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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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What do you mean by $\ker f$? – Felipe Voloch Nov 8 '12 at 18:15
If the map is generically finite, I think the degree is $m^n$. It certainly is that for "most" $f$. – Felipe Voloch Nov 8 '12 at 19:47
@ Felipe it seems like that is true if $f$ is a morphism never true in general. – Ongaro Nyang' Nov 8 '12 at 22:06
If $f$ is not a morphism, what does "defined by polynomials" mean? – Laurent Moret-Bailly Nov 9 '12 at 6:54
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. – Daniel Loughran Nov 9 '12 at 11:54

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

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.

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