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Let $F(x,y), A(x,y), B(x,y)$ be homogeneous polynomials with integer coefficients (a.k.a. binary forms) such that the degrees of $A(x,y)$ and $B(x,y)$ match. Define

$$R(x,y) := F\left(A(x,y), B(x,y)\right).$$

Then $R(x,y)$ is a homogeneous polynomial with integer coefficients of degree $dr$, where $d := \deg(F)$ and $r := \deg(A) = \deg(B)$. We identify the discriminant $D(R)$ of $R(x,y)$ with the discriminant of the polynomial $R(x,1)$. Similarly, we define the discriminants of $F, A, B$.

I would like to obtain the formula for $D(R)$. Alternatively, I need to bound $D(R)$ from below non-trivially in terms of the invariants of $F, A, B$, such as their discriminants. Has this question been explored at all?

It seems to me that John Cullinan in his short article The discriminant of a composition of two polynomials considers a somewhat similar problem, though restricting himself to the composition of univariate polynomials, not homogeneous polynomials in two variables.

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If I did not mess up the adaptation to your special case, Theorem 3.31 for $n=2$ in "A computational approach to the discriminant of homogeneous polynomials" by Busé and Jouanolou says that $$ {\rm Disc}(R)=({\rm Disc}(F))^r\times {\rm Res}(A,B)^{d(d-1)}\times K(F,A,B) $$ where $K$ is homogeneous of degree $2d(r-1)$ in the coefficients of $A$ or $B$ and of degree $2(r-1)$ in the coefficients of $F$.

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  • $\begingroup$ 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? $\endgroup$ – Anton May 21 '17 at 17:02
  • $\begingroup$ 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". $\endgroup$ – Abdelmalek Abdesselam May 21 '17 at 18:13
  • $\begingroup$ I should also mention that they show that $K$ is irreducible in $char\neq 2$. $\endgroup$ – Abdelmalek Abdesselam May 21 '17 at 18:43
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    $\begingroup$ 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. $\endgroup$ – Anton May 23 '17 at 13:31
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    $\begingroup$ @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 $\endgroup$ – Abdelmalek Abdesselam May 23 '17 at 14:00
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When $R = h(f,g)$ with $h,f,g$ quadratic forms, with coefficients $h_i, f_i, g_i$ respectively for $i = 0,1,2$, the $K$ polynomial is given explicitly as

$$\displaystyle K = g_1^4h_0^2 - 8 g_0 g_1^2 g_2 h_0^2 + 16 g_0^2 g_2^2 h_0^2 - 4 f_2 g_0 g_1^2 h_0 h_1 + 2 f_1 g_1^3 h_0 h_1 + 16 f_2 g_0^2 g_2 h_0 h_1 - 8 f_1 g_0 g_1 g_2 h_0 h_1 - 4 f_0 g_1^2 g_2 h_0 h_1 + 16 f_0 g_0 g_2^2 h_0 h_1 + f_1^2 g_1^2 h_1^2 - 4 f_0 f_2 g_1^2 h_1^2 - 4 f_1^2 g_0 g_2 h_1^2 + 16 f_0 f_2 g_0 g_2 h_1^2 + 16 f_2^2 g_0^2 h_0 h_2 - 16 f_1 f_2 g_0 g_1 h_0 h_2 + 2 f_1^2 g_1^2 h_0 h_2 + 8 f_0 f_2 g_1^2 h_0 h_2 + 8 f_1^2 g_0 g_2 h_0 h_2 - 16 f_0 f_1 g_1 g_2 h_0 h_2 + 16 f_0^2 g_2^2 h_0 h_2 - 4 f_1^2 f_2 g_0 h_1 h_2 + 16 f_0 f_2^2 g_0 h_1 h_2 + 2 f_1^3 g_1 h_1 h_2 - 8 f_0 f_1 f_2 g_1 h_1 h_2 - 4 f_0 f_1^2 g_2 h_1 h_2 + 16 f_0^2 f_2 g_2 h_1 h_2 + f_1^4 h_2^2 - 8 f_0 f_1^2 f_2 h_2^2 + 16 f_0^2 f_2^2 h_2^2.$$

Not only is this homogeneous in the variables here, it is bi-homogeneous in $(h_2,h_1,h_0)$ and $(f_2, f_1, f_0, g_2, g_1, g_0)$. With $h_2, h_1, h_0$ fixed, $K$ is the product of two quadratic forms in $f_2, f_1, f_0, g_2, g_1, g_0$. I suspect in general it might be very hard to study the nature of $K$.

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  • $\begingroup$ thanks for your answer! How did you derive the formula for K? Did you refer to the same paper as Abdelmalek mentioned? $\endgroup$ – Anton May 21 '17 at 21:34
  • $\begingroup$ I discovered this independently of that paper for my work on binary quartic forms $\endgroup$ – Stanley Yao Xiao May 21 '17 at 21:46
  • $\begingroup$ Nice explicit formula. I wonder what could be the geometric interpretation of this beast. $\endgroup$ – Abdelmalek Abdesselam May 23 '17 at 14:01

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