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

I know thar $\forall p$ prime, $\forall n>0$, it exists the finite field $GF(p^n)$. Can you help me proving this theorem? I do not need a formal proof, just an intuition, an idea...

Thank you

share|cite|improve this question
hint:you can use polynomial $x^{p^n}-x\in Z_p[X] $ – Babak Miraftab Jan 11 '12 at 16:26
up vote 2 down vote accepted

Hint: prove that the set of all $x$ such that $x^{p^n}-x=0$ is a field.

share|cite|improve this answer


First, constructing a finite field with $p = p^1$ elements is easy.

For $n > 1$: Consider the ring $R = (GF(p))[x]/E(x)$ where $E(x)$ is degree $n$ polynomial (you may even assume $E$ is monic). How many elements does $R$ have? Under what conditions on $E(x)$ is $R$ a field?

Note: As Steven points out, showing the existence of a polynomial $E(x)$ with the required properties is quite nontrivial. I am just hoping this is a fruitful direction for you to think about.

share|cite|improve this answer
This pushes the problem down one level, but it should be noted that there's still definitely a non-trivial assertion to be proved from here. – Steven Stadnicki Jan 11 '12 at 16:24
@Steven: Yes, I will make a note of that. In spite of that caveat, I am hoping that this is a fruitful direction to pursue. :=) – Srivatsan Jan 11 '12 at 16:26
Yes, but how can i prove that such $E(x)$ exists for every n? – Aslan986 Jan 11 '12 at 16:27
After we determine that $R$ is a field of order $p^n$ exactly when $E(x)$ is irreducible over $F_p$, we have to prove that there is at least one irreducible polynomial of degree $n$ over $F_p$ to show that the requirement can be met. – Dilip Sarwate Jan 11 '12 at 16:27
@DilipSarwate, but how to prove it? – Aslan986 Jan 11 '12 at 16:30

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.