Bezout theorem for Nash functions (Q916744)

Let \(V=\{x\in {\mathbb{R}}^ n:\) \(f_ 1(x)=...=f_ p(x)=0\}\). Milnor and Thom gave an upper bound for the sum of the Betti numbers of V in the case where V is compact and \(f_ i\) denotes a polynomial map such that for each \(i=1,...,p\), \(\deg (f_ i)\leq d:\sum^{n}_{i=0}b_ i(V) \leq d(2d-1)^{n-1}.\) Our aim is to obtain a similar relation in a larger class of functions than the polynomials: Nash functions \((C^{\infty}\quad semi-algebraic\) functions, defined in an open semi-algebraic U of \(R^ n\), where R denotes a real closed field). It is a well-known result that Nash functions verify a polynomial equation: \(P(x,f(x))=0\) where \(P\in R[X,Y]\) does not vanish identically. So we define the complexity of a Nash function as being the total degree of its minimal polynomial. This notion will replace the degree of polynomial in Milnor-Thom's formula. We give a Bezout theorem which uses this complexity. Then, we obtain an upper bound for the sum of the Betti numbers of a compact Nash set V:\(\sum^{n}_{i=0}b_ i(V) \leq d^{2n-1}\) where \(f_ i\) is a Nash function of complexity less than or equal to d. In the case where V is not compact, we have: \(\sum^{n}_{i=0}b_ i(V) \leq (2^{p+1}d^ p)^{2n-1}\).