The question concerns algebraic varieties.
I just read the question
The degree of an algebraic curve in higher dimensions
and great answer by user M P. One of the thing he says is that if a curve in $\mathbb{P}^n$ is given by $n-1$ equations (which is often not the case of course), then its degree is in fact the product of the degree of the polynomials defining it.
I expect this not to hold if there are more than $n-1$ polynomials necessary to define the curve. Could someone tell me if this is true, and why?
I do expect it to hold for general varieties. If a variety in $\mathbb{P}^n$ is of codimension $r$, and also given by exactly $n-r$ polynomials, is its degree in fact the product of the degrees of the polynomials defining it? Could you tell me if this is the case, and more importantly: why?
By the way, degree as in generic number of intersection points with a variety of complementary dimension..
Thanks a lot in advance,
Joachim
P.S. Georges E. i owe you one for your effort on my question on étale morphisms. I know this is not the place to say such things, but i'm doing it anyway.
Best Answer
Andrew's answer is correct, but here is a more geometric way to think about the same answer.
If a codimension $r$ variety is cut out by $r$ equations, then each equation genuinely cuts out a new dimension, i.e. $V(f_1,\ldots,f_{i+1})$ is a divisor in $V(f_1,\ldots,f_i)$ cut out by $f_{i+1} = 0$. Thus the intersection of $V(f_1,\ldots,f_i)$ and $V(f_{i+1})$ is a proper intersection, and so the degrees multiply.
Here is one way to think about this last statement:
We can assume that the degree of $V(f_1,\ldots,f_i)$ is equal to $d,$ the product of the degrees of $f_1, \ldots,f_i$, by induction. Geometrically, this means that a generic linear subspace $L$ of dimension $i$ meets $V(f_1,\ldots,f_i)$ in $d$ points.
To compute the degree of $V(f_1,\ldots,f_{i+1})$ we intersect with a generic linear subspace $L'$ of dimension $i+1$.
Suppose that $d'$ is the degree of $f_{i+1}$. Then you can deform the equation $f_{i+1} = 0$ to an equation of the form $l_1\ldots l_{d'}$, where each $l_j$ is a generically chosen linear equation. Thus $V(f_{i+1})$ can be deformed to the union of the hyperplanes $V(l_1)\cup \cdots \cup V(l_{d'})$, and so
$$V(f_1,\ldots,f_{i+1}) \cap L' = V(f_1,\ldots,f_i) \cap V(f_{i+1}) \cap L',$$ which can be deformed to $$V(f_1,\ldots,f_i) \cap \bigl(V(l_1)\cup \cdots \cup V(l_{d'})\bigr) \cap L' = \bigl(V(f_1,\ldots , f_i) \cap V(l_1) \cap L'\bigr) \cup \cdots \cup \bigl(V(f_1,\ldots,f_i)\cap V(l_1)\cap L'\bigr) \bigr).$$ Now each $V(l_j)\cap L'$ is the intersection of a generic hyperplane and the generic $i+1$-dimensional linear subspace $L'$, and so is a generic linear subspace of dimension $i$. Thus each intersection $V(f_1,\ldots,f_i)\cap V(l_j)\cap L'$ consists of $d$ points, and so their union consists of $dd'$ points. QED
If $V$ is an $r$-codimensional variety cut out by more than $r$ equations, then we can write it in the form $V(f_1,\ldots,f_r, f_{r+1},\ldots,f_s)$ for some $s > r$, where $V(f_1,\ldots,f_r)$ is already of codimension $r$ (but not irreducible). The closed subscheme $V(f_1,\ldots,f_r)$ of $\mathbb P^n$ is then of degree equal to the product of the degrees of the $f_1,\ldots,f_r$, but (by assumption) the additional equations $f_{r+1}, \ldots, f_s$ do not cut down the dimension any further. Instead, they cut out some particular irreducible component of $V(f_1,\ldots,f_r)$. In particular, the intersections $V(f_1,\ldots,f_r) \cap V(f_j)$ for $j > r$ are not proper, and the degree of a non-proper intersection is not given as the product of the degrees.
So the basic phenomenon here is as follows: if $V$ is a reducible algebraic set of (equi)codimension $r$, a union of components $V_1,\ldots,V_m$, then $\deg V = \sum_i \deg V_i$, but there is no immediate relationship between the degrees of the additional polynomials needed to cut out the various $V_i$ and the degrees of those $V_i$.
A good example to think about is the case of a twisted cubic curve in $\mathbb P^3$ (with hom. coords. $X,Y,Z,W$). It is obtained by first intersecting the quadrics $V(X^2 - YW)$ and $V(XZ - Y^2)$. This intersection is a degree $4$ curve which is reducible; it is the union of a line $L$ (the line cut out by $X = Y = 0$) and the twisted cubic curve $C$. To cut out $C$ we have to impose the additional equation $X^3 - ZW^2 = 0$.