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I want to compute the discriminant of the following polynomial $$ F(X,Y)=X^mY^n+\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}c_{ij}X^iY^j. $$ Here the discriminate means the equation $D(c_{i,j})$ in the variables $c_{ij}$ such that $F(X,Y)=0$ has singular locus if and only if $D(c_{ij})=0$.

Is an explicit formula for $D(c_{ij})$ known? Can one compute the degree of $D(c_{ij})$?

Any useful comments and references will be greatly appreciated. Thank you.

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  • $\begingroup$ I am not sure there is a general formula for the one-variable case, that is explicit in the coefficients (there is of course the determinant version). $\endgroup$ Oct 17, 2014 at 16:27

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One can rehomogenize your problem to deal with the polynomial $f(x,y,z)=z^{m+n}F(x/z,y/z)$. Then what you want is the discriminant of the degree $m+n$ homogeneous polynomial $f$. This is the same as the resultant of $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$. There is a complicated determinant formula for that due to Sylvester and Morley, see this paper. There are quite a few monomials missing in your $F$ so there might be better formulas based on $A$-discriminants and resultants. For that I would look up the book by Gelfand, Kapranov and Zelevinsky.

Update: I just looked up my copy of "Using Algebraic Geometry" by Cox Little and O'Shea. Exercise 15 of Ch. 3, Sec. 4 explains Sylvester's construction in a more pedagogical way. Ch. 7 should help with the degree determination, although the absence of $c$ in front of the leading monomial $X^m Y^n$ could complicate things.

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