Bezout's bound and resultants - reference request In Terry Tao's blog post about Bezout's inequality, he writes:

In our notation*, this theorem states the following:
Theorem 1 (Bezout’s theorem) Let $d=m=2$. If $V$ is finite, then it has cardinality at most $D_1 D_2$.
This result can be found in any introductory algebraic geometry textbook; it can for instance be proven using the classical tool of resultants. 

*Here, $V$ is the zero set of $m=2$ real polynomials in $d=2$ variables with degrees $D_1$, $D_2$.
I am looking for a reference for the proof of this version of the theorem (only an upper bound, not exact equality) with $d=m \ge 2$, preferably using resultants, that will be accessible to someone with no background in algebraic geometry. I prefer a textbook or standard reference that I can later use as a reference in a paper.
I've looked in Fulton's book Algebraic Curves and it is a bit too complicated for me, and it does not use resultants in the proof.
 A: The point is the following.
If you have two polynomials $f(x), g(x)\in A[x]$ where $A$ is a domain (including $A=k[y],$ for $k$ a field), and their respective degrees are $m$ and $n$, you wonder how to determine quickly whether they have a common factor (over the field of fractions $K$ of $A$) or not.  A quick way to realise this is that, if they had such common factor, then there would be a relation of this kind:
$$q(x)f(x)+r(x)g(x)=0,$$
where $deg\ q(x)\neq n-1, deg\ r(x)\leq m-1$ are polynomials in $A[x]$ or $K[x]$.
This is equivalent to showing that, over $K$, the following polynomials are linearly dependent:
$$f(x), xf(x), \cdots, x^{n-1}f(x), g(x), xg(x), \cdots , x^{m-1}g(x).$$
This is equivalent to saying that the determinant defining the resultant is $0$.
If $A=k[y]$, then the resultant is a polynomial in $y$ that is $0$ precisely when there is a common solution of $f(x,y_0)=0, g(x,y_0)=0$ for that particular value of $y$.
If you tilt the axes enough, you may suppose that the solutions $(x_i,y_i)$ in an algebraic closure of $k$ take different values of $y_i$, so the degree of the resultant is a precise bound for the system of equations given, namely:
$$f(x,y)=0, g(x,y)=0.$$
By using the appropriate combination, operating on the last row, one sees that the resultant is a suitable combination of polynomials 
$$R(y)=a(y)f(x,y)+b(y)g(x,y).$$
Thus, indeed the degree of the resultant is an upper bound for the number of solutions of the above system, assuming $f$ and $g$ have no common factors.
(It gets tedious if I write the matrices involved, but see the link).
This link is rather self-contained and accessible.
http://www.math.us.edu.pl/~pgladki/inedita/referat.pdf
LATER ADDITION (plane curves):  I do not know if you got my point.  The idea is, if you have a finite number of points in the plane, you can always choose the coordinates so that if the points are $P_i=(x_i,y_i)$, then all coordinates $y_i$ are different.  Thus you can count your points only through the resultant, so you need only count the number of zeros the resultant gives you.
INTERSECTION PRODUCT:  If you are familiar with homology theory, then you can view this as a problem of intersection in the complex projective plane, but I will omit this explanation for now.
THE CASE OF THREE VARIABLES (SKETCH):  Assume the intersection of the three hypersurfaces is finite.  This explanation is rather 'cooked-up', but let's go for it.  You have to assume, or to check, the following claim (always over the complex numbers).
{\textbf Claim:}  Let $F$ be a homogeneous polynomial in the variables $Z_0, \cdots ,Z_3$, and let $d$ be its degree.  Let $H$ be a generic plane in ${\mathbb P}$, and let $p$ be a general point (not on $F=0$ or on $H$).  Consider the projection of centre $p$ onto the plane $H$.
The projection is onto, and has degree $d$ (which is precisely the degree of intersection of a line through $p$ with the hypersurface $F=0$.)
If you have (dehomogenized) polynomials $f(x,y,z)=g(x,y,z)=0$ in ${\mathbb A}^3$, you can see that (after a suitable change of variables) for general $z$ fixed the above algebraic system of equations has at most $\leq \deg f \deg g$ solutions (unless $f, g$ are very special).
The idea is that the intersection (now we view them as homogenised) $F_1=F_2=0$ is a bunch of curves $C_i$, whose total degree sums up to $d_1 d_2$ (you can reduce to the situation of plane curves, explained above, by cutting with a general plane).
Now, we intersect those curves $C_i$ with our third hypersurface ($F=F_3$) $F_3=0$, and project to $H$ from the point $p$. It's time for counting.
An argument could be written down justifying the point count, but it's tedious.  It basically says that, counting with multiplicities, if the intersection of three homogeneous polynomials $F_i$ of degrees $d_i$ is finite, then it gives rise to a cycle of intersection
$$\sum e_i P_i,$$ where $\sum e_i=d_1 d_2 d_3.$
If we consider the equation $L$ of the plane $H$, one can write a function $\varphi=F_3/L^{d_3}$, which is meromorphic, and one can study each one of the lines of the projection over each one of the curves. The idea is that the total sum of zeros - poles of the restriction of $\varphi$ is zero over each line.
It's all indeed very cooked up, but I can't think of anything nicer ATM.
PASSING TO PROJECTIVE SPACE: It is true that you don't really need it unless you want to calculate the precise intersection, and in your case you just want a bound.
REFERENCE: This graduation thesis contains the definition of resultant of $n$ polynomials in $n$ variables, though it is rather tedious.
http://www.math.rwth-aachen.de/~kuenzer/diploma_deissler.pdf
