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I have a final tomorrow, and I was looking over some exercises in my textbook. However, I can't seem to work this problem out.

Let $F$ be a field of $p^n$ elements and let $\alpha \in F^*$, where $F^*=F-\{0\}$. Show $(x- \alpha)(x- \alpha^p)(x- \alpha^{p^2})\cdots(x- \alpha^{p^{n-1}}) \in \mathbb{Z}_p[x]$. Also, my text has some weird notation, $\mathbb{Z}_p$ is shorthand for $\mathbb{Z}/p\mathbb{Z}$.

It hints that I need to show both $\alpha+\alpha^{p}+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\in\mathbb{Z}_p$ and $\alpha\alpha^{p}\alpha^{p^2}\cdots\alpha^{p^{n-1}}\in\mathbb{Z}_p$, but I'm having trouble with it. Help would be much appreciated.

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@Berci I already know how the hint will allow me to solve the whole problem, but first I need to prove the hint is true, which is where I'm stuck. –  Nathan May 20 '13 at 1:22
    
Your exponents don't seem right. Is the last term $(x-\alpha^{p^n-1}$ supposed to be $(x-\alpha^{p^{n-1}})$? If so, the polynomial with $n$ factors is the minimal polynomial $M_\alpha(x)$ of $\alpha$ over $\mathbb F_p$ or $\mathbb Z/p\mathbb Z$, or a power $[M_\alpha(x)]^k$ thereof. –  Dilip Sarwate May 20 '13 at 1:29
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2 Answers

up vote 1 down vote accepted

In a field of characteristic $p$, $(a-b)^p = a^p-b^p$.

Now, $$\left[\prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right)\right]^p = \prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right)^p = \prod_{i=0}^{n-1}\left(x^p-\alpha^{p^{i+1}}\right) = \prod_{i=1}^{n}\left(x^p-\alpha^{p^i}\right) = \prod_{i=0}^{n-1}\left(x^p-\alpha^{p^i}\right)$$ where the rightmost equality follows from the fact that $\alpha^{p^n} = \alpha$ for all $\alpha \in \mathbb F_{p^n}$. Thus we have that $$f(x) = \prod_{i=0}^{n-1}\left(x-\alpha^{p^i}\right) = \sum_{j=0}^n f_jx^j$$ enjoys the property that $[f(x)]^p = f(x^p)$. But, $\displaystyle [f(x)]^p = \left[\sum_{j=0}^n f_jx^j\right]^p = \sum_{j=0}^n f_j^px^{jp}$ equals $\displaystyle f(x^p) = \sum_{j=0}^n f_jx^{jp}$ if and only if $f_j^p = f_j, 0 \leq j \leq n$, that is, $f_j \in \mathbb F_p$ and hence $f(x) \in \mathbb F_p[x]$.

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The easiest way to do this problem is the following: the map $\phi:\alpha \mapsto \alpha^p$ is an automorphism of $F$ whose fixed field is the prime subfield $\mathbb{Z}/p$. Furthermore, $\phi^n=1$ is the identity (since $F$ has $p^n$ elements). Your polynomial is therefore fixed by $\phi$, meaning that its coefficients are in $\mathbb{Z}/p$.

Here are the details if you want to do this from scratch (assuming Lagrange's theorem counts as "scratch"): first, for any $\alpha \in F$, we have $\phi^n(\alpha)=\alpha^{p^n}=\alpha$ by Lagrange's theorem applied to the group $F^\times$. This implies the fact mentioned above, that the coefficients of your polynomial are fixed by $\phi$.

In particular, taking $n=1$ shows that $\phi(\alpha)=\alpha$ for all $\alpha \in \mathbb{Z}/p$. In other words, every element of the prime field $\mathbb{Z}/p$ is a zero of the polynomial $x^p-x$. Now a polynomial of degree $p$ can have at most $p$ roots in any field, which shows that the prime subfield is exactly the set of $\alpha \in F$ with $\alpha=\phi(\alpha)$.

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We've never discussed fixed fields, so I'm stuck trying to show that this hint is true. –  Nathan May 20 '13 at 1:26
    
@CollegeKid, Well, do you know what the maximum number of roots for a polynomial of degree $p$ is? –  S123 May 20 '13 at 1:33
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