# 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 $$\delta_f = \sum_{i and $$D(f) = \sum_{i 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 $$[S_n:A_n]=2$$).