I am looking at Theorem 7.7 of Hartshorne where he states the general form of Bezout's Theorem. The hypotheses of the theorem are as follows. Let $H$ be a hypersurface of degree $d$ and $Y \subseteq \Bbb{P}^n$ a projective variety of dimension $r$. If $Z_1,\ldots,Z_s$ are the irreducible components of $Y \cap H$, then we have
$$\sum_{i=1}^s i(Y,H;Z_i)\deg Z_i = (\deg Y)(\deg H)$$
where $i(Y,H;Z_j)$ is the length of $S/(I_Y + I_H)_{\mathfrak{p}_j}$ as a $S_{\mathfrak{p}_j}$ module. $S = k[x_0,\ldots,x_n]$, $\mathfrak{p}_j = I(Z_j)$ and $I_Y,I_H$ the homogeneous ideals of $Y$ and $Z$ respectively.
My questions are:
- Is it possible to deduce the degree of the intersection $Y \cap H$ from this theorem? I could if I knew that $I_Y + I_H = I(Y \cap H)$ but this may not be true here.
- What do we know about $\dim Y \cap H$? At the moment I only know that every irreducible component of $Y \cap H$ has dimension $r-1$ but not necessarily $Y \cap H$ itself.
- Is there any relation between the dimension of a projective variety and its degree?
Best Answer
However it is not true that $I_Y + I_H = I_{Y \cap H}$ in the provisional context of Hartshorne's Chapter I, devoted to classical algebraic varieties:
For example if in $\mathbb P^2$ you consider the conic $H=V(yz-x^2)$ and the line $Y=V(y)$, you get $I_H=(yz-x^2), I_Y=(y)$ but $I_{H\cap Y}=(x,y)\neq I_H+I_Y=(x^2,y)$.
This regrettable inequality of ideals is remedied by a more sophisticated definition of intersection in scheme theory, a theory you will soon meet in Chapter II of Hartshorne's book.
The dimension of a topological space having finitely many irreducible components is the maximum of the dimensions of those components (this follows from the definitions).
So here $Y\cap H$ has dimension $r-1$.
No: there are linear subspaces $L_m\subset \mathbb P^n$ of any dimension $0\leq m\leq n $ , but they all have degree one.