Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.

Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$.

A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.

It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial $V(X) = \prod_{i<j} (X_j-X_i)$, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.

Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.

That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.

I already figured out what is the "elementary" alternating polynomial $u/V$ and what is the Artin-Schreir equation it satisfies: $u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$ and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.)

My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$?

reallywant the definition of alternating to be $f(sX)-f(X)=\mathrm{add.sign(s)}$? When $f$ is $u$, the left hand side vanishes when two of the variables are zero, yet the right hand side of the equation does not if $u$ is odd. $\endgroup$ – Mariano Suárez-Álvarez Jan 1 '10 at 21:51