# Does Bezout theorem provide a bound for the number of connected components of the zero set?

Let $$f_1,\dots,f_n$$ be homogeneous polynomials in an $$n$$-dimensional complex projective space. Is it true that the number of connected components of the set of their common zeroes doesn't exceed the product of their degrees?

Wille Liou already provided a proof, but here is a reference in case you need one: Fulton, Intersection Theory (2d ed, 1998), example 8.4.6 on page 146:

Let $$V_1,\ldots,V_r$$ be subvarieties of $$\mathbb{P}^n$$. Let $$Z_1,\ldots,Z_t$$ be the irreducible components of $$V_1\cap\cdots\cap V_r$$. Then: $$\sum_{i=1}^t \deg(Z_i) \leq \prod_{j=1}^r \deg(V_j)$$ In particular, the number of irreducible components of [the intersection of the] $$V_j$$ is at most the product of the degrees.

See also example 12.3.1 on page 233 op. cit. The same result is contained in ¶2.3 (page 10) of the older (shorter, and perhaps more readable) Introduction to Intersection Theory (1984) by the same author.

Note that this gives a bound on the number of irreducible components, which also works for the number of connected components since connected components are finite unions of irreducible components. I don't know how Fulton's proof and the one given in Wille Liou's answer are related, and, in particular, whether the latter can be adapted to give the bound on the number of irreducible components.

• Thank you! Having a reference is very helpful. Both because I can cite it and because it's much easier to learn all the definitions from a single book. Commented Jun 19, 2023 at 9:12

Yes, it is a consequence of Bézout's theorem and the Stein factorisation.

Let $$d_i$$ be the degree of $$f_i$$ for $$i = 1, \ldots, n$$. Consider the vector space $$S = \bigoplus_{i = 1}^n \mathbb{C}[z_0, \cdots, z_n]_{d_i}$$ as complex algebraic variety, where $$\mathbb{C}[z_0, \cdots, z_n]_{d_i}$$ is the space of homogeneous polynomials of degree $$d_i$$. There are tautological sections $$g_i\in \Gamma(S, \mathcal{O}_S[z_0, ..., z_n]_{d_i})$$ for $$i = 1,\ldots, n$$. Let $$\mathcal{X} = \mathrm{V}_+(g_1, \ldots, g_n)\subseteq \mathbb{P}^n_S$$ be the universal family over $$S$$. It is easy to see that the total space $$\mathcal{X}$$ is non-singular and irreducible (projecting it onto $$\mathbb{P}^n_{\mathbb{C}}$$). The projective variety in question can be identified with its fibre $$\mathcal{X}_f$$ at $$f = (f_1, \ldots, f_n)\in S(\mathbb{C})$$. By the Stein factorisation, the projection $$\mathcal{X}\to S$$ can be factorised as $$\mathcal{X}\xrightarrow{\alpha} \bar{\mathcal{X}}\xrightarrow{\beta} S$$ such that $$\alpha$$ is proper with connected fibres and $$\beta$$ is finite. For a generic $$\mathbb{C}$$-point $$s\in S(\mathbb{C})$$, the fibre $$\mathcal{X}_s$$ is a disjoint union of $$d = \prod_i d_i$$ copies of $$\operatorname{Spec} \mathbb{C}$$, so $$\beta^{-1}(s)$$ has $$d$$ points. The semi-continuity of the fibre of $$\beta$$ implies that $$\beta^{-1}(f)$$ has $$\le d$$ points. It follows that $$\mathcal{X}_f = \alpha^{-1}(\beta^{-1}(f))$$ has $$\le d$$ connected components.

• Thank you very much! I have to learn a lot of new words to read this answer :) Commented Jun 19, 2023 at 9:09