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Math Help - Elements in a finite field

  1. #1
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    Elements in a finite field

    Hello.

    I have to proof the following proposition:

    Let K be a finite field. Then K has q=p^n elements, whereas p is the characteristic of K and n=[K:\mathbb{F}_p].
    Furthermore I have to proof, that x^q = x for all x \in K.

    Could anybody of you explain, how to proof the proposition?

    Bye,
    Lisa
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  2. #2
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    Quote Originally Posted by lisa View Post
    Hello.

    I have to proof the following proposition:

    Let K be a finite field. Then K has q=p^n elements, whereas p is the characteristic of K and n=[K:\mathbb{F}_p].
    Furthermore I have to proof, that x^q = x for all x \in K.

    Could anybody of you explain, how to proof the proposition?

    Bye,
    Lisa
    I'll assume p is a prime number. A finite field K of order q=p^n is the splitting field of x^q - x over F_p (see here). Since a finite field is perfect, K is a separable extension of F_p. It follows that K is a Galois extension of a field F_p. By the fundamental theorem of Galois theory, [K:F] = |Gal(K/F)|=n, where F=F_p and Gal(K/F) is a cyclic group of order n generated by a Frobenius automorphism.
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  3. #3
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    Quote Originally Posted by lisa View Post
    Hello.

    I have to proof the following proposition:

    Let K be a finite field. Then K has q=p^n elements, whereas p is the characteristic of K and n=[K:\mathbb{F}_p].
    Furthermore I have to proof, that x^q = x for all x \in K.

    Could anybody of you explain, how to proof the proposition?

    Bye,
    Lisa

    One of the easiest proofs is, perhaps, to note that any field is a vector space over any of its subfields, and then: a finite field K of characteristic p is a v.s. over the prime field of char. p, \mathbb{F}_p\cong \mathbb{Z}\slash p\mathbb{Z}, obviously of finite dimension n, and choosing any basis for this v.s. a simple combinatoric reasoning gives that the number of elements in K must be p^n

    Take the polynomial f(x)=x^{p^n}-x\,\,\,over\,\,\mathbb{F}_p\left[x\right] . Now you can go with the argument from fields extensions and Galois theory that was given to you by Aliceinwonderland or else use a little group theory: the multiplicative group K^{*} has order p^n-1\,\Longrightarrow \omega^{p^n-1}=1\,\,\forall\,\,\omega\,\in\,K\,\Longrightarrow  \,\omega^{p^n}=\omega , and we're done.

    Tonio
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  4. #4
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    Thank you. Your explanations were very helpful for me.

    Best greets,
    Lisa
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