4:48 AM
Yeah, that should be right. $[KL, F] = [K, F][L, F]$ implies $K \cap L = \{0\}$, because otherwise choose an $F$-basis $\{\alpha_1, \cdots, \alpha_k\}$ for $K \cap L$, extend that to a basis $\{\alpha_1, \cdots, \alpha_k, \gamma_1, \cdots, \gamma_{n-k}\}$ of $K$ and a basis $\{\alpha_1, \cdots, \alpha_k, \beta_1, \cdots, \beta_{m-k}\}$ of $L$.
Then you can make an $F$-generating set of $KL$ with less than $nm = [K, F][L, F]$ elements by taking mutual products of $\alpha_i, \beta_j, \gamma_k$.
With $mn - k^2$ elements.
Then choose some primitive element $\alpha \in K$, so $K \cong F(\alpha) \cong F[x]/(f(x))$. Then $K \otimes_F L \cong L[x]/(f(x)) \cong L(\alpha)$.
But if $\{\beta_1, \cdots, \beta_m\}$ is an $F$-basis for $L$, then as $\{1, \alpha, \cdots, \alpha^n\}$ is an $F$-basis for $K$, jointly they form an $F$-basis for $KL$, which shows $KL \cong F \langle \beta_1, \cdots, \beta_m, 1, \alpha, \cdots, \alpha^n \rangle = L(\alpha)$.
I guess I used the primitive element theorem, so separability, somewhere. Unsure how to argue cleanly without that.
OK, same stuff. If $K = F(\alpha_1, \cdots, \alpha_n)$, then $K \cong F[x_1, \cdots, x_n]/(f_1(x_1), \cdots, f_n(x_n))$ where $f_i$ is the minimal polynomial of $\alpha_i$.
Then $K \otimes_F L = L[x_1, \cdots, x_n]/(f_1(x_1), \cdots, f_n(x_n))$, which is still a field because of given hypothesis, $K \otimes_F L \cong L(\alpha_1, \cdots, \alpha_n)$ which is in turn $KL$.
For completion: $[KL, F] = [K, L][L, F]$ implies $K \otimes_F L$ is a field because an irreducible polynomial over $F$ which has a root in $K$ cannot be reducible in $L$, otherwise $[KL, L] \leq [K, F]$. Then you run the same argument as before.
I think disconnected covers correspond to product of fields. For example, if you take the double cover $S^2 \to \Bbb{RP}^2$ and do fibered product of it with itself, $S^2 \to \Bbb{RP}^2 \leftarrow S^2$, then you get the disconnected $2$-sheeted cover $S^2 \times \{0, 1\} \to S^2$.
Similarly, $\Bbb C \otimes_{\Bbb R} \Bbb C \cong \Bbb C \times \Bbb C$.
Can we classify finite dimensional $F$-algebras? They are not always product of field extensions of $F$ because of weird things like $F[x]/(x^n)$.
If $A$ is a finite dimensional $F$-algebra then $A$ has finitely many prime ideals, because $A \otimes_F F$ is a finite $F$-vector space, and since Spec of that is fiber of the map $\text{Spec} A \to \text{Spec} F$ over $(0)$, this implies this is a finite-to-one map. So $A$ has finitely many prime ideals, and $A/\text{nil}(A)$ is a product of finite extensions of $F$ by CRT.
That's all I can say I suppose.