# Question about Bezout's theorem between varieties and schemes.

Exercise 18.6.K in Vakil's Foundations of Algebraic geometry is:

Let $$X$$ be a projective scheme of dimension $$\geq 1$$ over a field $$k$$, with a fixed closed immersion $$i : X \rightarrow \mathbb{P}^n_k$$. Let $$H= V(f)$$ be a hypersurface not containing any associated points of $$X$$. Then, $$\deg(H \cap X) = \deg(H) \cdot \deg(X)$$

Theorem 1.7.7 in Hartshorne's Algebraic Geometry is : Let $$Y$$ be a variety of dimension $$\geq 1$$ in $$\mathbb{P}^n_k$$. Let $$H$$ be a hypersurface not containing $$Y$$. Let $$Z_1, \dots, Z_s$$ be the irreducible components of $$Y \cap H$$, corresponding to prime ideals $$p_1, \dots, p_s$$. Define $$i(Y, H ; Z_j)$$ to be the length of $$k[x_0, \dots, x_n] / (I(Y) + I(H))$$ at $$p_j$$. Then, $$\sum_{j=1}^s i(Y, H ; Z_j) \cdot \deg(Z_j) = \deg(Y) \cdot \deg(H)$$.

I want to use theorem 1.7.7 to solve exercise 18.6.K. The difficulty that I'm having is that in chapter 1 of Hartshorne, only integral varieties over algebraically closed fields are considered. But in Vakil, $$X$$ can be reduce or have multiple components.

My attempt at solving this problem is: if I take $$X = \operatorname{Proj}(k[x_0, \dots, x_n] / I)$$, I can then take a primary decomposition of $$I = \cap Q_i$$ to consider each individual irreducible component separately, therefore I reduce to where $$X$$ is irreducible. But then I'm not sure how to relate the degree of $$k[x_0, \dots, x_n] / I$$ with $$k[x_0, \dots, x_n] / Q_i$$. Furthermore, $$Q_i$$ is just primary , and not prime. So I can't apply theorem 1.7.7 directly. I also don't know how relate the degree of $$k[x_0, \dots, x_n] / Q_i$$ with its length over $$p_i$$.

• I think you'll find it instructive to dig in to the proof of I.7.7 - the technique of the proof is the thing to copy more than the result. (I also fixed up your formatting a bit - check the edit summary if you're interested in what exactly happened.) Commented May 29, 2021 at 7:29

Let me expand my comments in to an answer. Writing $$S$$ for the homogeneous coordinate ring of $$X\subset\Bbb P^n$$, $$f$$ for the defining equation of $$H$$, and $$d=\deg f$$, we have the exact sequence $$0\to S(-d)\stackrel{\cdot f}{\to} S\to S/(f)\to 0,$$ where the first map is injective because $$V(f)$$ contains no associated points of $$X$$. Now we can take Hilbert polynomials to see that $$H_X(n)-H_X(n-d)=H_{X\cap H}(n)$$. Writing $$H_X(n)= c_0n^a+c_1n^{a-1}+\cdots$$, we see that $$H_X(n-d)= c_0(n-d)^a+c_1(n-d)^{a-1}+\cdots = c_0(n^a-adn^{a-1}+\cdots)+c_1n^{a-1}+\cdots$$. So $$H_X(n)-H_X(n-d)= c_0adn^{a-1}+\cdots$$, giving that $$\deg X\cap H = (a-1)!c_0ad=a!c_0d=d\cdot \deg X$$, and we're done.