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Let $k$ be an algebraically closed field. Suppose that $f\colon \mathbb{P}^n(k)\to \mathbb{P}^n(k)$ is a morphism of the form $f = [f_0:\cdots: f_n]$ where the $f_i$ are homogeneous polynomials of degree $d$ with no nontrivial common zeros. In this case, the degree of the morphism $f$ is $d^n$. The only way I know how to compute this is via the intersection theory of $\mathbb{P}^n(k)$. Since the degree of $f$ is defined (very concretely) as the degree of the field extension $f^*k(\mathbb{P}^n)\subseteq k(\mathbb{P}^n)$, I wonder if there is a less high-tech way of computing that $\deg f = d^n$. Does anyone know a way?

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What about the Finiteness Theorem (if $k$ has characteristic $0$)? The pre-image of a generic point has size $d^n$ – Jonathan May 8 '12 at 2:47
up vote 2 down vote accepted

Sketch of alternative: Consider the ring map $k[Y_0, \ldots, Y_n] \to k[X_0, \ldots, X_n]$, $Y_i \mapsto f_i(X_0, \ldots, X_n)$. Looking at Hilbert functions (thinking of these as graded rings) we see that $k[\underline{X}]$ has rank $d^{n + 1}$ as a $k[\underline{Y}]$-module (hint: compute leading coefficients of Hilbert pols and use that a graded module supported in a proper closed subscheme of $\mathbf{A}^{n + 1}$ has Hilbert pol of lower degree). The degree of $f$ is the degree of the field extension $k(Y_i/Y_0) \subset k(X_i/X_0)$. The reason this is $d^n$ and not $d^{n + 1}$ is because we are taking degree $0$ parts in the extension $k(Y_i/Y_0)[Y_0^{\pm 1}] \subset k(X_i/X_0)[X_0^{\pm 1}]$ which has degree $d^{n + 1}$ by the above.

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