Finding primitive element of field extension in characteristic 2 corresponding under Galois correspondence to the group $G_f\cap A_n$ Let $F$ be a field and let $f(X)\in F[X]$ be a separable polynomial over $F$ of degree $n$. Let $F_f$ be the splitting field of $f$. Then the Galois group $Gal(F_f/F)=G_f$ acts as a group of permutations on the roots of $f$ and we can consider $G_f$ as a subgroup of the symmetric group $S_n$. Define
$$
SG_f = G_f\cap A_n,
$$
where $A_n$ is the alternating group. Let
$$
\Delta(f) = \prod_{1\leq i < j\leq n} (\alpha_i - \alpha_j),
$$
where $\alpha_1,\dots,\alpha_n$ are the distinct roots of $f(X)$ in $F_f$. 
It is easy to prove that $\sigma\Delta(f)=\mbox{sgn}(\sigma)\Delta(f)$ for every $\sigma\in S_n$, and thus we have, if $\mbox{char}(F)\neq 2$, that $\sigma\in S_n$ fixes $\Delta(f)$ if and only if $\sigma\in A_n$. Thus under the Galois correspondence, we have that the group $SG_f = G_f\cap A_n$ corresponds to the field $F[\Delta(f)]$.
The condition that $\mbox{char}(F)\neq 2$ is essential, for if $\mbox{char}(F)=2$ then every element of $S_n$ fixes $\Delta(f)$.
Here is my question: How can I find some element $\delta\in F_f$ such that, under the Galois correspondence, the group $SG_f=G_f\cap A_n$ corresponds to the field $F[\delta]$ when $\mbox{char}(F)=2$?
Added: It is easy to prove that such $\delta_f$ exists: $A_n$ is a normal subgroup of $S_n$ and thus $SG_f$ is a normal subgroup of $G_f$. If $L=F_f^{SG_f}$ is the fixed subfield of $F_f$ corresponding to de group $SG_f$, then we have that $L/F$ is a Galois extension and by the primitive element theorem there we have that $L=F[\delta_f]$ for some $\delta_f\in L$. But it is an existence proof, I need an explicit expression for such $\delta_f$. Even by following the proof of the primitive element theorem, I am unable to determine such an expression for $\delta_f$.

Assume $\mbox{char}(F)=2$ and let $\alpha_1,\dots,\alpha_n$ be the distinct roots of $f(X)$. Define
$$
\delta_f = \sum_{i<j} \frac{\alpha_i}{\alpha_i + \alpha_j}
$$
and
$$
D(f) = \sum_{i<j} \frac{\alpha_i\alpha_j}{\alpha_i^2 + \alpha_j^2}.
$$
The element $D(f)$ is called the Berlekamp discriminant of $f(X)$. It is straighforward to verify that
$$
\delta_f^2 + \delta_f + D(f) = 0. 
$$
If $\tau\in G_f$ is the transposition $(k \; \; k+1)$, we obtain
$$
\sigma\delta_f = \delta_f + \frac{\alpha_{k}}{\alpha_k + \alpha_{k+1}} + \frac{\alpha_{k+1}}{\alpha_{k+1} + \alpha_{k}} = \delta_f + 1.
$$
Thus $\sigma\delta_f = \delta_f$ if and only if $\sigma\in A_n\cap G_f=SG_f$. By the Galois correspondence you obtain that 
$$
F(\alpha_1,\dots,\alpha_n)^{SG_f} = F(\delta_f)
$$
as desired (it is necessary to use the fact that $[S_n:A_n]=2$).
