Let $L$ and $M$ be intermediate fields in the extension $K \subset F$.
(a) $[LM : K]$ is finite if and only if $[L : K]$ and $[M : K]$ are finite.
(b) If $[LM : K]$ is finite, then $[L : K]$ and $[M : K]$ divide $[LM : K]$ and $[ LM : K] \le [L : K][ M : K]$.
(c) If $[L : K]$ and $[M : K]$ are finite and relatively prime, then $[LM : K] = [L : K][M : K]$.
(d) If $L$ and $M$ are algebraic over $K$, then so is $LM$.
I think $LM$ is supposed to be the subfield generated by $L \cup M$. What I've done so far:
(a) Theorem V.1.2 in Hungerford's Algebra is that if $E$ is an intermediate field of the extension $K \subset F$, then $[F : K] = [F : E][E : K]$ and $[F : K]$ is finite if and only if $[F : E]$ and $[E : K]$ are finite. I'm genuinely trying to apply this, but I am drawing a blank.
(b) Suppose $\{a_1,...,a_n\}$ is a basis for $L$ over $K$ and $\{b_1,...,b_m\}$ is a basis for $M$ over $K$. I know I should should show that $\{a_ib_j: 1 \le i \le n, 1 \le j \le m\}$ generates $LM$ because that way $[LM:K] \le mn = [L:K][M:K]$. I'm not sure how to do that though.
I think I have (c) and (d):
(c) Since $[L: K]$ and $[M: K]$ both divide $[LM : K]$, we see that $[L : K][M : K]$ divides $[LM : K]$ so $[L : K][M : K] \le [LM : K]$. Which with (b) implies $[L : K][M : K] = [LM : K]$
(d) Let $x \in LM$, then $$x = \sum_{i} r_is_i$$ where $r_i \in L$ and $s_i \in M$. Then $x \in K(r_1,...,r_k,s_1,...,s_k)$ which is a finite extension of $K$ since each $r_i$ and $s_i$ is algebraic over $K$. Then $x$ is also algebraic over $K$.