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I've tried to endow $\mathbb{F}^{n}$, where $\mathbb{F}=\mathbb GF(p)$ with a field structure, but I was not able to do it. Could you please help me with this question?

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There is really no point in several question marks!!!!! :-) –  Asaf Karagila Jan 24 '12 at 19:46
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$\mathbf{GF}(p)^n$ is a vector space. By adding structure to it in a canonical way, you can use it to represent the field $\mathbf{GF}(p^n)$ --- notice the placement of the exponent. What background are you starting from? –  Niel de Beaudrap Jan 24 '12 at 19:46
    
Hello! Sorry, but my English needs some improvements! I am studying Golan's book: The Linear Algebra a Beginning Student OUght to Know. –  spohreis Jan 24 '12 at 20:01

1 Answer 1

up vote 4 down vote accepted

The classification theorem for finite fields says that for each $n\geq1$ there's just one field $\Bbb F_{p^n}$ with $p^n$ elements up to isomorphism. It's the field made up with the roots of the polynomial $X^{p^n}-X$ in an algebraic closure of $\Bbb F_p=\Bbb Z/p\Bbb Z$.

Concretely, $\Bbb F_{p^n}$ can be realized as the quotient $\Bbb F_p[X]/(P(X))$ where $P(X)$ is any irreducible polynomial of degree $n$.

So, if you start with ${\Bbb F_p}^n$ and you want to endow it with a field structure you can do it in the following two steps:

  • Pick a monic irreducibe polynomial $P(X)\in\Bbb F_p[X]$ (monic is not really essential).

  • Write ${\Bbb F_p}^n=\Bbb F_p+\Bbb F_px+\cdots+\Bbb F_px^{n-1}$ and (1) define the usual product for the $x^k$; (2) use $P(x)=0$ to decide what $x^n$ should be.

The theory says that you get the "same thing" whatever irreducible polynomial you choose.

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