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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View attachment 6648
My questions are as follows:
Question 1In the above text from D&F we read the following:

" ... ... If $$\mathbb{F}$$ is of dimension $$n$$ over its prime subfield $$\mathbb{F}_p$$, then $$\mathbb{F}$$ has precisely $$p^n$$ 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 $$\mathbb{F}^{ \times }$$ is (in fact cyclic) of order $$p^n - 1$$, we have $$\alpha^{ p^n - 1} = 1$$ for every $$\alpha \ne 0 $$ in $$\mathbb{F}$$ ... ... "Can someone give me the exact reasoning concerning why $$\mathbb{F}^{ \times }$$ being of order $$p^n - 1$$ implies that $$\alpha^{ p^n - 1} = 1$$ for every $$\alpha \ne 0$$ in $$\mathbb{F}$$ ... ... ?(I am guessing that for some reason I cannot explain, that $$\mathbb{F}^{ \times }$$ being of order $$p^n - 1$$ implies that the characteristic is $$p^n - 1 $$ ... ... 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 $$\mathbb{F}$$ is of dimension $$n$$ over its prime subfield $$\mathbb{F}_p$$, then $$\mathbb{F}$$ has precisely $$p^n$$ elements. ... ... "

Can someone please explain why, exactly, this follows?

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

Question 2

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

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

This is just Fermat's little theorem.

 

[Math Amateur]

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

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

... appreciate your help and support ...

Peter