Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I'm hoping someone can give me a nudge in the right direction...

Let $F$ be a finite field, and let $f(x)$ be a nonconstant polynomial whose derivative is the zero polynomial. Prove that $f$ cannot be irreducible over $F$.

I've got that every root of $f$ is a multiple root and that for $F=\mathbb{F}_{p^r}$, the exponent of every term of $f$ is a multiple of $p$.

share|cite|improve this question
up vote 4 down vote accepted

Hint $\ $ prime $\rm\:P\equiv 0,\ \ A^P\equiv A,\ B^P\equiv B\ \ \Rightarrow\ \ A\:X^{JP} +\!\: B\:X^{KP}\equiv\: (A\:X^J +\!\: B\:X^K)^P$

share|cite|improve this answer

Hint: there is a polynomial $g$ over $F$ such that $f=g^p$. Do you see what it is?

share|cite|improve this answer
+1 Also, all the coefficients are $p^\text{th}$ powers. Hopefully the OP knows why that is true? – Jyrki Lahtonen Apr 2 '12 at 20:03

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.