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I am wondering if there is a generalisation of Bézout theorem in higher dimension. By Bézout theorem I mean:

If $f_1,f_2\in\mathbb{C}[X_1,X_2]$ such that $\dim V(f_1,f_2)=0$, then $\#V(f_1,f_2)\leq \deg f_1 \cdot \deg f_2$.

The generalisation I am expecting (and which may be false) is:

If $f_1,\ldots, f_n\in\mathbb{C}[X_1,\ldots,X_n]$ such that $\dim V(f_1,\ldots,f_n)=0$, then $\#V(f_1,\ldots,f_n)\leq \deg f_1 \cdots \deg f_n$.

Of course we could consider the same question in the more precise setting of projective space, intersection multiplicity and equality instead of inequality ; but my interest for the moment is only to have a bound for $\#V(f_1,\ldots,f_n)$.

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  • $\begingroup$ I think the Bezout bound still applies. See researchgate.net/post/… (which doesn't prove anything, except that someone who seems to know what he's talking about says the Bezout bound applies). $\endgroup$ – Gerry Myerson Sep 1 '13 at 12:13
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There is the Berstein's theorem that answer this question. This bound is right and we can be more precise.

Theorem (Bernstein's theorem): Let $\Delta_1,\ldots,\Delta_n$ be lattice polytope in $\mathbb{R}^n$ (i.e. polytope whose edges have integer coordinates). Let $f_1,\ldots,f_n \in \mathbb{C}[X_1,X_1^{-1},\ldots,X_n,X_n^{-1}]$ be Laurent polynomials with Newton polytope $\Delta_1,\ldots,\Delta_n$ in general position. Then the number of solution of $f_1=\cdots=f_n=0$ is $n!\mathrm{Vol}(\Delta_1,\ldots,\Delta_n)$.

I recall that the Newton polytope of a Laurent polynomial $f=\sum_{\alpha\in\mathbb{Z}^n}c_\alpha x^\alpha$ is the convex hull in $\mathbb{R^n}$ of $\{\alpha\in\mathbb{Z}^n\mid c_\alpha\neq0\}$ ; and the mixed volume $\mathrm{Vol}(\Delta_1,\ldots,\Delta_n)$ is the unique 'multilinear' symmetric operator such that $\mathrm{Vol}(\Delta,\ldots,\Delta)=\mathrm{Vol}(\Delta)$.

The bound of the question is a corollary: $f_i\in\mathbb{C}[X_1,\ldots,X_n]$ and $\deg f_i=d_i$, so the Newton polytope of $f_i$ is asubset of the elementary simplex of vertices $(0,0,\ldots,0), (d_i,0,\ldots,0), (0,d_i,\ldots,0),\ldots,(0,0,\ldots,d_i)$. Hence the mixed volume of the Newton polytopes of the $f_i$ is smaller then the mixed volume of these simplices, which is $\frac{d_1\cdots d_n}{n!}$ (can be computed using the definition of Wikipedia of mixed volume ).

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