# Thread: trouble understanding a theorem on finite fields.

1. ## trouble understanding a theorem on finite fields.

THEOREM: Let $\displaystyle E$ be a field of $\displaystyle p^n$ elements contained in an algebraic closure $\displaystyle \bar{\mathbb{Z}_p}$ of $\displaystyle \mathbb{Z}_p$. The elements of $\displaystyle E$ are precisely the zeros in $\displaystyle \bar{\mathbb{Z}_p}$ of the polynomial $\displaystyle x^{p^n}-x$ in $\displaystyle \mathbb{Z}_p[x]$.

PROOF: The set $\displaystyle E^*$ of non-zero elements of $\displaystyle E$ forms a multiplicative group of order $\displaystyle p^n-1$ under the field multiplication. For $\displaystyle \alpha \in E^*$, the order of $\displaystyle \alpha$ in this group divides $\displaystyle |E^*|=p^n-1$. Thus for $\displaystyle \alpha \in E^*$ we have $\displaystyle \alpha^{p^n-1}=1,$ so $\displaystyle \alpha^{p^n}=\alpha$. Therefore every element of $\displaystyle E$ is a zero of $\displaystyle x^{p^n}-x$. Since $\displaystyle x^{p^n}-x$ can have atmost $\displaystyle n$ zeros, we see that $\displaystyle E$ contains precisely the zeros of $\displaystyle x^{p^n}-x$ in $\displaystyle \bar{\mathbb{Z}_p}$.

QUERY: Where have we used the algebraic closureness of $\displaystyle \bar{\mathbb{Z}_p}$. Wouldn't the proof still work if instead of $\displaystyle \bar{\mathbb{Z}_p}$ we had any field $\displaystyle K$ which is an extension field of the field $\displaystyle \mathbb{Z}_p$ containing the field $\displaystyle E$.
May be i am missing the obvious but can't figure it out.

2. ## Re: trouble understanding a theorem on finite fields.

The roots of $\displaystyle x^{p_n}- x= x(x^{p_n-1}- 1)= 0$, other than the trivial root, x= 0, form a cyclic set. That is, there exist a root $\displaystyle x_i$ such that $\displaystyle \left{x_i^j\right}$ consists of all of the non-zero roots of that equation. If all of those roots are not in K, then K is not closed under multiplication.

3. ## Re: trouble understanding a theorem on finite fields.

Originally Posted by HallsofIvy
The roots of $\displaystyle x^{p_n}- x= x(x^{p_n-1}- 1)= 0$, other than the trivial root, x= 0, form a cyclic set. That is, there exist a root $\displaystyle x_i$ such that $\displaystyle \left{x_i^j\right}$ consists of all of the non-zero roots of that equation. If all of those roots are not in K, then K is not closed under multiplication.
I still can't see how it answers the "QUERY" i have posted in post#1. please explain a bit more.

4. ## Re: trouble understanding a theorem on finite fields.

the point of using the algebraic closure of Zp is to guarantee that we have such a splitting field for EVERY value of n. every "E" can be shown to be the roots of $\displaystyle x^{p^n} - x$ only given some "K" containing that "E", but if we want to consider every possible n, we need a lot of "different" K's.

5. ## Re: trouble understanding a theorem on finite fields.

Originally Posted by Deveno
the point of using the algebraic closure of Zp is to guarantee that we have such a splitting field for EVERY value of n. every "E" can be shown to be the roots of $\displaystyle x^{p^n} - x$ only given some "K" containing that "E", but if we want to consider every possible n, we need a lot of "different" K's.
well then can the theorem be stated like this..

THEOREM: Let $\displaystyle M=\bar{\mathbb{Z}_p}$ be the algebraic closure of $\displaystyle \mathbb{Z}_p$. Then for all $\displaystyle n \in \mathbb{Z}^+$ there exists a subfield $\displaystyle E$ of $\displaystyle M$ such that each element of $\displaystyle E$ is a zero of $\displaystyle x^{p^n}-x$ with $\displaystyle |E|=p^n$

6. ## Re: trouble understanding a theorem on finite fields.

why, yes, it can! in fact, that is sort of the point, to show that there does in fact exist some finite field of order $\displaystyle p^n$ for every prime p and every positive integer n. one can even go further, because any two such fields are isomorphic, so one often sees the terminology "the field of order $\displaystyle p^n$".

7. ## Re: trouble understanding a theorem on finite fields.

Originally Posted by Deveno
why, yes, it can! in fact, that is sort of the point, to show that there does in fact exist some finite field of order $\displaystyle p^n$ for every prime p and every positive integer n. one can even go further, because any two such fields are isomorphic, so one often sees the terminology "the field of order $\displaystyle p^n$".
so does this also concludes that the algebraic closure of $\displaystyle \mathbb{Z}_p$ has infinite elements since n could be chosen to be any positive integer?

indeed.