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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?
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  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.

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    $\begingroup$ Dear @Georges, I am a bit confused by your point 1: since at this point in Hartshorne we are talking only about set-theoretic intersections, the claim that deg($Y \cap H$) equals deg Y . deg H seems to be false, for instance in your example of a line and a conic meeting at a point. Please let me know if I am misunderstanding you. $\endgroup$ – user64687 Jun 24 '13 at 18:48
  • $\begingroup$ Dear @Asal, I meant that in scheme-theoretic intersection theory we do indeed have equality: the intersection of a line and a conic always has degree two. However you are right that I didn't check Hartshorne and I don't remember exactly what the point of view on intersection theory is in chapter I. In general I remember all the theorems of algebraic geometry in the context of scheme theory and, unfortunately, I sometimes forget that I have to adapt them here for beginners who momentarily only know classical varieties. I have edited my answer in order to suppress this misunderstanding. $\endgroup$ – Georges Elencwajg Jun 24 '13 at 19:20
  • $\begingroup$ Georges: yes, sometimes it can be tricky to answer questions like this using the appropriate language; Bezout's theorem wants to be stated in the language of schemes! $\endgroup$ – user64687 Jun 24 '13 at 19:30
  • $\begingroup$ @Asal(continued) ...and most importantly: thanks for your comment! $\endgroup$ – Georges Elencwajg Jun 24 '13 at 19:30
  • $\begingroup$ Dear @GeorgesElencwajg, thanks for your answer. Perhaps I should have been more specific with 3: What if we have non-degenerate projective varieties (i.e. those not contained in any hyperplane), is there a relationship between the degree and the dimension? $\endgroup$ – user38268 Jun 25 '13 at 3:14
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  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.
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