Undergrad Existence and Uniqeness of Finite Fields ....

[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:

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

 

[fresh_42]

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. ... ... "

What does it mean "of dimension $n$"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?

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 $\mathbb{F}$ ... ... ?

If you have any finite group $G$, say $|G|=m$ and take an element $g \in G$, then consider the subgroup $U:=\langle g \rangle \subseteq G$. What can be said about the order of $U$ and therewith the order of $g$? Now with that, what do you get for $g^m$?

(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 ...? )

No. The characteristic of a field always has to be either $0$ or prime, and $p^n-1$ is only in one single case prime whereas the statement is generally valid. The characteristic is the smallest number $n$ such that $\underbrace{1+ \ldots + 1}_{n-times} =0$ or zero, if $n$ isn't finite. The order of $\mathbb{F}^\times$ is $m=|\mathbb{F}|-|\{0\}|=p^n-1$ not the characteristic.

 

[Math Amateur]

What does it mean "of dimension $n$"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?

If you have any finite group $G$, say $|G|=m$ and take an element $g \in G$, then consider the subgroup $U:=\langle g \rangle \subseteq G$. What can be said about the order of $U$ and therewith the order of $g$? Now with that, what do you get for $g^m$?

No. The characteristic of a field always has to be either $0$ or prime, and $p^n-1$ is only in one single case prime whereas the statement is generally valid. The characteristic is the smallest number $n$ such that $\underbrace{1+ \ldots + 1}_{n-times} =0$ or zero, if $n$ isn't finite. The order of $\mathbb{F}^\times$ is $m=|\mathbb{F}|-|\{0\}|=p^n-1$ not the characteristic.

Hi fresh_42 ... ... thanks for the help ...

Regarding Question 1, you write the following"

"What does it mean "of dimension n" role="presentation">n"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?"

Well, dimension $n$ means there are $n$ basis vectors, and each linear combination of the basis vectors has $n$ coefficients with $p$ possibilities for each ... ... so we get a total of $p^n$ unique elements in the field $\mathbb{F}$ ... ... Is that the correct analysis/reasoning?

Still thinking about Question 2 ...

Peter

 

[fresh_42]

Hi fresh_42 ... ... thanks for the help ...

Regarding Question 1, you write the following"

"What does it mean "of dimension n" role="presentation">n"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?"

Well, dimension $n$ means there are $n$ basis vectors, and each linear combination of the basis vectors has $n$ coefficients with $p$ possibilities for each ... ... so we get a total of $p^n$ unique elements in the field $\mathbb{F}$ ... ... Is that the correct analysis/reasoning?

Still thinking about Question 2 ...

Peter

Yes. And they are all different, because $\alpha_1 b_1 + \ldots + \alpha_n b_n = \beta_1 b_1 + \ldots + \beta_n b_n$ means, that each $\alpha_i=\beta_i$ if $\{ b_1,\ldots ,b_n\}$ is a basis. So there cannot be two different combinations resulting in the same vector, i.e. all $p^n$ combinations are different elements.