I have a rational map $f:\mathbb C^n\longrightarrow \mathbb C^n,$ all I know $f$ is defined by homogenious polynomials of degree $m$ and $f$ not necessarily a morphism. Computer packages aside, I am wondering if the passonate algebraic geometers have a general scheme of computing $\deg f$ explicitly?
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3$\begingroup$ What do you mean by $\ker f$? $\endgroup$– Felipe VolochCommented Nov 8, 2012 at 18:15
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1$\begingroup$ If the map is generically finite, I think the degree is $m^n$. It certainly is that for "most" $f$. $\endgroup$– Felipe VolochCommented Nov 8, 2012 at 19:47
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$\begingroup$ @ Felipe it seems like that is true if $f$ is a morphism never true in general. $\endgroup$– Ongaro Nyang'Commented Nov 8, 2012 at 22:06
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4$\begingroup$ If $f$ is not a morphism, what does "defined by polynomials" mean? $\endgroup$– Laurent Moret-BaillyCommented Nov 9, 2012 at 6:54
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1$\begingroup$ So you actually have a rational map from projective space to itself and you want to know the degree of this map? As Felipe asks, do you know that if map is generically finite? The degree will not be defined if every fibre is infinite. $\endgroup$– Daniel LoughranCommented Nov 9, 2012 at 11:54
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1 Answer
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An upper estimate can be obtained from Proposition 1.1 in the following paper:
Rusek, Kamil; Winiarski, Tadeusz Polynomial automorphisms of $\mathbb{C}^n$. Univ. Iagel. Acta Math. No. 24 (1984), 143–149
http://www2.im.uj.edu.pl/actamath/PDF/24-143-149.pdf
Let $F=(F_1,...,F_n):\mathbb{C}^n \mapsto \mathbb{C}^n$, where $F_1,...,F_n$ are polynomials (none of them identically zero). Assume that $F^{-1}(0)=\{a_1,...a_k\}$. Then $\nu_F:= \sum_i m_{a_i}F \leq {\rm deg }F_1\cdot ...\cdot {\rm deg }F_n$, where $m_{a_i}F$ is the multiplicity of $F$ at the point $a_i$.
Note that the polynomials in the proposition are not necessarily homogeneous. The assumption that $F^{-1}(0)$ be finite ensures that the multiplicities are well defined.
answered Nov 8, 2012 at 22:52