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 $f(X) = X^n + T_1 X^{n-1} + \cdots + T_n$, be the generic polynomial, that is, $T = (T_1, \ldots, T_n)$ is a tuple of algebraically independent variables over $K$.

Let $\Omega = $ { $\omega_1, \ldots, \omega_m$} be a finite subset of $K$, let $f_i(X) = f(X) - \omega_i$, and let $F_i$ be the splitting field of $f_i$ over $K(T)$ ($i=1,\ldots, m$).

Question: For which $\Omega$ the splitting fields $F_1, \ldots, F_m$ are linearly disjoint over $K(T)$?


  1. 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.

  2. The answer cannot be ALWAYS, as in the previous remark. Indeed, one can show that if $p=n=2$, $m=4$, and $\omega_1 + \omega_2 + \omega_3 + \omega_4 = 0$, 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.

  3. 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 $H^1(K,\mathbb{Z}/2\mathbb{Z})$. If $p\nmid n$, then one can use ramification theory to achieve this, if $p\neq 2$ 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

  • $\begingroup$ When you say discriminant of a polynomial, you mean an element of the coefficient field that is zero if and only if the polynomial has a repeated root, and is "universal" in some sense? Also, have you tried reducing the left-over case to the one you already know how to treat by multiplying each of your polynomials by a general linear factor? If the linear factors are independent enough, this should not affect the linear disjointness... $\endgroup$ – damiano Apr 13 '10 at 9:19
  • $\begingroup$ By discriminant in characteristic $2$ I mean an element in the field, say $K$, such that the field of degree at most $2$ that corresponds to the even Galois permutations is generated by a root of $X^2 + X + a$. In fact, one should consider the Artin-Schreier coset of $a$. $\endgroup$ – Lior Bary-Soroker Apr 14 '10 at 19:10

Lior, you say:

If $p=2$, I know of no formula for the discriminant in terms of the coefficients.

However, this link says that the Sylvester matrix works the same over every field, whether $p=2$ or not. The article on PlanetMath about determinants seems to support this, not including a requirement of $p\neq2$. Using the determinant Sylvester matrix you get a formula in terms of the polynomial's coefficients like you wanted; I don't see why counting a determinant over a field with $p=2$ should be any different than in any other field, of course other than the fact that your definition of '$+$' is different. It doesn't seem you need anything other than to know how to differentiate polynomials over your field.

On the other hand, the Wikipedia article on the determinant also says that the determinant of quadratic forms cannot be generalized for fields of characteristic $p=2$. Is this what you were thinking of?

I have a feeling an answer to your question might be in the book "Generic Polynomials. Constructive Aspects of the Inverse Galois Problem" By Jensen, Ledet, Yui.

You might also find inspiration in the paper "Computing Galois Groups of Polynomials through ODEs", by Cormier, Singer, Trager, Ulmer, 2000, wherein they construct a linear differential operator that takes a polynomial and spits out information about its Galois group over the field as well as its algebraic closure; you might be able to come up with some form of chain rule for composite polynomials, solving your original problem. The paper was submitted to Journal of Symbolic Computation.

  • $\begingroup$ Thanks, I will look on the references you suggest. Discriminant in char 2 is an element a in the field such that the quadratic (or trivial) extension that corresponds to the even permutation in the Galois group is generated by a root of the equation X^2 + X = a. This $a$ is NOT the product of the differences of the roots of the polynomial. There is a formula for the discriminant in terms of the roots, and as Poonen mentioned to me, one can take the discriminant in characteristic zero of a lifting subtract 1 and divide by 4 (or something similar) however it didn't helped me in my calculations. $\endgroup$ – Lior Bary-Soroker Apr 14 '10 at 18:04
  • $\begingroup$ Lior, I'm not sure why you are mentioning the roots. The Sylvester matrix is made out of polynomial coefficients - not roots. You do know the coefficients of your polynomial, right? Just plug them in and try calculating the determinant of the matrix. $\endgroup$ – cheater Apr 14 '10 at 20:17
  • $\begingroup$ OK, I have just read your other comment. I see you have a different definition of 'discriminant'. Can't you just use the usual definition of 'discriminant' (i.e. a number that is zero when there are repeated roots, etc, etc)? On the other hand, I bet your definition is equivalent, or almost equivalent, to the one in common use. $\endgroup$ – cheater Apr 14 '10 at 20:20
  • $\begingroup$ Let x_1,..., x_n be the roots. The usual discriminant is $\prod_{i\neq j} (x_i-x_j)$. Then the square root of it, $\prod_{i<j} (x_i-x_j)$ is the expression fixed by even permutations but not by odd ones. In characteristic $2$, since $-1=1$, the square root of the permutation is symmetric. So you need to find a different element fixed just by the Alternating subgroup. There is some formula relating the zero characteristic and non-zero characteristic. But I couldn't used it to prove the above. $\endgroup$ – Lior Bary-Soroker Apr 17 '10 at 18:16
  • $\begingroup$ But why insist on expressing the discriminant with the roots which doesn't work, when you can express it with the polynomial coefficients which works? They're the same numbers, just computed with different formulas. Or do you need it because you need the number that you get from the second product (i<j)? If so, why do you need that number - what properties does it have that you need to use? Do you just need some property of polynomials that does not change through even permutations? $\endgroup$ – cheater Apr 17 '10 at 21:51

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