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Assume that $E\subset F$ and that $E$ and $F$ are fields. Also, say $[E:F]=p$, where $p$ is a prime. If $a$ is any element of $F\setminus E$. Show that $F=E(a)$.

I'm pretty unsure of how to approach this.

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1 Answer 1

Degrees are multiplicative: $$ [F : E] = [F : E(a)][E(a) : E]. $$ See Proposition 1.20 of Milne's notes for a proof. What can you say about $[E(a) : E]$ now?

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