Newton's identities over finite fields The Newton identities (including over finite fields) are given by
$$
ke_k = \sum_{i=1}^k (-1)^{i-1} e_{k-i}p_i,
$$
where the $e_k$ is the $k$-th elementary symmetric polynomials and the $p_k$ is the $k$-th power sum (see here http://en.wikipedia.org/wiki/Newton%27s_identities).
If we work modulo a prime $\ell$ (over the field $\mathbb{F}_e\\$) and $\ell \mid k$, obviously the identity becomes $0 = 0$. In this case nothing useful is said of $e_k$. My question is whether it is implicit in these identities (but not mentioned) that 
$$
e_k = \left(\dfrac{1}{k}\sum_{i=1}^k (-1)^{i-1} e_{k-i}p_i \right)?
$$
That is, $\ell \nmid \frac{1}{k}\sum_{i=1}^k (-1)^{i-1} e_{k-i}p_i$? Thanks.
 A: As Robert Israel explained, consequences of Freshman's dream ruin your day. Relations like
$$
p_\ell=p_1^\ell
$$
make it clear that the power sums may not be algebraically independent even when the exponent is $<n$, if $\ell\le n$. The ring of symmetric polynomials in $n$ variables with coefficients in a field $K$ is always generated by $e_1,e_2,\ldots,e_n$. But if $n!$ is divisible by $\operatorname{char} K$, then, as we saw, it is not generated by $p_1,p_p,\ldots,p_n$.  $K[p_1,p_2,\ldots,p_n]$ is then a proper subring of $K[e_1,e_2,\ldots,e_n]$.
It is still often possible to recover the missing $e_i$:s, but we need to use power sums $p_i, i>n$, and we need to work with fields of rational functions. The simplest case is that from Robert's answer, so let's continue with $n=2=\ell$. We get (by Newton, or just a straightforward characteristic two calculation)
$$
p_3=X_1^3+X_2^3=(X_1+X_2)^3-3X_1X_2(X_1+X_2)=e_1^3+e_2e_1.
$$
We can solve for $e_2$ here
$$
e_2=\frac{p_3+e_1^3}{e_1}=\frac{p_3+p_1^3}{p_1}.
$$
So we see that $e_2\in K(p_1,p_3)$. So in this case we can conclude that the fields of symmetric rational functions
$$
K(e_1,e_2)=K(p_1,p_3).
$$
IIRC this generalizes to any number of variables $n$ and any characteristic $\ell$ in the sense that the fields $K(e_i\mid 1\le i\le n)$ and $K(p_i\mid 1\le i)$ have the same transcendence degree, so the former is a finite extension of the latter. I'm ashamed to admit that at the moment I don't know/remember whether they are always equal.
To close I cannot resist the temptation to bring up a point near and dear to my heart. When decoding a received word of a double-error-correcting binary BCH code the receiver's task is to figure out the error locations $X_1$ and $X_2$ based on syndromes that are exactly the power sums $p_i$. Because we are in characteristic two, the above observations imply that the receiver can successfully calculate the (error locator) polynomial
$$
(T-X_1)(T-X_2)=T^2+e_1T+e_2,
$$
if they know $p_1$ and $p_3$. That BCH-code is defined in terms of power sums $p_1$ and $p_3$, so this is just what the doctor ordered! We also notice that the case $p_1=0$, $p_3\neq0$ is a nasty special case, because our formula for $e_2$ then attempts to divide by zero.
A: For example, take $\ell = 2$, and two variables.  The identity for $k=2$: 
$$2 e_2 = e_1 p_1 - p_2 = e_1^2 - p_2$$
i.e.
$$ 2 X_1 X_2 = (X_1 + X_2)^2 - (X_1^2 + X_2^2)$$
Mod $2$, this is just the identity 
$$ (X_1 + X_2)^2 = X_1^2 + X_2^2$$
$e_1 = X_1 X_2$ is not in the ideal generated by $X_1 + X_2$ and $X_1^2 + X_2^2$ over $\mathbb F_2$.  For example, with $X_1 = X_2 = 1$ you have 
$X_1 + X_2 = X_1^2 + X_2^2 = 0$, but $X_1 X_2 = 1$. 
