Find a field with $a,b$ in an extension field such that $F(a, b) \neq F(a), F(a, b) \neq F(b)$, and $[F(a, b):F] \lt [F(a):F][F(b):F]$

Question

Find an example of a field $F$ and elements $a$ and $b$ from some extension field such that $F(a, b) \neq F(a), F(a, b) \neq F(b)$, and $[F(a, b):F] \lt [F(a):F][F(b):F]$.

I can't think of a single field that this would work for, every case that I've tried ends up failing the third condition.

Anyone have any ideas?

Answer 1

Hint You know that $$[F(a, b):F] = [F(a,b):F(b)][F(b):F] \\ [F(a,b):F(b)] \leq [F(a):F]$$

Therefore, you need to find some $b$ such that the minimal polynomial of $b$ over $F(a)$ is of smaller degree than over $F$.

Hint 2 $$\left( \sqrt[4]{18}\right)^2 \in \mathbb Q[\left( \sqrt[4]{2}\right)^2 ]$$

Answer 2

$$F=\Bbb Q\;,\;\;a=\sqrt[3]2\,w\;,\;\;b=\sqrt2\,i$$

with $\;w=e^{2\pi i/3}\;$. Check that:

$$\Bbb Q(\sqrt[3]2\,w,\,\sqrt2)\neq\Bbb Q(\sqrt[3]2\,w),\,\Bbb Q(\sqrt2)$$

and

$$\left[\Bbb Q(\sqrt[3]2\,w,\,\sqrt2):\Bbb Q\right]=\left[\Bbb Q(\sqrt[3]2\,w,\,\sqrt2):\Bbb Q(\sqrt2)\right]\left[\Bbb Q(\sqrt2):\Bbb Q\right]=$$

$$=2\cdot2<\left[\Bbb Q(\sqrt[3]2\,w):\Bbb Q\right]\left[\Bbb Q(\sqrt2):\Bbb Q\right]=3\cdot2$$

Further hint:

$$(\sqrt[3]2\,w)^2+\sqrt[3]2\,(\sqrt[3]2\,w)+(\sqrt[3]2)^2=0$$