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


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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.