Let me start by rigorously pose my question.
Let K be an algebraically closed field of characteristic 2, let n be an even integer number, let f(X)=Xn+T1Xn−1+cdots+Tn, be the generic polynomial, that is, T=(T1,ldots,Tn) is a tuple of algebraically independent variables over K.
Let Omega= { omega1,ldots,omegam} be a finite subset of K, let fi(X)=f(X)−omegai, and let Fi be the splitting field of fi over K(T) (i=1,ldots,m).
Question: For which Omega the splitting fields F1,ldots,Fm are linearly disjoint over K(T)?
Remarks:
If the characteristic of K is NOT 2, or if n is odd, then the splitting fields are linearly disjoint for arbitrary Omega. Thus, I pose the question the specific case of p=2 and n even.
The answer cannot be ALWAYS, as in the previous remark. Indeed, one can show that if p=n=2, m=4, and omega1+omega2+omega3+omega4=0, then the splitting fields are not linearly disjoint. In fact, if p=n=2, the answer is that the splitting fields are linearly disjoint if and only if the sum of any even number of elements of Omega does not vanish.
How one proves 1 + 2: The linear disjointness of the splitting fields can be reduced to the linear independent of the discriminant as elements in H1(K,mathbbZ/2mathbbZ). If pnmidn, then one can use ramification theory to achieve this, if pneq2 but divides n, one can calculate this by hand using the formula given by the determinant of the Sylvester matrix. If p=2, I know of no formula for the discriminant in terms of the coefficients. However when p=n=2 situation is simple enough to do calculations and hence get 2.
Motivation: The linear disjointness of the splitting fields allows one to calculate a Galois group of a composite of polynomials, which in turn yields arithmetic features of the ring of polynomials over large finite fields. Let me not elaborate on that here
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