Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F=\mathbb Z/(p)$, where $p$ a prime number, $f(x)$ a monic irreducible polynomial in $K=F[x]$ of degree $n$, $K=F[x]/(f(x))$, and $E$ the multiplicative group of nonzero elements of $K$. Then it is easy to see that $K=F(x+f(x))$. Is $x+(f(x))$ a generator of $E$ as a multiplicative group?

share|cite|improve this question
up vote 2 down vote accepted

It depends on the polynomial $f$. If $x + (f(x))$ is a generator, $f$ is called primitive. For each finite base field and each degree, a primitive polynomial always exists.


There are 3 irreducible polynomials of degree 4 over $\Bbb F_2$: $$x^4 + x + 1,\quad x^4 + x^3 + 1,\quad x^4 + x^3 + x^2 + x + 1$$

The first two polynomials are primitive, but the third one is not. This can be checked as follows:

In $\Bbb F_2[x]/(x^4 + x^3 + x^2 + x + 1)$, polynomial long division gives $$ x^5 = (x + 1)\cdot(\underbrace{x^4 + x^3 + x^2 + x + 1}_{=0}) + 1 = 1\text{,} $$ so $x + (f(x))$ has multiplicative order $5$. To generate the multiplicative group, its order must be $15$.

share|cite|improve this answer
i want to know under what condictions , $x + (f(x))$ is a generator – Aimin Xu Dec 3 '13 at 8:18
@AiminXu You have to compute the multiplicative order of $x + (f(x))$, like I did above. Like for irreducibility, in general there is no "easier" way to check if a polynomial is primitive. – azimut Dec 3 '13 at 8:21
thanks a lot, a good example! – Aimin Xu Dec 3 '13 at 10:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.