Basic question regarding degrees of algebraic sets

I am learning about degrees of algebraic sets at the moment, and in an article I am reading I came across the following:

Let $V_i \subseteq \mathbb{C}^n$ be a hypersurface of degree at most $D$ for each $1 \leq i \leq l$. Let $V = \cap_{1 \leq i \leq l} V_i$. Then by the basic properties of degree, $V$ has at most $D^l$ irreducible components.

Could someone please clarify to me what "basic properties of degree" the author is referring to? Thank you very much!

• Perhaps they are referring to Bézout's theorem? (My deleted answer was written a bit too quickly... I misread your intersection for a union and kept writing intersection everywhere anyway!) You should check out Hartshorne's last section of Chapter 1, they discuss degrees of projective varieties using the Hilbert polynomial. Jul 3 '15 at 20:45

There are two ways to approach the topic. The elementary way I'm afraid is rather ugly. But if we bring in the cavalry you might have to wait a while before you fill in all the details. I chose the second way but briefly mentioned how you could bring this down to earth.

First some motivation. In the plane, a line should be degree one whatever degree is supposed to mean. But two lines sometimes intersect at a point giving the count $1\times 1 =1$ but sometimes the two lines are parallel not giving a common solution. When the degrees are greater and the dimension is greater, we lose much more information and the number of cases to consider become intractable. This is why, the projective space—and not the affine space—is the right place for intersection theory. Observe that two distinct lines in the projective space always intersect at a single point—and this is just the beginning of all the good things to come.

Another observation is that if we want to count number of solutions, we can linearly deform our objects without changing our count. Again, take the notion of "linear deformation" as a black box for now.

Putting the two together, we form an intersection ring (called the Chow ring) for the projective space $\mathbb{P}^n$, lets call it $R$. Elements in this ring are linear equivalence classes of subvarieties of the n-dimensional projective space. Addition works so that: $$[X] + [Y] := [X\cup Y]$$ And multiplication works so that: $$[X]\cdot [Y] := [\tilde X \cap Y]$$ where $\tilde X$ is an element in the linear equivalence class of $X$ that is more suitable to perform the intersection. For instance, two hyperplanes $H$ and $H'$ in $\mathbb{P}^n$ are linearly equivalent and if they are distinct $H\cap H'$ is a codimension 2 linear space. Now it makes sense to talk about 'self intersection of $H$ with itself'. What we mean is $[H]\cdot [H]$ which is $[H' \cap H]$.

A miracle happens here and we can prove that $R \simeq \mathbb{Z}[h]/(h^{n+1})$ where $\mathbb{Z}[h]$ is the polynomial ring over $\mathbb{Z}$ with variable $h$. This isomorphism carries $[H]$ to $h$. More significantly, a hypersurface cut out by a degree $d$ polynomial gets sent to $dh$.

For a subvariety of smaller dimension this isomorphism provides a way to define degree. If $X \subset \mathbb{P}^n$ is of codimension $m$ then $[X] = d_X [H]^m$. We may define the degree of $X$ to be this integer $d_X$.

A more pedestrian way to define degree is to intersect $X$ by as many general hyperplanes as the dimension of $X$ an then to count points. Realizing that the class of a point is $[H]^n$, you can see that these two definitions agree. But the pedestrian way of doing it would work also for your varieties in affine space.

Now that we have everything in its place we may tackle your question. By degree of $V$ I will mean the degree of its closure $\bar V$ in $\mathbb{P^n}$. Once again, this agrees with the pedestrian definition, because the general hyperplanes can be chosen so as not to simulatneously intersect the parts at infinity.

Then $\bar V$ is a subset of the intersection $W := \bigcap_i \bar V_i$, as there may be new things as a result of intersecting the closure of the $V_i$'s that appear in the complement $\mathbb{P}^n \setminus \mathbb{A}^n$. In fact, $\bar V$ is formed as the union of all the components of $W$ that are not completely in the boundary. If we denote the rest of the components of $W$ by $U$ then: $$[W] = [\bar V] + [U]$$ Now we can calculate. Each $\bar V_i$ being a degree $d_i$ hypersurface, $[\bar V_i]$ is equivalent to $d_i[H]$. The intersection $W$ therefore is equivalent to $\prod_i d_i [H]$. Hence the degree of $W$, $d_W$ equals $\prod_i d_i$.

The addition formula above gives: $$d_W [H]^l = d_{\bar V}[H]^l + d_U[H]^l$$ hence $$d_V = d_{\bar V} \le d_W = \prod_i d_i \le D^l.$$ The basic properties of degree the author is referring to is essentially the sum and product rules described above.

Edit: If $V$ has the components $Z_1,\dots,Z_k$ then $[Z_i] = d_{Z_i} [H]^l$. Recall each of these degrees are positive, this follows also from the pedestrian approach of counting points. Then $d_V = \sum_i d_{Z_i}$. Because each term in the sum is a positive integer, and due to our bound on $d_V$, we can not have more than $D^l$ terms in the sum. That is the bound on the number of components of $[V]$.

• Thank you very much for this very thorough answer! I greatly appreciate it!! :D Sep 21 '15 at 15:00
• You are welcome =)
– Emre
Sep 23 '15 at 4:26

AS Patrick Da Silva suggests, this follows from Bézout's theorem.

A weak form of Bézout says that for closed subsets $V$ and $W$ in $\mathbf C^n$, we have $$\operatorname{deg} V \cap W \leq \operatorname{deg} V \cdot \operatorname{deg} W$$.

Applying this repeatedly, your intersection $V$ has degree at most $D^l$. But any irreducible component contributes at least 1 to the degree, so there are at most $D^l$ components.

• Could you possibly explain a little more on what you mean by "any irreducible component contributes at least $1$ to the degree"? Jul 4 '15 at 17:25