In an endeavor to prove Bezout's theorem for algebraic plane curves, I am finding myself stuck on one particular result that does not have a good reference in any literature. I am trying to prove weak Bezout - that any real algebraic plane curves have at most $\deg(f)\deg(g)$ points of intersection - as opposed to the version taking account for intersection multiplicities. I am also trying to avoid visiting the projective spaces.
There are several documents available online, most of which outline the same method of proof. Throughout $\mathrm{Res}(f,g,x)$ will denote the Sylvester resultant (i.e. the determinant of the relevant Sylvester matrix), and $\deg(f)$ (sans subscript) will denote the total degree of $f$.
A crucial step in proving the bound is to prove that for coprime bivariate polynomials $f(x,y),g(x,y)$ we have $$\deg_y\mathrm{Res}(f,g,x) \leq \deg(f)\deg(g) \tag{1}$$ This is actually most of the way towards Bezout.
In every proof I find online and in physical books however, the authors make an odd observation about the degrees of these polynomials. Suppose one writes as usual $$f(x,y) = \sum_{i=0}^n a_i(y)x^i$$ where $a_i(y)$ is a polynomial in $y$. Then the authors assert that $$\deg_y(a_i) \leq n-i \tag{2}$$ from which the proof of (1) follows easily by counting the degree of the determinant of the Sylvester matrix over all permutations. My issue is with the assumption that $\deg_y(a_i) \leq n-i$.
There seems to be two competing notions of how to write bivariate polynomials: one in which $n = \deg_x(f)$ and one in which $n = \deg(f)$. If we use the former then the bound (2) fails trivially by considering a polynomial such as $yx^2 + 2x + y$ as we should expect the coefficient of $x^2$ (i.e. $y$) to have degree 0.
If one instead more reasonably assumes that $n = \deg(f)$ then the bound indeed holds as we have by definition $\displaystyle\deg(f) = \max_{0\leq i \leq n} \{i+\deg(a_i)\}$. But then the resultant no longer has the property that $\mathrm{Res}(f,g,x) = 0 \iff f,g \text{ share a factor}$. To see this consider the polynomials $$f(x,y) = yx^2 + yx + 1, \qquad g(x,y) = yx^2+x+2$$ Clearly $\deg(f) = \deg(g) = 3$, so to rewrite these in the form where we sum to the total degree we must write $$f(x,y) = 0x^3 + yx^2 + yx + 1, \qquad g(x,y) = 0x^3 + yx^2+x+2$$
Then the appropriate Sylvester matrix is the 6x6 matrix \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ y & 0 & 0 & t & 0 & 0 \\ y & y & 0 & 1 & y & 0 \\ 1 & y & y & 2 & 1 & y \\ 0 & 1 & y & 0 & 2 & 1 \\ 0 & 0 & 1 & 0 & 0 & 2 \end{bmatrix} But this matrix has a zero row, so its determinant must be zero as well. But then the resultant is zero, despite the fact that the two polynomials are clearly coprime.
What am I doing wrong here?
EDIT:
I did not make it clear why proof of (1) fails using $(\deg_x(f)+\deg_x(g)) \times (\deg_x(f)+\deg_x(g))$ with $n = \deg(f)$ (total degree).
Ideally, one would like to use some bound on each term in the Sylvester matrix, for instance $$\deg(p_{ij}) \leq n+i-j, \deg(p_{ij}) \leq m+i-j$$ for $1 \leq i \leq n$ or as Darij has done in these comments. This works perfectly fine if one assumes bound (2) with $n=\deg_x(f)$. However, this bound only works when we assume that we indeed have the bound $\deg_y(a_i) \leq \deg_x(f) - i$ as otherwise these methods give us a bound for (1) as $m\deg(f)+n\deg(g) - nm$. (n,m here are $\deg_x(f)$ and $\deg_x(g)$ respectively.) which (I think) isn't quite enough to prove what we want. It seems thus that the supposition that $\deg_y(a_i) \leq \deg_x(f) - i$ is crucial.
In particular it seems we need to have the dimensions of the matrix to coincide with the $n$ from bound (2). However if we try to extend the matrix to $(n+m) = \deg(f) + \deg(g)$ we end up with a zero row as above.
