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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1This must be a duplicate but I can't find a good target at the moment. For now, I'll leave you with https://en.wikipedia.org/wiki/Primitive_element_theorem – Eric Wofsey May 27 '19 at 23:34
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. – dantopa May 27 '19 at 23:48
1 Answers
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:
- The extension $E/F$ is simple.
- 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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1In 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 '19 at 00:33
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