When to read of the degree of a variety from its defining polynomials 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.
 A: 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 $X^3 - ZW^2 = 0$. 
A: A variety $X$ in $\mathbb P^n_k$ of codimension $r$ generated by $r$ forms is called a complete intersection. Equivalently, $X$ is defined by a regular sequence $f_0,\ldots,f_r$ of (homogeneous) elements of $k[x_0,\ldots,x_n].$ It is well-known that the Koszul complex associated to a regular sequence $(f_0,\ldots,f_r)$ is a free resolution of the coordinate ring $k[x_0,\ldots,x_n]/(f_0,\ldots,f_r)$ and so enables us to compute the Hilbert polynomial of $X.$ Moreover, it can be shown that the Hilbert polynomial depends only on the degrees of $f_0,\ldots,f_r$, and its leading coefficient is $d_0\cdots d_r/(n-r)!,$ where $d_i$ is the degree of $f_i,$ so in this case the degree of $X$ is $d_0\cdots d_r.$
If $X$ is not a complete intersection, then the Koszul complex of defining equations will not be exact, and thus does not determine the Hilbert polynomial, though I'm not sure how badly behaved the degree can be in this case.
