# trouble understanding a theorem on finite fields.

• July 13th 2011, 02:21 AM
abhishekkgp
trouble understanding a theorem on finite fields.
THEOREM: Let $E$ be a field of $p^n$ elements contained in an algebraic closure $\bar{\mathbb{Z}_p}$ of $\mathbb{Z}_p$. The elements of $E$ are precisely the zeros in $\bar{\mathbb{Z}_p}$ of the polynomial $x^{p^n}-x$ in $\mathbb{Z}_p[x]$.

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

QUERY: Where have we used the algebraic closureness of $\bar{\mathbb{Z}_p}$. Wouldn't the proof still work if instead of $\bar{\mathbb{Z}_p}$ we had any field $K$ which is an extension field of the field $\mathbb{Z}_p$ containing the field $E$.
May be i am missing the obvious but can't figure it out.
• July 13th 2011, 04:37 AM
HallsofIvy
Re: trouble understanding a theorem on finite fields.
The roots of $x^{p_n}- x= x(x^{p_n-1}- 1)= 0$, other than the trivial root, x= 0, form a cyclic set. That is, there exist a root $x_i$ such that $\left{x_i^j\right}$ 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.
• July 13th 2011, 04:50 AM
abhishekkgp
Re: trouble understanding a theorem on finite fields.
Quote:

Originally Posted by HallsofIvy
The roots of $x^{p_n}- x= x(x^{p_n-1}- 1)= 0$, other than the trivial root, x= 0, form a cyclic set. That is, there exist a root $x_i$ such that $\left{x_i^j\right}$ 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.
• July 13th 2011, 06:03 AM
Deveno
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 $x^{p^n} - x$ only given some "K" containing that "E", but if we want to consider every possible n, we need a lot of "different" K's.
• July 13th 2011, 06:10 AM
abhishekkgp
Re: trouble understanding a theorem on finite fields.
Quote:

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 $x^{p^n} - x$ 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 $M=\bar{\mathbb{Z}_p}$ be the algebraic closure of $\mathbb{Z}_p$. Then for all $n \in \mathbb{Z}^+$ there exists a subfield $E$ of $M$ such that each element of $E$ is a zero of $x^{p^n}-x$ with $|E|=p^n$
• July 13th 2011, 07:14 AM
Deveno
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 $p^n$ 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 $p^n$".
• July 13th 2011, 07:21 AM
abhishekkgp
Re: trouble understanding a theorem on finite fields.
Quote:

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 $p^n$ 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 $p^n$".

so does this also concludes that the algebraic closure of $\mathbb{Z}_p$ has infinite elements since n could be chosen to be any positive integer?
• July 13th 2011, 11:41 AM
Deveno
Re: trouble understanding a theorem on finite fields.
indeed.