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 $K$ be a finite field. Let us define a primitive polynomial as an $f \in K[X]$ s.t. the multiplicative order of $X$ in $K[X]/(f)$ is equal to $|K|^{\deg f} - 1$. I want to show that $f \in K[X]$ is primitive if and only if $f$ is irreducible and $X$ generates the multiplicative group $(K[X]/(f))^\times$.

I would like to ask how to show this. I already showed that if $f$ is primitive and irreducible the latter half of the condition holds, but I cannot figure out the rest. I would also like to know if it is customary to talk about the multiplicative order of an element of a ring whose multiplicative part is not necessarily a group.

share|cite|improve this question
I think you mean "multiplicative order of $X$." Also, the multiplicative group has $|K|^{\deg f}-1$ elements at most, so $X$ can never have the order you give in your problem. – Thomas Andrews Mar 10 '13 at 4:10
up vote 2 down vote accepted

I assume that the definition of primitive includes that $X$ is relatively prime to $f$ (since if not, $X$ has no well-defined multiplicative order in $K[X]/(f)$.)

If $f$ has a nontrivial divisor $g$, there are at least two elements of $K[X]/(f)$, $0$ and $g$, which do not have multiplicative inverses, so $K[X]/(f)$ has at most $|K|^{\deg f}-2$ invertible elements. However, if $f$ is primitive, then, by definition, $K[X]/(f)$ contains $|K|^{\deg f}-1$ powers of $X$, all of which are invertible. Therefore, if $f$ is primitive, then it is also irreducible. This reduces the problem to showing that if $f$ is irreducible, then $f$ generates the multiplicative group of $K[X]/(f)$ if and only if $f$ has order $|K|^{\deg f}-1$. But this is obvious because the group has order $|K|^{\deg f}-1$.

share|cite|improve this answer

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.