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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?

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  • $\begingroup$ Hence one of the extensions has degree one. Conclusion? $\endgroup$ – user26857 Sep 26 '16 at 14:43
  • $\begingroup$ @user26857 So since $[E_1 : E_1 \cap E_2] = 1$ or $p_1$, if $[E_1 \cap E_2] =1$, then one $E$ is a subset of the other $E'$, which causes contradictions, therefore the only other option is $E_1 \cap E_2 = p_1$. Therefore $[E_1 \cap E_2 : F] = 1$, which implies that the extension $E_1 \cap E_2$ over $F$ has a minimal polynomial of degree $1$. And a degree $1$ minimal polynomial is of the form $x - a \in (E_1 \cap E_2) [x]$. But from here I can't see how this shows that $E_1 \cap E_2 = F$, because can't there still be points in $E_1 \cap E_2$ that are in the extension field and not in $F$? $\endgroup$ – Oliver G Sep 26 '16 at 14:54
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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.

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