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In other words, does there exist any element in $E$ such that its minimal polynomial over $E$ is of degree $n$?

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What you are asking is whether any finite field extension has a primitive element; that is, if $[E:F]=n$, does there exist an $a\in E$ such that $E=F(a)$.

To see that this is equivalent to what you are asking, note that if such an element exists, then its minimal polynomial must have degree $[F(a):F] = [E:F] = n$; conversely, if there is an $a$ with minimal polynomial of degree $n$, then $[E:F(a)][F(a):F] = [E:F]=n$, but $[F(a):F]=n$, and therefore $[E:F(a)]=1$ so $E=F(a)$.

Such an extension is called a simple extension.

The answer is that such an element does not always exist, but it exists in most standard situations.

The most common situation is separability.

Primitive Element Theorem If $E$ is finite and separable over $F$, then $E/F$ is simple.

More generally, we have:

Theorem. Let $E$ be a finite extension of $F$. Then the following are equivalent:

  1. The extension $E/F$ is simple.
  2. There are only finitely many intermediate fields $L$, $F\subseteq L\subseteq E$.

You can find proofs of this in this site, for example here.

You can also find examples of finite extensions where it does not happen; they have to be inseparable, which means they have to be in characteristic $p$ and infinite fields. You can an example here

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    $\begingroup$ In particular, the Primitive Element Theorem holds if $F$ has characteristic zero; in particular in particular, it holds if $F$ is the rationals. $\endgroup$ – Gerry Myerson May 28 at 0:33
  • $\begingroup$ Thanks a lot, Gerry! $\endgroup$ – mathcuriosity May 28 at 9:56

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