I want to prove that if $K\leq L,M\leq E$ are fields and $[LM:K]=[L:K][M:K]<\infty$ then $L\cap M=K$.
I did prove it and used the proof of the following theorem, in which I gave a field isomorphism between $LM$ and $L\otimes_K M$.
If $K\leq L,M\leq E$ are fields with finite degree over $K$, then $L\otimes_K M$ is a field iff $[LM:K]=[L:K][M:K]$.
The proof involves that in the isomorphism given $f:L\otimes_K M\to LM$, we have that the inverse image of $U,V$ is identified with $U,V$ in the tensor product, their intersection is $K$ (The space generated for $1\otimes 1$) and its direct image is a field extension of $K$, isomorphic to $K$, so it's obviously $K$ (Since it's finite dimension and, under bijections direct image of intersections is intersection of direct image).
The problem is that even if I know a bit about using tensor products and can use it here, we haven't ever seen that topic in any class. So it's obvious that we won't be able to use tensor products to prove the proposition. So I need to find another proof.
Is there any idea about how to give an elementary proof which only uses basic facts of field theory (Previous to Galois)? Something like Dummit until 13.3 without part III.
For example, we have the tower $$K\leq L\cap M\leq L,M\leq LM$$ With all extension over any field below is finite and the following relations hold, given that $[L:K]=n, [M:K]=m$:
- $[LM:K]=mn=[LM:L]n=[LM:M]m=[LM:L\cap M][L\cap M:K]$
- $[LM:L\cap M]\leq mn$. If I prove the equality I think I'l have the proof.
- From 1, we have $[LM:L]=m,[LM:M]=m$.
Can I use this to prove it, or must I use another approach?