Let $K$ be an extension of $F$. Suppose that $E_1$ and $E_2$ are contained in $K$ and are extensions of $F$. If $[E_1:F]$ and $[E_2:F]$ are both prime, show that $E_1 = E_2$ or $E_1 \cap E_2 = F$.
I start by assuming that $E_1 \neq E_2$ and proving $E_1 \cap E_2 = F$.
I know by the fact that $[E_1:F] = p_1$, that $[E_1:F] = [E_1:E_1 \cap E_2][E_1 \cap E_2:F]$, where $[E_1:E_1 \cap E_2][E_1 \cap E_2:F] = 1\cdot p_1$ or $p_1\cdot 1$ respectively, but I can't come up with a good reason how $E_1 \neq E_2$ effects either possibility. I see that one can't be a subset of the other because that leads to the contradictions, but otherwise I can't think of a reason.
Anyone have any ideas?
Recall that $[F:E] =1 \iff F=E$. This is because this is the dimension as a vector space. Since any non-zero element of $F$ generates it as a vector space over $E$, we see that $1$ is such a choice. But then every element of $F$ is of the form $1_F\cdot e$ for some $e\in e$, but $1_F=1_E$ so this just means $F\subseteq E$ showing equality.