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May 23, 2017 at 14:00 comment added Abdelmalek Abdesselam @Anton: You are right. I just fixed the formula. I think you are also right that there are quite a few typos in the article regarding $d_1\ldots d_n$ instead of $d_1\ldots d_{n-1}$. Unfortunately, these errors are also in the published version link.springer.com/article/10.1007/s11786-014-0188-7
May 23, 2017 at 13:58 history edited Abdelmalek Abdesselam CC BY-SA 3.0
corrected formula
May 23, 2017 at 13:31 comment added Anton I think there is a mistake in the degree of K w.r.t. A or B. It should be $2d(r-1)$, not $2dr(r-1)$. This matches the example that Stanley Yao Xiao wrote (case d=r=2). Also, I think there is a typo in the paper that is a little bit confusing: sometimes they write a product of $d_i$ from i to n, which is wrong, since $d_i$ are defined only up to $n-1$. Just wanted to confirm with you that my intuition is correct.
May 21, 2017 at 21:36 comment added Anton Thanks Abdelmalek, this is clearly what I was looking for. What really puzzled me is that K is a polynomial in F,A,B, but then I saw in your answer that it is actually a polynomial in coefficients of F,A,B.
May 21, 2017 at 21:33 vote accept Anton
May 21, 2017 at 18:43 comment added Abdelmalek Abdesselam I should also mention that they show that $K$ is irreducible in $char\neq 2$.
May 21, 2017 at 18:13 comment added Abdelmalek Abdesselam They do not give more info than that. $K$ is some (multi) homegeneous polynomial in $F$ and $(A,B)$. There are nice (base change) formulas for resultants of compositions but the authors say that the discriminant case is "much more involved".
May 21, 2017 at 17:02 comment added Anton thanks a lot for your respond! I don't quite understand what K is. You say tat it is homogeneous. Homogeneous what? Is this some kind of invariant?
May 21, 2017 at 16:51 history answered Abdelmalek Abdesselam CC BY-SA 3.0