Let $\alpha\in \overline {\mathbb F_2}$ (the algebraic closure of $\mathbb F_2$ ) be such that $\mathbb F_2[\alpha]$ is a field of order $2^n$ (where $n>1$).

Then is it true that $\alpha \in \mathbb F_2[\alpha]^{\times}$ generates the multiplicative group $\mathbb F_2[\alpha]^{\times}$ i.e. is $2^n-1$ the multiplicative order of $\alpha$ ?

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    $\begingroup$ This is essentially the difference between irreducible polynomials and primitive polynomials (supposing we look at the minimal polynomial of $\alpha$). $\endgroup$
    – Erick Wong
    Oct 17, 2018 at 2:04
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    $\begingroup$ The answer to the title question depends on what you mean by primitive. If you adopt the field theory convention, then the answer is no, see Lord Shark's answer for the smallest counterexample. But, you will be misunderstood in the finite field community. There the notion of a primitive element extends the notion of a primitive root modulo $p$. So, if a paper dealing with finite fields discusses primitive elements you can bet that the author means a generator of the multiplicative group. $\endgroup$ Oct 17, 2018 at 12:30

3 Answers 3


Not necessarily. For instance $f(x)=x^4+x^3+x^2+x+1$ is irreducible over $\Bbb F_2$, so a solution $\alpha$ of $f(x)=0$ generates $\Bbb F_{16}$. But $\alpha$ has multiplicative order $5$ and does not generate $\Bbb F_{16}^\times$.

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    $\begingroup$ Depends on the definitions. This correctly answers the question in the body, but is wrong for the question in the title. Caveat reader. $\endgroup$ Oct 17, 2018 at 12:40

Yes. Every primitive element of a finite field $K$ is a generator of the multiplicative group $K^*$.


  • In general field theory an element $\alpha\in L$, $L$ an extension field of $K$, is called primitive, if $L=K(\alpha)$. In other words, all we need is that $L$ is the smallest extension field of $K$ that also contains the element $\alpha$.
  • But, in the context of finite fields, a different convention has been adopted. An element $g$ of a finite field $K$ is called primitive if it is a generator of the multiplicative group $K^*$. In the context of finite fields an element does not gain the attribute primitive simply by generating $K$ as an extension of the prime field.
  • You don't have to take my word alone for this. The bible of finite fields, Lidl & Niederreiter also uses this convention, see their definition 2.9. from page 51

A generator of the cyclic group $\Bbb{F}_q^*$ is called a primitive element of $\Bbb{F}_q$.

I have tried to explain this difference in the tag wiki.

I believe (my impressions only) the difference in conventions comes from the following sources:

  1. The finite field community usage is generalizing the notion of a primitive root modulo a prime $p$. A coset $\overline{a}$ of an integer $a$ modulo $p$ is a primitive root modulo $p$ if and only if it is a generator of the multiplicative group $\Bbb{Z}_p^*$. In that setting only requiring $\overline{a}$ to generate the field extension is utterly non-interesting because we never leave the prime field.
  2. In coding theoretical settings the finite field usage is the most useful one. Comes up much more frequently (here you may take my word for it). For example, we often view elements of a code as functions from a cyclic subgroup of the multiplicative group of the underlying field, and do things like discrete Fourier analysis on the subgroup. As often long codes are preferred, a natural thing is to go to the maximal cyclic subgroup, hence a primitive generator is required.

Here is a systematic way to find counterexamples. Note that $\mathbb{F}_{2^n}$ has a primitive $m$th root of unity iff $m\mid 2^n-1$, since its multiplicative group is cyclic of order $2^n-1$. This means that if you take a primitive $m$th root of unity, the field extension of $\mathbb{F}_2$ that it generates will have $2^n$ elements for the least $n$ such that $m\mid 2^n-1$. (Note that if $m$ is odd then such an $n$ always exists, since $2$ is invertible mod $m$ and so some power of $2$ is $1$ mod $m$.)

In particular, if $m$ is odd and not one less than a power of $2$, then a primitive $m$th root of unity will be a primitive element of $\mathbb{F}_{2^n}$ for this value of $n$ but will not have order $2^n-1$. Lord Shark the Unknown's example takes $m=5$, the smallest odd number that is not one less than a power of $2$, for which $n=4$.


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