2
$\begingroup$

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

  • 3
    hint:you can use polynomial $x^{p^n}-x\in Z_p[X] $2012-01-11

2 Answers 2

2

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

3

HINT:

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.

  • 2
    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.2012-01-11
  • 0
    @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. :=)2012-01-11
  • 0
    Yes, but how can i prove that such $E(x)$ exists for every n?2012-01-11
  • 2
    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.2012-01-11
  • 0
    @DilipSarwate, but how to prove it?2012-01-11