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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 $u$ 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$? 2 Fix an equation # ExcplicitExplicit expression of an alternating polynomial in characteristic$2$? 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$u$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$T$? 1 # Excplicit expression of an alternating polynomial in characteristic$2$? 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

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 $u$ 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 $T$?