by mistake), then both $30$ and $90$ are possible. Commented Apr 18, 2022 at 4:03

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.

• How do you get $60$? Commented Apr 17, 2022 at 3:24
• If $\alpha$ and $\beta$ are respectively $6$th and $15$th roots of $2$, then I believe you get $30$ if $\alpha^2$ and $\beta^5$ are the same cube root of $2$, and $60$ if they're different. Commented Apr 17, 2022 at 3:36

$$F[\alpha,\beta]\ge F[\beta]\ge F$$, 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 $$\alpha=2^{1/6}, \beta=(e^{2\pi i/15})(2^{1/5})$$.

• Thank you. Would you mind explaining why it must divide $15\cdot6=90$? Commented Apr 17, 2022 at 3:03
• $60$ doesn't divide $90$. Commented Apr 17, 2022 at 3:20
• Actually I'm not convinced by your argument that it divides $90$. Certainly we don't always have $[F(\alpha, \beta) : F(\beta)] | [F(\alpha) : F]$: consider $F = \mathbb Q$ and $\alpha, \beta$ two different cube roots of $2$. Commented Apr 17, 2022 at 3:24
• The first statement. It's not always true that $[F(\alpha, \beta) : F(\beta)]$ divides $[F(\alpha) : F]$; in my example above, the first is $2$ and the second is $3$. Commented Apr 17, 2022 at 3:29