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 closure***ness* 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.

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.

Re: trouble understanding a theorem on finite fields.

Quote:

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.

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.

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 $\displaystyle x^{p^n} - x$ 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 $\displaystyle |E|=p^n$

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 $\displaystyle p^n$".

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 $\displaystyle p^n$ 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?

Re: trouble understanding a theorem on finite fields.