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I'm reading David R. Finston and Patrick J. Morandi's book Abstract Algebra: Structure and Application and in section 7.2 page 109 it mentions

It’s true that every finite field has a primitive element. However, the proof of this fact involves more complicated ideas that we will consider, so we do not give the proof.

I understand some proof might be hard, based on more complex concepts, but leaving this dangling it’s a bit itchy .. May I ask where can I find a relative simple proof?

athos
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    You need to know that the multiplicative group of a finite field is cyclic, which involves combining the Fundamental Theorem of Finitely Generated Abelian Groups, with the fact that a polynomial of degree $k$ with coefficients in a field has at most $k$ roots. – Arturo Magidin Aug 29 '19 at 21:51
  • If "primitive element" means generator of its multiplicative group, then https://math.stackexchange.com/q/59903/96384. – Torsten Schoeneberg Aug 29 '19 at 21:51
  • On the other hand, if primitive element means an element $\alpha\in F$ such that $F=\Bbb{F}_p(\alpha)$ then this might be easier to show. This of course depends a lot on the context. – Servaes Aug 29 '19 at 21:53
  • This is not the general case, but here's a good exposition of seven proofs that $(\mathbb{Z}/p)^{\times}$ is cyclic. – Dzoooks Aug 29 '19 at 21:54
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    Most likely it relies on the fact that if $F$ is a finite field, there is at most one subgroup of given size, and as a consequence, in $F^$, there are at most $\phi(d)$ elements with order $d$. It follows from this, Lagrange’s theorem, and $\sum_{d|n}{\phi(d)}=n$ that any subgroup of order $n$ of $F^$ must have one element of order $n$ so is cyclic. In particular, $F^*$ is a cyclic group: if $\alpha$ is its generator, $\alpha$ is primitive. – Aphelli Aug 29 '19 at 21:55
  • See https://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic – lhf Aug 30 '19 at 01:00
  • I rather like https://math.stackexchange.com/a/837571/403337. This uses precisely what @Arturo mentioned. –  Aug 30 '19 at 03:34

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Here is an elementary proof, in the sense that it avoids the fundamental theorem of finite abelian groups.

The proof is based on the following elementary statement:

Let $F$ be a finite field and $a$ an element of order $m$ and $b$ an element of order $n$. Then $F$ contains an element $c$ of order $lcm(m,n)$.

Let's begin the proof. Let $a$ be an element of maximal order, say $n$. To obtain a contradiction, we assume that $n<|F|-1$. Then we can pick an element $b\notin \langle a \rangle\cup \{0\}$. Denote the order of $b$ with $m$. By maximality of $n$, we know $m\leq n$. If $m$ divides $n$, then the polynomial $X^n-1$ has strictly more than $n$ roots ($b$ and the elements of $\langle a\rangle$). Thus we must have that $m$ does not divide $n$. But then $lcm(m,n)>n$ and the statement above gives an element with order $lcm(m,n)$, contradicting maximality of $n$.

This shows that $n=|F|-1$ and thus $a$ is a primitive element, i.e. a generator for the multiplicative group.

J. De Ro
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