Can a primitive root of a polynomial over $GF(2)$ ever not generate a multiplicative group?

I have some notes from my review of finite field extensions a while ago that I've been rereading. It's the last statement that's throwing me. I've included some preceding notes for context.

If $p(x)$ is an irreducible polynomial of degree $n$, then adjoining a root of $p$ to $GF(2)$ generates an extension of degree $n$, which is necessarily a field, $E$, with $2^n$ elements.

The multiplicative group of nonzero elements in $E$ has order $2^n - 1$.
Thus by Lagrange's Theorem, every nonzero element $a$ of $E$ satisfies $a^{2^n - 1} = 1$. Thus every element $a$ in $E$ is a root of $g(X) = > X^{2^n} - X$.

In other words, $E$ is exactly the set of all roots of $g(X)$. Now the roots of the original $p(x)$ are also roots of $g(X)$, and so $p(x)$ divides $g(X)$ (after making the variables the same).

Conversely, if $f(x)$ is any polynomial that divides $g(x)$, then the roots of $f(x)$ lie in $E$, so they generate a subfield of $E$. If they generate all of $E$, and if $f(x)$ is irreducible, then they must have degree $n$.

Now, let $\rho$ be a primitive root of $f$, where f is irreducible. So $\rho$ will generate $E$ as a field, but not necessarily generate $E-\{0\}$ as a multiplicative group.

  • $\begingroup$ Please put the question in the body of the message, not just the title. $\endgroup$ – Arturo Magidin May 18 '11 at 4:31
  • $\begingroup$ @Arturo, thanks for edit. $\endgroup$ – ThomasMcLeod May 18 '11 at 4:33
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    $\begingroup$ What is a primitive root of a polynomial, in some general context? $\endgroup$ – KCd May 18 '11 at 4:36
  • $\begingroup$ I also don't see what the problem is. For example, looking at fields of size 9, we can use F_9 = F_3[i] = F_3[x]/(x^2+1). In this field, i generates the field F_9 over F_3 but i is not a generator of the group F_9*. Big deal. There is no paradox. $\endgroup$ – KCd May 18 '11 at 4:38
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    $\begingroup$ @KCd, the problem is that someone new to the topic may expect that the question, "Is $\alpha$ a generator?" would have a yes/no answer, and is surprised to learn that both "yes" and "no" are correct, depending on whether one is generating the field or the multiplicative group of the field. A rose is a rose is a rose, but a generator is not a generator, pace Gertrude Stein. $\endgroup$ – Gerry Myerson May 18 '11 at 6:32

I think this is an example of what you're after. The polynomial $x^4+x^3+x^2+x+1$ is irreducible over the field of 2 elements, so any root $\alpha$ generates the field of 16 elements. But $\alpha^5=1$ (note that the given polynomial is a factor of $x^5-1$), so $\alpha$ doesn't generate the 15-element multiplicative subgroup as a group.

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    $\begingroup$ And of course, the smallest example occurs in the field of 16 elements, since all smaller fields have multiplicative group that is either trivial or of prime order. $\endgroup$ – Arturo Magidin May 18 '11 at 4:36
  • $\begingroup$ Um, a smaller example occurs in the field of 9 elements, as in my comments to the original question. The group F_9* is not trivial or of prime order. $\endgroup$ – KCd May 18 '11 at 4:44
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    $\begingroup$ @KCd: The original was only asking about fields of order $2^n$, and the question asks specifically about extensions of $GF(2)$, so I wasn't counting $F_{p^k}$ for odd primes. Though the comment does not make that clear. $\endgroup$ – Arturo Magidin May 18 '11 at 4:46
  • $\begingroup$ This is precisely the answer. Was head scratching for an hour on this. So, can we say then that $x^4+x^3+x^2+x+1$ is irreducible but not primitive? $\endgroup$ – ThomasMcLeod May 18 '11 at 4:58
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    $\begingroup$ The one meaning of "primitive" only applies over finite fields, the other is useful only over non-fields (since over a field all the non-zero coefficients are units). $\endgroup$ – Gerry Myerson May 18 '11 at 6:37

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