Existence and Uniqeness of Finite Fields .... Example from D&F ....

[Math Amateur]

I am reading David S. Dummit and Richard M. Foote : Abstract Algebra ...

I am trying to understand the example on Finite Fields in Section 13.5 Separable and Inseparable Extensions ...The example reads as follows:
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My questions are as follows:
Question 1In the above text from D&F we read the following:

" ... ... If \(\displaystyle \mathbb{F}\) is of dimension \(\displaystyle n\) over its prime subfield \(\displaystyle \mathbb{F}_p\), then \(\displaystyle \mathbb{F}\) has precisely \(\displaystyle p^n\) elements. ... ... "Can someone please explain why, exactly, this follows?

Question 2

In the above text from D&F we read the following:

" ... ... Since the multiplicative group \(\displaystyle \mathbb{F}^{ \times }\) is (in fact cyclic) of order \(\displaystyle p^n - 1\), we have \(\displaystyle \alpha^{ p^n - 1} = 1\) for every \(\displaystyle \alpha \ne 0 \) in \(\displaystyle \mathbb{F}\) ... ... "Can someone give me the exact reasoning concerning why \(\displaystyle \mathbb{F}^{ \times }\) being of order \(\displaystyle p^n - 1\) implies that \(\displaystyle \alpha^{ p^n - 1} = 1\) for every \(\displaystyle \alpha \ne 0\) in \(\displaystyle \mathbb{F}\) ... ... ?(I am guessing that for some reason I cannot explain, that \(\displaystyle \mathbb{F}^{ \times }\) being of order \(\displaystyle p^n - 1\) implies that the characteristic is \(\displaystyle p^n - 1 \) ... ... but why does it mean this is the case ...? )

Hope someone can help ...

Peter

 

[Opalg]

Question 1

In the above text from D&F we read the following:

" ... ... If \(\displaystyle \mathbb{F}\) is of dimension \(\displaystyle n\) over its prime subfield \(\displaystyle \mathbb{F}_p\), then \(\displaystyle \mathbb{F}\) has precisely \(\displaystyle p^n\) elements. ... ... "

Can someone please explain why, exactly, this follows?

If \(\displaystyle \mathbb{F}\) is of dimension \(\displaystyle n\) as a vector space over \(\displaystyle \mathbb{F}_p\) then a basis for this vector space will contain $n$ elements. An element of \(\displaystyle \mathbb{F}\) is then uniquely specified by its coordinates with respect to this basis. Each coordinate is an element of \(\displaystyle \mathbb{F}_p\), and can therefore take $p$ possible values, since that is the number of elements of \(\displaystyle \mathbb{F}_p\). There are $n$ such coordinates, each taking $p$ values, which gives you a total of $p^n$ possible expressions to specify an element of \(\displaystyle \mathbb{F}\).

Question 2

In the above text from D&F we read the following:

" ... ... Since the multiplicative group \(\displaystyle \mathbb{F}^{ \times }\) is (in fact cyclic) of order \(\displaystyle p^n - 1\), we have \(\displaystyle \alpha^{ p^n - 1} = 1\) for every \(\displaystyle \alpha \ne 0 \) in \(\displaystyle \mathbb{F}\) ... ... "Can someone give me the exact reasoning concerning why \(\displaystyle \mathbb{F}^{ \times }\) being of order \(\displaystyle p^n - 1\) implies that \(\displaystyle \alpha^{ p^n - 1} = 1\) for every \(\displaystyle \alpha \ne 0\) in \(\displaystyle \mathbb{F}\) ... ... ?

This is just Fermat's little theorem.

 

[Math Amateur]

If \(\displaystyle \mathbb{F}\) is of dimension \(\displaystyle n\) as a vector space over \(\displaystyle \mathbb{F}_p\) then a basis for this vector space will contain $n$ elements. An element of \(\displaystyle \mathbb{F}\) is then uniquely specified by its coordinates with respect to this basis. Each coordinate is an element of \(\displaystyle \mathbb{F}_p\), and can therefore take $p$ possible values, since that is the number of elements of \(\displaystyle \mathbb{F}_p\). There are $n$ such coordinates, each taking $p$ values, which gives you a total of $p^n$ possible expressions to specify an element of \(\displaystyle \mathbb{F}\).This is just Fermat's little theorem.

Thanks Opalg ... very clear and very helpful ...

... appreciate your help and support ...

Peter