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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?

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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.

Example

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 $15$.

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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

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