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!