# Prove that $GF(p^n)$ exists

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

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hint:you can use polynomial $x^{p^n}-x\in Z_p[X]$ –  Babak Miraftab Jan 11 '12 at 16:26

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

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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.

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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