# Let $E$ be an extension field of $F$, of dimension $n$. Is there an element of degree $n$ in $E$ over $F$?

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

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