Degree and dimension of intersection of projective variety and hypersurface

Question

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:

1. 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.
2. 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.
3. Is there any relation between the dimension of a projective variety and its degree?

Answer 1

1. The degree of $Y\cap H$ is indeed $(\deg Y)(\deg H)$ in the correct context of scheme theory.
   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.

2. 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 $r-1$.

3. 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.

Answer 2

1. Suppose you have $s$ subvarieties $Y_1,\dots,Y_s$ of $\mathbb P^n$, with $Y_i$ of codimension $c_i$. Then Bézout is saying that if they meet transversely (and $\sum c_i\leq n$), then $$ \deg \,(Y_1\cap\dots \cap Y_s)=\prod \deg Y_i. $$ Now if $H$ and $Y$ are transverse then the degree of $Y\cap H$ is the product of the degrees computed by your (and my) formula. The formula you quoted is remarkable in the sense that the coefficients $i(Y,H;Z_i)$ are exactly those that make the ring structure on $A^\ast(\mathbb P^n)$ act as we desire: thanks to the (definitely nontrivial) definition of those coefficients, we have that "the class of the intersection is the product of the classes". Hence the degree is the product of the degrees.

2. The dimension is $r-1$: if you know that the components of a variety $V$ have the same dimension $m$, then $m=\dim V$ (for a variety, a possible definition of dimension is: maximum between the dimensions of all the irreducible components).

3. I don't think there is any reasonable relation between dimension and degree. Here is the motivation: the dimension is something rigidly attached to a scheme, while the degree depends on where you are embedding that scheme. For instance, $\mathbb P^1$ has degree 1 inside itself but becomes a conic in $\mathbb P^2$, and more generally for every $d$ it becomes a curve of degree $d$ in $\mathbb P^d$. But once you fix an ambient space, say $\mathbb P^n$, and you look at your (projective) variety $Y$ inside this $\mathbb P^n$, there is a relation between dimension and degree. The algebraic object that allows you to see it is the (leading term of the) Hilbert polynomial of $Y$. Its degree is the dimension of $Y$, while you recover $\deg Y$ as $(\dim Y)!$ times the leading coefficient.