Upper and lower bound on the degree of a field extension generated by two elements

Question

Let $K$ be a field and $A=K(\alpha), B=K\alpha$, with $[A:K]=a, [B:K]=b$. Can we put upper and lower bounds on the degree of the field extenstions $K(\alpha,\beta)/K, K(\alpha,\beta)/K(\alpha \beta), K(\alpha \beta)/K$?

Answer 1

Well, we can put an upper bound on $K(\alpha,\beta)/K$ certainly, since $[K(\alpha,\beta):K(\alpha)]\le [K(\beta):K]=b$, so $$[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha)][K(\alpha):K]\le ab.$$ Similarly, a lower bound is $a$, since $[K(\alpha,\beta):K(\alpha)]\ge 1$. $b$ is of course also a lower bound, so we get a lower bound of $\max\{a,b\}$. Note that these are sharp. For the upper bound, take $K=\newcommand{\QQ}{\Bbb{Q}}\QQ$, $\alpha=\sqrt{2}$, $\beta=\sqrt{3}$, and for the lower bound take $K=\QQ$, $\alpha=\beta=\sqrt{2}$. Thus we have $$\max\{a,b\} \le [K(\alpha,\beta):K]\le ab.$$

Now it's basically the same idea for $K(\alpha\beta)$. Notice that $K(\alpha\beta,\alpha)=K(\alpha,\beta)$, so $[K(\alpha,\beta):K(\alpha\beta)]\le a$, and similarly $[K(\alpha,\beta):K(\alpha\beta)]\le b$, so $$1\le [K(\alpha,\beta):K(\alpha\beta)]\le \min \{a,b\}.$$ Note that we can achieve the lower bound of 1, for example with $\QQ$, $\alpha=\sqrt{2}$, and $\beta=\zeta_3$. Since $(\alpha\beta)^2=2\zeta_3^2$, and $(\alpha\beta)^3=2\sqrt{2}$, so $\QQ(\alpha\beta)=\QQ(\alpha,\beta)$ already in this case.

Then we observe that we thus have $1\le [K(\alpha\beta):K] \le ab$, we have already seen that we might have $K(\alpha\beta)=K(\alpha,\beta)$, and in that example, we had $\QQ(\sqrt{2}\zeta_3)=\QQ(\sqrt{2},\zeta_3)=\QQ(\sqrt{2},\sqrt{-3})$, which is a degree $4=ab$ extension. Thus the upper bound is sharp. It is trivial to produce an example where the lower bound is sharp, namely take $\beta=1/\alpha$. Thus we have the (rather weak) bounds $$1\le [K(\alpha\beta):K] \le ab.$$

Admittedly, I went for obvious bounds only involving $a$ and $b$. If you allow yourself other degrees, you might get better bounds.