Can [F: F ∩ E] Differ from 2 in Quadratic Field Extensions?

[minibear]

if K/E is a quadratic extension and field F is contained in K
such that FE=K and [K:F] is finite,
how do I give a non-example to show
[F: F intersects E] might not be 2?

Thanks a lot!

 

[Take_it_Easy]

Very EASY!
Do you have Galios theory in your hands?

Well consider $$K$$ the splitting field of $$x^3 - 2$$ over $$\mathbb Q$$.

By Galois theory you should know the lattice of inter-fields between $$Q$$ and
$$K$$ is isomorphic to the lattice subgroup of the group $$\mathbb S_3$$
of the permutations on 3 elements.

Now this lattice as a unique subgroup on 3 elements and 3 distinct subgroups of 2 elements. Choose two distinct of these and call them $$G_1, G_2$$.
Let's call $$e$$ the trivial subgroup (just one element: the identity permutation).

Call $$\prime$$ the Galois corrispondence and you have the fields

$$K = e \prime$$
$$E = G_1 \prime$$
$$F = G_2\prime$$
and $$E \cap F = (G_1\cdot G_2)\prime = \mathbb S_3\prime = \mathbb Q$$.

You have $$[K:E] = [G_1:e] = 2$$ and $$K/E$$ is a quadratic extension
You have $$[K:F] = [G_2:e] = 2$$ and $$K/F$$ is a finite extension
You have $$F\cdot E = (G_2 \cap G_3)\prime = e\prime = K$$
You have $$[F:E\capF] = [\mathbb S_3:G_2] = 3 \not = 2$$.

 

[minibear]

Thank you so much!

 

[Take_it_Easy]

You are WELCOME!

Well I also noticed I made a 'print' mistake...

in the last row

I wrote $$[F:E]$$ instead of
$$[F: F \cap E]$$

but I guess you noticed the mistake and you got the right meaning.

See you next time!