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Let $F$ be a splitting field of $x^{p^{n}} - x \in \mathbb{Z}_p[x]$.

How is it that the nonzero elements multiply to $-1$ and sum to $0$? I don't get how we get that result.

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That follows from Vieta's formulas (en.wikipedia.org/wiki/Vieta%27s_formulas). –  Yury Aug 11 '12 at 15:03
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Do you know that the elements of the field of size $p^n$ are exactly the zeros of that polynomial? If so, look here for the product. Also the usual argument works: if $$x^r+p_{r-1}x^{r-1}+\cdots+p_0=(x-a_1)(x-a_2)\cdots(x-a_r),$$ then $p_{r-1}=-\sum_i a_i$ and $p_0=(-1)^r\prod_i a_i.$ –  Jyrki Lahtonen Aug 11 '12 at 15:06
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What if $p=2$? The sum shouldn't be 0 in that case. –  Miha Habič Aug 11 '12 at 15:10
    
@MihaHabič That's what I thought at first, and then I realised that there had to be some non-trivial cancellations happening. –  Zhen Lin Aug 11 '12 at 15:10
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@ZhenLin I guess the trivial case $p=2,n=1$ should be excluded then. –  Miha Habič Aug 11 '12 at 15:14

4 Answers 4

up vote 2 down vote accepted

We appeal to the structure theorem for finite fields: the field $\mathbb{F}_{p^n}$ is precisely the splitting field of the polynomial $x^{p^n} - x$ over the prime field $\mathbb{F}_p$. We thus have a factorisation in $\mathbb{F}_{p^n}$ $$x^{p^n} - x = x (x-a_1) \cdots (x-a_{p^n - 1})$$ and expanding the right hand side and comparing coefficients, assuming $p^n > 2$, we find \begin{align} a_1 + \cdots + a_{p^n - 1} & = 0 \\ a_1 \cdots a_{p^n - 1} & = (-1)^{p^n} \end{align} which is what we wanted. (Note that the polynomial $x^{p^n} - x$ is separable because its derivative is the constant $1$, so its zeros are all distinct and constitute all the elements of $\mathbb{F}_{p^n}$.)

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Your answer does not answer the OP's question if you don't show $\,a_1,...,a_{p^n-1}\,$ are all the nonzero elements of the splitting field $\,\Bbb F\,$, and for that I think you need the structure theorem of finite fiels. –  DonAntonio Aug 11 '12 at 16:07

Here's an answer that avoids using the polynomial $X^{p^n}-X$ and uses the fact that you are computing in a finite field.

So, you are asking why $$ \sum_{x\in\Bbb F_q}x=0 \qquad\text{and}\qquad \prod_{0\neq x\in\Bbb F_q}x=-1. $$ in any finite field $\Bbb F_q$ with $q=p^n$ elements.

For the first one: let $s=\sum_{x\in\Bbb F_q}x$. Pick any $z\in\Bbb F_q$, $z\notin\{0,1\}$. Multiplication by $z$ defines a bijection $\Bbb F_q\rightarrow\Bbb F_q$. Thus $$ zs=\sum_{x\in\Bbb F_q}zx=\sum_{y\in\Bbb F_q}y=s, $$ from which $s=0$.

For the other one, consider first the case $p\neq2$. Then you have an even number of $x$'s different from $0$. Pair them together considering pairs $(x,x^{-1})$ where $x\neq x^{-1}$. What is left out is the elements such that $x^2=1$ of which there are exactly two: the roots of the polynomial $X^2-1$, i.e. $1$ and $-1$ (distinct!). Now the product of each previous pair is $1$ and of the last two is $-1$.

Finally, consider the case $p=2$. It goes as above, except that now there are an odd number of elements to be multiplied together but also there's only one repeated root of $X^2-1=(X-1)^2$. Thus, $1$ is the only element left out from the pairings and $\prod_{0\neq x\in\Bbb F_q}x=1=-1$ (last equality holds since $p=2=0$ in $\Bbb F_q$).

WARNING : All of the above works with one exception, namely $q=2$. In this case $\sum_{x\in\Bbb F_2}x=0+1\neq0$. The above arguments breaks down because there's no element $z$ than can be chosen as requested.

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Let $\theta_1,\theta_2,...,\theta_{p^{n-1}}$ be the non-zero roots of the polynomial in a splitting field. Then, in the splitting field, we have $$x(x^{p^n-1} - 1) = x\prod_{k=1}^{p^n-1}(x-\theta_k) = x(x^{p^n-1} + ... + x(\theta_1 + ... + \theta_{p^n-1}) + \theta_1\theta_2...\theta_{p^n-1}).$$ Hence mathching up coefficients on the left and right we find that the product of the non-zero roots is $-1$ while the sum is $0$.

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There is another approach. For note that the only elements which are their own multiplicative inverses are $1$ and $-1$ (if $p =2,$ these are the same, but the argument still applies). Hence when you multiply all non-zero elements of the field together, the terms other than $1$ and $-1$ cancel out in mutually inverse pairs, leaving a product $-1$ (ok even if $p=2$!). Similarly, when you add all elements of the field, when $p \neq 2$, every element except zero is different from its additive inverses, the mutually inverse pairs cancel each other out, and only zero is left. When $p =2,$ the case $n = 1$ is a genuine exception to the additive property, for the sum of the element of the field is $1$ in that case. It is slightly unsatisfactory, but one way to deal with the case $p = 2, n > 1$ is to use the fact that the multiplicative group of the field of $p^{n}$ elements is cyclic. Let $\alpha$ be an element of multiplicative order $p^{n}-1.$ Then the non-zero elements of the field are $1, \alpha, \alpha^{2}, \ldots, \alpha^{p^{n}-2}.$ Their sum is $\frac{1 - \alpha^{p^{n}-1}}{1-\alpha}$ which is zero since $\alpha \neq 1.$

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