Let $H$ be a degree $d$ hypersurface in $\mathbb CP^n$ defined by an explicit equation $F=0$. Let $\varphi: \mathbb P^n \to \mathbb P^n$ be an explicit degree $m$ morphism. In this case $\varphi(H)$ is a degree $d^{n-1}m$ hypersurface. Is there an algorithm to calculate the coefficients of the degree $d^{n-1}m$ polynomial that defines this hypersuface?
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1$\begingroup$ I think this is (part of) what is called elimination theory: you spell everything out and eliminate the old coordinates. In principle, this is algorithmical, but, as far as I understand, absolutely unpractical. $\endgroup$– Alex DegtyarevCommented Feb 25, 2014 at 22:09
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$\begingroup$ The degree of the Cartier divisor $\phi_*H$ is $d^{n-1}m$, not $dm$. You can compute it explicitly using norms. You can read about this in the section of Mumford's, "Lectures on Curves on an Algebraic Surface" that discusses norms and pushforwards of effective Cartier divisors. There may also be a discussion in the appendix of Fulton's "Intersection Theory" using Herbrand quotients (need to check that one). $\endgroup$– Jason StarrCommented Feb 26, 2014 at 3:53
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$\begingroup$ Jason, thank you for this comment, you are right of course. $\endgroup$– aglearnerCommented Feb 26, 2014 at 8:23
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$\begingroup$ I think that is not correct. Probably only we can say that $f_*H \in \mathcal{O}(d^{n-1}m)$, as a divisor. For example, consider the map $f: \mathbb{P}^n \to \mathbb{P}^n$ defined by $(x_0, x_1, ...x_n) \to (x_0^m, x_1^m, ...x_n^m)$, then the image of the hyperplane defined by $x_i=0$ is itself. $\endgroup$– LAPRASCommented Aug 14 at 6:30
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1 Answer
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As suggested by Alex in his comment, this is a classical problem involving elimination theory.
The relevant result is the following, see [Greuel - Pfister, A Singular introduction to commutative algebra, Proposition A.7.12 p. 505].
Proposition. Let $f=(f_0: \ldots :f_m)\colon \mathbb{P}^n \longrightarrow \mathbb{P}^m$ be a morphism, with $f_i \in \mathbb{K}[x_0, \ldots, x_n]$ homogeneous polynomials of the same degree without common zeroes. Moreover, let $I \subset \mathbb{K}[x_0, \ldots, x_n]$ be a homogeneous ideal, defining the projective variety $X=V(I) \subset \mathbb{P}^n$, and set $$J= \langle I, \, f_0-y_0, \ldots, f_m-y_m \rangle \cap \mathbb{K}[y_0, \ldots, y_m].$$ Then $$f(X)=V(J)\subset \mathbb{P}^m.$$
This can be explicitly determined by using Groebner bases. The computation is usually unpractical by hands, but it can be done with all the most common Computer Algebra Systems. For instance, many explicit computations with $\textrm{Singular}$ can be found in the book by Greuel and Pfister quoted above.
answered Feb 25, 2014 at 22:54